Question

Đã trả lời bởi chuyên gia

Answer 1

$a . {(2 x - 1)}^{2} - 2 x - 1 = 2 - > {(2 x - 1)}^{2} - 2 x - 1 - 2 = 0 - > {(2 x - 1)}^{2} - 2 x - 3 = 0 - > {(2 x - 1)}^{2} - (2 x - 1) - 4 = 0 đ ặ t : 2 x - 1 = t - > t^{2} - t - 4 = 0 - > (t - \frac{1 + \sqrt{17}}{2}) (t - \frac{1 - \sqrt{17}}{2}) = 0 T H 1 : t - \frac{1 + \sqrt{17}}{2} = 0 - > t = \frac{1 + \sqrt{17}}{2} - > 2 x - 1 = \frac{1 + \sqrt{17}}{2} - > x = \frac{3 + \sqrt{17}}{4} T H 2 : t - \frac{1 - \sqrt{17}}{2} = 0 - > t = \frac{1 - \sqrt{17}}{2} 2 x - 1 = \frac{1 - \sqrt{17}}{2} - > x = \frac{3 - \sqrt{17}}{4} v ậ y p t c ó n g h i ệ m x = \frac{3 + \sqrt{17}}{4} v à x = \frac{3 - \sqrt{17}}{4}$

$b . (x$ ²-3x)2+5(x2-3x)+6=0đặt x2-3x=t-> t2+5t +6 = 0-> t2+2t+3t+6=0-> (t2 +2t)+(3t+6)=0-> t(t+2)+3(t+2)=0-> (t+2)(t+3)= 0TH1: t+2=0 -> t=-2 -> x2-3x=-2-> x2-3x+2=0-> x2-x-2x+2=0-> (x2-x)-(2x-2)=0-> x(x-1)-2(x-1)=0-> (x-1)(x-2)=0-> x=1, x=2TH2: t+3= 0 -> t=-3 -> x2-3x=-3-> x2-3x+3=0-> (x2-2.32x+ 94) -94+3=0->(x-32)2 +34=0 pt vô nghiemvậy pt có nghiem x=1, x=2 $c . (x^{2} - x - 1) (x^{2} - x) - 2 = 0 đ ặ t x^{2} - x = t - > (t - 1) t - 2 = 0 - > t^{2} - t - 2 = 0 - > t^{2} + t - 2 t - 2 = 0 - > (t^{2} + t) - (2 t + 2) = 0 - > t (t + 1) - 2 (t + 1) = 0 - > (t + 1) (t - 2) = 0 T H 1 : t + 1 = 0 - > t = - 1 - > x^{2} - x = - 1 - > x^{2} - x + 1 = 0 - > (x^{2} - 2 x \frac{1}{2} + \frac{1}{4}) - \frac{1}{4} + 1 = 0 - > {(x - \frac{1}{2})}^{2} + \frac{3}{4} = 0 p t v ô n g h i ệ m T H 2 : t - 2 = 0 - > t = 2 - > x^{2} - x = 2 - > x^{2} - x - 2 = 0 - > x^{2} + x - 2 x - 2 = 0 - > (x^{2} + x) - (2 x + 2) = 0 - > x (x + 1) - 2 (x + 1) = 0 - > (x + 1) (x - 2) = 0 - > x = - 1, x = 2 v ậ y p t c ó n g h i e m x = - 1 x = 2$

$d . {(5 - 2 x)}^{2} + 4 x - 10 = 0 - > 25 - 20 x + 4 x^{2} + 4 x - 10 = 0 - > 4 x^{2} - 16 x + 15 = 0 - > 4 x^{2} - 10 x - 6 x + 15 = 0 - > (4 x^{2} - 10 x) - (6 x - 15) = 0 - > 2 x (2 x - 5) - 3 (2 x - 5) = 0 - > (2 x - 5) (2 x - 3) = 0 - > x = \frac{5}{2}, x = \frac{3}{2} v ậ y p t c ó n g h i ệ m x = \frac{5}{2}, x = \frac{3}{2}$

$e . (x^{2} + 2 x + 3) (x^{2} + 2 x + 1) = 3 đ ặ t x^{2} + 2 x = t - > (t + 3) (t + 1) = 3 - > t^{2} + 4 t + 3 - 3 = 0 - > t^{2} + 4 t = 0 - > t (t + 4) = 0 - > t = 0, t = - 4 T H 1 : t = 0 - > x^{2} + 2 x = 0 - > x (x + 2) = 0 - > x = 0; x = - 2 T H 2 : t = - 4 - > x^{2} + 2 x = - 4 - > x^{2} + 2 x + 4 = 0 - > x^{2} + 2 x + 1 + 3 = 0 - > {(x + 1)}^{2} + 3 = 0 - > p t v ô n g h i ệ m^{} v ậ y p t c ó n g h i ệ m x = 0; x = - 2$ $f . x (x - 1) (x^{2} - x + 1) - 6 = 0 - > (x^{2} - x) (x^{2} - x + 1) - 6 = 0 đ ặ t x^{2} - x = t - > t (t + 1) - 6 = 0 - > t^{2} + t - 6 = 0 - > t^{2} + 3 t - 2 t - 6 = 0 - > (t^{2} + 3 t) - (2 t + 6) = 0 - > t (t + 3) - 2 (t + 3) = 0 - > (t + 3) (t - 2) = 0 - > t = - 3, t = 2 T H 1 : t = - 3 - > x^{2} - x = - 3 - > x^{2} - x + 3 = 0 - > (x^{2} - 2 x \frac{1}{2} + \frac{1}{4}) - \frac{1}{4} + 3 = 0 - > {(x - \frac{1}{2})}^{2} + \frac{11}{4} = 0 - > p t v ô n g h i ệ m T H 2 : t = 2 - > x^{2} - x = 2 - > x^{2} - x - 2 = 0 - > x^{2} - 2 x + x - 2 = 0 - > (x^{2} - 2 x) + (x - 2) = 0 - > x (x - 2) + (x - 2) = 0 - > (x - 2) (x + 1) = 0 - > x = 2, x = - 1 v ậ y p t c ó n g h i e m x = 2, x = - 1$

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Answer 2

$a . {(2 x - 1)}^{2} - 2 x - 1 = 2 - > {(2 x - 1)}^{2} - 2 x - 1 - 2 = 0 - > {(2 x - 1)}^{2} - 2 x - 3 = 0 - > {(2 x - 1)}^{2} - (2 x - 1) - 4 = 0 đ ặ t : 2 x - 1 = t - > t^{2} - t - 4 = 0 - > (t - \frac{1 + \sqrt{17}}{2}) (t - \frac{1 - \sqrt{17}}{2}) = 0 T H 1 : t - \frac{1 + \sqrt{17}}{2} = 0 - > t = \frac{1 + \sqrt{17}}{2} - > 2 x - 1 = \frac{1 + \sqrt{17}}{2} - > x = \frac{3 + \sqrt{17}}{4} T H 2 : t - \frac{1 - \sqrt{17}}{2} = 0 - > t = \frac{1 - \sqrt{17}}{2} 2 x - 1 = \frac{1 - \sqrt{17}}{2} - > x = \frac{3 - \sqrt{17}}{4} v ậ y p t c ó n g h i ệ m x = \frac{3 + \sqrt{17}}{4} v à x = \frac{3 - \sqrt{17}}{4}$

$b . (x$ ²-3x)2+5(x2-3x)+6=0đặt x2-3x=t-> t2+5t +6 = 0-> t2+2t+3t+6=0-> (t2 +2t)+(3t+6)=0-> t(t+2)+3(t+2)=0-> (t+2)(t+3)= 0TH1: t+2=0 -> t=-2 -> x2-3x=-2-> x2-3x+2=0-> x2-x-2x+2=0-> (x2-x)-(2x-2)=0-> x(x-1)-2(x-1)=0-> (x-1)(x-2)=0-> x=1, x=2TH2: t+3= 0 -> t=-3 -> x2-3x=-3-> x2-3x+3=0-> (x2-2.32x+ 94) -94+3=0->(x-32)2 +34=0 pt vô nghiemvậy pt có nghiem x=1, x=2 $c . (x^{2} - x - 1) (x^{2} - x) - 2 = 0 đ ặ t x^{2} - x = t - > (t - 1) t - 2 = 0 - > t^{2} - t - 2 = 0 - > t^{2} + t - 2 t - 2 = 0 - > (t^{2} + t) - (2 t + 2) = 0 - > t (t + 1) - 2 (t + 1) = 0 - > (t + 1) (t - 2) = 0 T H 1 : t + 1 = 0 - > t = - 1 - > x^{2} - x = - 1 - > x^{2} - x + 1 = 0 - > (x^{2} - 2 x \frac{1}{2} + \frac{1}{4}) - \frac{1}{4} + 1 = 0 - > {(x - \frac{1}{2})}^{2} + \frac{3}{4} = 0 p t v ô n g h i ệ m T H 2 : t - 2 = 0 - > t = 2 - > x^{2} - x = 2 - > x^{2} - x - 2 = 0 - > x^{2} + x - 2 x - 2 = 0 - > (x^{2} + x) - (2 x + 2) = 0 - > x (x + 1) - 2 (x + 1) = 0 - > (x + 1) (x - 2) = 0 - > x = - 1, x = 2 v ậ y p t c ó n g h i e m x = - 1 x = 2$

$d . {(5 - 2 x)}^{2} + 4 x - 10 = 0 - > 25 - 20 x + 4 x^{2} + 4 x - 10 = 0 - > 4 x^{2} - 16 x + 15 = 0 - > 4 x^{2} - 10 x - 6 x + 15 = 0 - > (4 x^{2} - 10 x) - (6 x - 15) = 0 - > 2 x (2 x - 5) - 3 (2 x - 5) = 0 - > (2 x - 5) (2 x - 3) = 0 - > x = \frac{5}{2}, x = \frac{3}{2} v ậ y p t c ó n g h i ệ m x = \frac{5}{2}, x = \frac{3}{2}$

$e . (x^{2} + 2 x + 3) (x^{2} + 2 x + 1) = 3 đ ặ t x^{2} + 2 x = t - > (t + 3) (t + 1) = 3 - > t^{2} + 4 t + 3 - 3 = 0 - > t^{2} + 4 t = 0 - > t (t + 4) = 0 - > t = 0, t = - 4 T H 1 : t = 0 - > x^{2} + 2 x = 0 - > x (x + 2) = 0 - > x = 0; x = - 2 T H 2 : t = - 4 - > x^{2} + 2 x = - 4 - > x^{2} + 2 x + 4 = 0 - > x^{2} + 2 x + 1 + 3 = 0 - > {(x + 1)}^{2} + 3 = 0 - > p t v ô n g h i ệ m^{} v ậ y p t c ó n g h i ệ m x = 0; x = - 2$ $f . x (x - 1) (x^{2} - x + 1) - 6 = 0 - > (x^{2} - x) (x^{2} - x + 1) - 6 = 0 đ ặ t x^{2} - x = t - > t (t + 1) - 6 = 0 - > t^{2} + t - 6 = 0 - > t^{2} + 3 t - 2 t - 6 = 0 - > (t^{2} + 3 t) - (2 t + 6) = 0 - > t (t + 3) - 2 (t + 3) = 0 - > (t + 3) (t - 2) = 0 - > t = - 3, t = 2 T H 1 : t = - 3 - > x^{2} - x = - 3 - > x^{2} - x + 3 = 0 - > (x^{2} - 2 x \frac{1}{2} + \frac{1}{4}) - \frac{1}{4} + 3 = 0 - > {(x - \frac{1}{2})}^{2} + \frac{11}{4} = 0 - > p t v ô n g h i ệ m T H 2 : t = 2 - > x^{2} - x = 2 - > x^{2} - x - 2 = 0 - > x^{2} - 2 x + x - 2 = 0 - > (x^{2} - 2 x) + (x - 2) = 0 - > x (x - 2) + (x - 2) = 0 - > (x - 2) (x + 1) = 0 - > x = 2, x = - 1 v ậ y p t c ó n g h i e m x = 2, x = - 1$

Answer 3

−> (2x−1)2 − 2x−1−2=0−> (2x−1)2− 2x−3=0−> (2x−1)2−(2x−1) −4=0đặt : 2x−1 = t−> t2−t−4=0−> (t−1+√172) (t−1−√172)=0TH1: t−1+√172= 0 −> t=1+√172 −>2x−1 = 1+√172 −> x= 3+√174TH2: t−1−√172=0 −> t= 1−√1722x−1 = 1−√172 −> x= 3−√174vậy pt có nghiệm x= 3+√174 và x= 3−√174

b. (x2-3x)2+5(x2-3x)+6=0đặt x2-3x=t-> t2+5t +6 = 0-> t2+2t+3t+6=0-> (t2 +2t)+(3t+6)=0-> t(t+2)+3(t+2)=0-> (t+2)(t+3)= 0TH1: t+2=0 -> t=-2 -> x2-3x=-2-> x2-3x+2=0-> x2-x-2x+2=0-> (x2-x)-(2x-2)=0-> x(x-1)-2(x-1)=0-> (x-1)(x-2)=0-> x=1, x=2TH2: t+3= 0 -> t=-3 -> x2-3x=-3-> x2-3x+3=0-> (x2-2.32x+ 94) -94+3=0->(x-32)2 +34=0 pt vô nghiemvậy pt có nghiem x=1, x=2c. (x2−x−1)(x2−x)−2=0đặt x2−x=t−> (t−1)t−2=0−> t2−t−2=0−> t2+t−2t−2=0−> (t2+t)−(2t+2)=0−> t(t+1) −2(t+1)=0−> (t+1)(t−2)=0TH1: t+1=0 −> t=−1−> x2−x=−1−> x2−x+1=0−> (x2−2x12+14) −14+1=0−> (x−12)2+34=0 pt vô nghiệmTH2: t−2=0 −> t=2−> x2−x=2 −> x2−x−2=0−> x2 +x−2x−2=0−> (x2 +x)−(2x+2)=0−>x(x+1) −2(x+1)=0−> (x+1)(x−2)=0−> x=−1, x=2vậy pt có nghiem x=−1 x=2

d. (5−2x)2+4x−10=0−> 25−20x+4x2+4x−10=0−> 4x2−16x+15=0−> 4x2−10x−6x+15=0−> (4x2−10x)−(6x−15)=0−> 2x(2x−5)−3(2x−5)=0−> (2x−5)(2x−3)=0−> x=52, x=32vậy pt có nghiệm x=52, x=32

e. (x2+2x+3)(x2+2x+1)=3đặt x2+2x=t−> (t+3)(t+1)=3−> t2+4t+3−3=0−> t2+4t=0−> t(t+4)=0−> t=0, t=−4TH1: t=0 −> x2 +2x=0−> x(x+2)=0−> x=0; x=−2TH2: t=−4 −>x2+2x=−4−> x2+2x+4=0−> x2+2x+1+3=0−> (x+1)2+3=0−>pt vô nghiệm vậy pt có nghiệm x=0; x=−2f. x(x−1)(x2−x+1)−6=0−> (x2−x)(x2−x+1)−6=0đặt x2−x=t−> t(t+1)−6=0−> t2+t−6=0−> t2 +3t−2t−6=0−> ( t2 +3t)−(2t+6)=0−>t(t+3)−2(t+3)=0−> (t+3)(t−2)=0−> t=−3, t=2TH1: t=−3 −> x2−x=−3−> x2−x+3=0−>( x2−2x12+14)−14+3=0−> (x−12)2+114=0 −> pt vô nghiệmTH2: t=2 −>x2−x=2−> x2−x−2=0−> x2−2x+x−2=0−>(x2−2x)+(x−2)=0−> x(x−2)+(x−2)=0−> (x−2)(x+1)=0−> x=2, x=−1vậy pt có nghiem x=2, x=−1

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Answer 4

−> (2x−1)2 − 2x−1−2=0−> (2x−1)2− 2x−3=0−> (2x−1)2−(2x−1) −4=0đặt : 2x−1 = t−> t2−t−4=0−> (t−1+√172) (t−1−√172)=0TH1: t−1+√172= 0 −> t=1+√172 −>2x−1 = 1+√172 −> x= 3+√174TH2: t−1−√172=0 −> t= 1−√1722x−1 = 1−√172 −> x= 3−√174vậy pt có nghiệm x= 3+√174 và x= 3−√174

b. (x2-3x)2+5(x2-3x)+6=0đặt x2-3x=t-> t2+5t +6 = 0-> t2+2t+3t+6=0-> (t2 +2t)+(3t+6)=0-> t(t+2)+3(t+2)=0-> (t+2)(t+3)= 0TH1: t+2=0 -> t=-2 -> x2-3x=-2-> x2-3x+2=0-> x2-x-2x+2=0-> (x2-x)-(2x-2)=0-> x(x-1)-2(x-1)=0-> (x-1)(x-2)=0-> x=1, x=2TH2: t+3= 0 -> t=-3 -> x2-3x=-3-> x2-3x+3=0-> (x2-2.32x+ 94) -94+3=0->(x-32)2 +34=0 pt vô nghiemvậy pt có nghiem x=1, x=2c. (x2−x−1)(x2−x)−2=0đặt x2−x=t−> (t−1)t−2=0−> t2−t−2=0−> t2+t−2t−2=0−> (t2+t)−(2t+2)=0−> t(t+1) −2(t+1)=0−> (t+1)(t−2)=0TH1: t+1=0 −> t=−1−> x2−x=−1−> x2−x+1=0−> (x2−2x12+14) −14+1=0−> (x−12)2+34=0 pt vô nghiệmTH2: t−2=0 −> t=2−> x2−x=2 −> x2−x−2=0−> x2 +x−2x−2=0−> (x2 +x)−(2x+2)=0−>x(x+1) −2(x+1)=0−> (x+1)(x−2)=0−> x=−1, x=2vậy pt có nghiem x=−1 x=2

d. (5−2x)2+4x−10=0−> 25−20x+4x2+4x−10=0−> 4x2−16x+15=0−> 4x2−10x−6x+15=0−> (4x2−10x)−(6x−15)=0−> 2x(2x−5)−3(2x−5)=0−> (2x−5)(2x−3)=0−> x=52, x=32vậy pt có nghiệm x=52, x=32

e. (x2+2x+3)(x2+2x+1)=3đặt x2+2x=t−> (t+3)(t+1)=3−> t2+4t+3−3=0−> t2+4t=0−> t(t+4)=0−> t=0, t=−4TH1: t=0 −> x2 +2x=0−> x(x+2)=0−> x=0; x=−2TH2: t=−4 −>x2+2x=−4−> x2+2x+4=0−> x2+2x+1+3=0−> (x+1)2+3=0−>pt vô nghiệm vậy pt có nghiệm x=0; x=−2f. x(x−1)(x2−x+1)−6=0−> (x2−x)(x2−x+1)−6=0đặt x2−x=t−> t(t+1)−6=0−> t2+t−6=0−> t2 +3t−2t−6=0−> ( t2 +3t)−(2t+6)=0−>t(t+3)−2(t+3)=0−> (t+3)(t−2)=0−> t=−3, t=2TH1: t=−3 −> x2−x=−3−> x2−x+3=0−>( x2−2x12+14)−14+3=0−> (x−12)2+114=0 −> pt vô nghiệmTH2: t=2 −>x2−x=2−> x2−x−2=0−> x2−2x+x−2=0−>(x2−2x)+(x−2)=0−> x(x−2)+(x−2)=0−> (x−2)(x+1)=0−> x=2, x=−1vậy pt có nghiem x=2, x=−1

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