Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
\;a){\rm{ }}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right){\rm{ }} - {\rm{ }}x\left( {x + 2} \right)\left( {x - 2} \right){\rm{ }} = {\rm{ }}5\\
\Leftrightarrow {x^3}{\rm{ - 1 - x}}\left( {{x^2} - 4} \right){\rm{ = 5}}\\
\Leftrightarrow {\rm{ }}{x^3}{\rm{ - 1 - }}{{\rm{x}}^3}{\rm{ + 4x = 5}}\\
\Leftrightarrow {\rm{ }}4x = 6\\
\Leftrightarrow {\rm{ }}x = \frac{3}{2}\\
{\rm{vay:x = }}\frac{3}{2}{\rm{ }}\\
b){\rm{ }}3\left( {x - 2} \right)\left( {x + 2} \right) + {\left( {x - 1} \right)^3}{\rm{ }} - {\rm{ }}\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) = {\rm{ }} - 46\\
\Leftrightarrow {\rm{ }}3\left( {{x^2} - 4} \right){\rm{ + }}{{\rm{x}}^3}{\rm{ - 3}}{{\rm{x}}^2}{\rm{ + 3x - 1 - }}\left( {{x^3} + 27} \right){\rm{ = - 46}}\\
\Leftrightarrow {\rm{ }}3{x^2}{\rm{ - 12 + }}{{\rm{x}}^3}{\rm{ - 3}}{{\rm{x}}^2}{\rm{ + 3x - 1 - }}{{\rm{x}}^3}{\rm{ - 27 = - 46}}\\
\Leftrightarrow {\rm{ 3x - 40 = - 46}}\\
\Leftrightarrow {\rm{ }}3x = - 6\\
\Leftrightarrow {\rm{ }}x = - 2\\
Vay:x = - 2{\rm{ }}\\
c){\rm{ }}\left( {2 + 5x} \right)\left( {5x - 2} \right){\rm{ }} - {\rm{ }}{\left( {3x + 1} \right)^2}{\rm{ }} = {\rm{ }}\left( {3 + 4x} \right)\left( {4x - 3} \right)\\
\Leftrightarrow {\rm{ }}\left( {25{x^2} - 4} \right){\rm{ - }}\left( {9{x^2} + 6x + 1} \right){\rm{ = 16}}{{\rm{x}}^2}{\rm{ - 9}}\\
\Leftrightarrow {\rm{ }}25{x^2}{\rm{ - 4 - 9}}{{\rm{x}}^2}{\rm{ - 6x - 1 - 16}}{{\rm{x}}^2}{\rm{ + 9 = 0}}\\
\Leftrightarrow {\rm{ }} - 6x + 4 = 0\\
\Leftrightarrow {\rm{ }}6x = 4\\
\Leftrightarrow {\rm{ }}x = \frac{2}{3}\\
{\rm{vay:x = }}\frac{2}{3}{\rm{ }}\\
d)\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) - x\left( {2 + x} \right)\left( {x - 2} \right) = 6\left( {4 + x} \right)\\
\Leftrightarrow {x^3} - 27 - x\left( {{x^2} - 4} \right) = 24 + 6x\\
\Leftrightarrow {x^3} - 27 - {x^3} + 4x - 6x = 24\\
\Leftrightarrow - 2x = 24 + 27\\
\Leftrightarrow - 2x = 51\\
\Leftrightarrow x = \frac{{ - 51}}{2}\\
vay:x = \frac{{ - 51}}{2}\\
{\rm{ }}e){\rm{ }}{\left( {x - 2} \right)^3} - \left( {x - 3} \right)\left( {{x^2} - 3x + 9} \right){\rm{ }} - {\rm{ }}x\left( {2 + x} \right)\left( {x - 2} \right){\rm{ }} = {\rm{ }}6\left( {4 + x} \right)\\
\Leftrightarrow {x^3} - 6{x^2} + 12x - 8 - \left( {{x^3} - 3{x^2} + 9x - 3{x^2} + 9x - 27} \right) - x\left( {{x^2} - 4} \right) = 24 + 6x\\
\Leftrightarrow {x^3} - 6{x^2} + 12x - 8 - {x^3} + 6{x^2} - 18x + 27 - {x^3} + 4x - 6x = 24\\
\Leftrightarrow - {x^3} - 8x + 19 = 24\\
\Leftrightarrow {x^3} + 8x + 5 = 0\\
\Leftrightarrow x \approx - 0,598\\
vay:x \approx - 0,598
\end{array}\]

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Answer 2

\[\begin{array}{l}
\;a){\rm{ }}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right){\rm{ }} - {\rm{ }}x\left( {x + 2} \right)\left( {x - 2} \right){\rm{ }} = {\rm{ }}5\\
\Leftrightarrow {x^3}{\rm{ - 1 - x}}\left( {{x^2} - 4} \right){\rm{ = 5}}\\
\Leftrightarrow {\rm{ }}{x^3}{\rm{ - 1 - }}{{\rm{x}}^3}{\rm{ + 4x = 5}}\\
\Leftrightarrow {\rm{ }}4x = 6\\
\Leftrightarrow {\rm{ }}x = \frac{3}{2}\\
{\rm{vay:x = }}\frac{3}{2}{\rm{ }}\\
b){\rm{ }}3\left( {x - 2} \right)\left( {x + 2} \right) + {\left( {x - 1} \right)^3}{\rm{ }} - {\rm{ }}\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) = {\rm{ }} - 46\\
\Leftrightarrow {\rm{ }}3\left( {{x^2} - 4} \right){\rm{ + }}{{\rm{x}}^3}{\rm{ - 3}}{{\rm{x}}^2}{\rm{ + 3x - 1 - }}\left( {{x^3} + 27} \right){\rm{ = - 46}}\\
\Leftrightarrow {\rm{ }}3{x^2}{\rm{ - 12 + }}{{\rm{x}}^3}{\rm{ - 3}}{{\rm{x}}^2}{\rm{ + 3x - 1 - }}{{\rm{x}}^3}{\rm{ - 27 = - 46}}\\
\Leftrightarrow {\rm{ 3x - 40 = - 46}}\\
\Leftrightarrow {\rm{ }}3x = - 6\\
\Leftrightarrow {\rm{ }}x = - 2\\
Vay:x = - 2{\rm{ }}\\
c){\rm{ }}\left( {2 + 5x} \right)\left( {5x - 2} \right){\rm{ }} - {\rm{ }}{\left( {3x + 1} \right)^2}{\rm{ }} = {\rm{ }}\left( {3 + 4x} \right)\left( {4x - 3} \right)\\
\Leftrightarrow {\rm{ }}\left( {25{x^2} - 4} \right){\rm{ - }}\left( {9{x^2} + 6x + 1} \right){\rm{ = 16}}{{\rm{x}}^2}{\rm{ - 9}}\\
\Leftrightarrow {\rm{ }}25{x^2}{\rm{ - 4 - 9}}{{\rm{x}}^2}{\rm{ - 6x - 1 - 16}}{{\rm{x}}^2}{\rm{ + 9 = 0}}\\
\Leftrightarrow {\rm{ }} - 6x + 4 = 0\\
\Leftrightarrow {\rm{ }}6x = 4\\
\Leftrightarrow {\rm{ }}x = \frac{2}{3}\\
{\rm{vay:x = }}\frac{2}{3}{\rm{ }}\\
d)\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) - x\left( {2 + x} \right)\left( {x - 2} \right) = 6\left( {4 + x} \right)\\
\Leftrightarrow {x^3} - 27 - x\left( {{x^2} - 4} \right) = 24 + 6x\\
\Leftrightarrow {x^3} - 27 - {x^3} + 4x - 6x = 24\\
\Leftrightarrow - 2x = 24 + 27\\
\Leftrightarrow - 2x = 51\\
\Leftrightarrow x = \frac{{ - 51}}{2}\\
vay:x = \frac{{ - 51}}{2}\\
{\rm{ }}e){\rm{ }}{\left( {x - 2} \right)^3} - \left( {x - 3} \right)\left( {{x^2} - 3x + 9} \right){\rm{ }} - {\rm{ }}x\left( {2 + x} \right)\left( {x - 2} \right){\rm{ }} = {\rm{ }}6\left( {4 + x} \right)\\
\Leftrightarrow {x^3} - 6{x^2} + 12x - 8 - \left( {{x^3} - 3{x^2} + 9x - 3{x^2} + 9x - 27} \right) - x\left( {{x^2} - 4} \right) = 24 + 6x\\
\Leftrightarrow {x^3} - 6{x^2} + 12x - 8 - {x^3} + 6{x^2} - 18x + 27 - {x^3} + 4x - 6x = 24\\
\Leftrightarrow - {x^3} - 8x + 19 = 24\\
\Leftrightarrow {x^3} + 8x + 5 = 0\\
\Leftrightarrow x \approx - 0,598\\
vay:x \approx - 0,598
\end{array}\]

Answer 3

a)(x−1)(x2+x+1)−x(x+2)(x−2)=5⇔x3−1−x(x2−4)=5⇔x3−1−x3+4x=5⇔4x=6⇔x=32vay:x=32b)3(x−2)(x+2)+(x−1)3−(x+3)(x2−3x+9)=−46⇔3(x2−4)+x3−3x2+3x−1−(x3+27)=−46⇔3x2−12+x3−3x2+3x−1−x3−27=−46⇔3x−40=−46⇔3x=−6⇔x=−2Vay:x=−2c)(2+5x)(5x−2)−(3x+1)2=(3+4x)(4x−3)⇔(25x2−4)−(9x2+6x+1)=16x2−9⇔25x2−4−9x2−6x−1−16x2+9=0⇔−6x+4=0⇔6x=4⇔x=23vay:x=23d)(x+3)(x2−3x+9)−x(2+x)(x−2)=6(4+x)⇔x3−27−x(x2−4)=24+6x⇔x3−27−x3+4x−6x=24⇔−2x=24+27⇔−2x=51⇔x=−512vay:x=−512e)(x−2)3−(x−3)(x2−3x+9)−x(2+x)(x−2)=6(4+x)⇔x3−6x2+12x−8−(x3−3x2+9x−3x2+9x−27)−x(x2−4)=24+6x⇔x3−6x2+12x−8−x3+6x2−18x+27−x3+4x−6x=24⇔−x3−8x+19=24⇔x3+8x+5=0⇔x≈−0,598vay:x≈−0,598

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