Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
a,(5x + 2)(3x - 9) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
5{\rm{x}} + 2 = 0\\
3{\rm{x}} - 9 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 2}}{5}\\
x = 3
\end{array} \right.\\
Vay:S = \{ \frac{{ - 2}}{5};3\} \\
b,3x(x - 2) - 5(x - 2) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {3{\rm{x}} - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
3{\rm{x}} - 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \frac{5}{3}
\end{array} \right.\\
Vay:S = \{ 2;\frac{5}{3}\} \\
c,(2x + 3)(x + 2) = {x^2} - 4\\
\Leftrightarrow 2{{\rm{x}}^2} + 4{\rm{x}} + 3{\rm{x}} + 6 = {x^2} - 4\\
\Leftrightarrow {x^2} + 7{\rm{x}} + 10 = 0\\
\Leftrightarrow {x^2} + 2{\rm{x}} + 5{\rm{x}} + 10 = 0\\
\Leftrightarrow x\left( {x + 2} \right) + 5\left( {x + 2} \right) = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
x + 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 2\\
x = - 5
\end{array} \right.\\
Vay:S = \{ - 2; - 5\} \\
d,{x^2} - 6x + 9 = {x^2} - 3x\\
\Leftrightarrow {\left( {x - 3} \right)^2} - x\left( {x - 3} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x - 3 - x} \right) = 0\\
\Leftrightarrow - 3\left( {x - 3} \right) = 0\\
\Leftrightarrow x - 3 = 0\\
\Leftrightarrow x = 3\\
Vay:S = \{ 3\} \\
e,{x^2} - 3x + 2 = {x^2} - 1\\
\Leftrightarrow - 3{\rm{x}} + 3 = 0\\
\Leftrightarrow 3{\rm{x}} = 3\\
\Leftrightarrow x = 1\\
Vay:x = 1
\end{array}\]

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Answer 2

\[\begin{array}{l}
a,(5x + 2)(3x - 9) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
5{\rm{x}} + 2 = 0\\
3{\rm{x}} - 9 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 2}}{5}\\
x = 3
\end{array} \right.\\
Vay:S = \{ \frac{{ - 2}}{5};3\} \\
b,3x(x - 2) - 5(x - 2) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {3{\rm{x}} - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
3{\rm{x}} - 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \frac{5}{3}
\end{array} \right.\\
Vay:S = \{ 2;\frac{5}{3}\} \\
c,(2x + 3)(x + 2) = {x^2} - 4\\
\Leftrightarrow 2{{\rm{x}}^2} + 4{\rm{x}} + 3{\rm{x}} + 6 = {x^2} - 4\\
\Leftrightarrow {x^2} + 7{\rm{x}} + 10 = 0\\
\Leftrightarrow {x^2} + 2{\rm{x}} + 5{\rm{x}} + 10 = 0\\
\Leftrightarrow x\left( {x + 2} \right) + 5\left( {x + 2} \right) = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
x + 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 2\\
x = - 5
\end{array} \right.\\
Vay:S = \{ - 2; - 5\} \\
d,{x^2} - 6x + 9 = {x^2} - 3x\\
\Leftrightarrow {\left( {x - 3} \right)^2} - x\left( {x - 3} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x - 3 - x} \right) = 0\\
\Leftrightarrow - 3\left( {x - 3} \right) = 0\\
\Leftrightarrow x - 3 = 0\\
\Leftrightarrow x = 3\\
Vay:S = \{ 3\} \\
e,{x^2} - 3x + 2 = {x^2} - 1\\
\Leftrightarrow - 3{\rm{x}} + 3 = 0\\
\Leftrightarrow 3{\rm{x}} = 3\\
\Leftrightarrow x = 1\\
Vay:x = 1
\end{array}\]

Answer 3

a) (5x+2)(3x-9)=0

$\Leftrightarrow 5 x + 2 = 0$ hoặc 3x- 9 = 0

$\Leftrightarrow$ x= -2/5 hoặc x= 3

b) 3x(x-2)-5(x-2)=0

$\Leftrightarrow$ ( x-2 )(3x-5)=0

$\Leftrightarrow$ x-2=0 hoặc 3x-5=0

$\Leftrightarrow$ x=2 hoặc x=5/3

c) (2x+3)(x+2)=x^2-4

$\Leftrightarrow$ x^2 +7x +10 =0

$\Leftrightarrow$ (x+2)(x+5)=0

$\Leftrightarrow$ x= -2 hoặc x=-5

d,x^2-6x+9=x^2-3x

$\Leftrightarrow$ 3x=9

$\Leftrightarrow$ x=3

e,x^2-3x+2=x^2-1

$\Leftrightarrow$ 3x=3

$\Leftrightarrow$ x=1

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Answer 4

a) (5x+2)(3x-9)=0

$\Leftrightarrow 5 x + 2 = 0$ hoặc 3x- 9 = 0

$\Leftrightarrow$ x= -2/5 hoặc x= 3

b) 3x(x-2)-5(x-2)=0

$\Leftrightarrow$ ( x-2 )(3x-5)=0

$\Leftrightarrow$ x-2=0 hoặc 3x-5=0

$\Leftrightarrow$ x=2 hoặc x=5/3

c) (2x+3)(x+2)=x^2-4

$\Leftrightarrow$ x^2 +7x +10 =0

$\Leftrightarrow$ (x+2)(x+5)=0

$\Leftrightarrow$ x= -2 hoặc x=-5

d,x^2-6x+9=x^2-3x

$\Leftrightarrow$ 3x=9

$\Leftrightarrow$ x=3

e,x^2-3x+2=x^2-1

$\Leftrightarrow$ 3x=3

$\Leftrightarrow$ x=1

Answer 5

a) 5x+2=0;3x-9=0

<=> x=-2/5;x=3

b) 3x $^{2}$ -6x-5x+10=0

x=2,x=5/3

c) 2 $x^{2}$ +7x+6= $x^{2}$ -4

<=>x=-2;x=-5

d)x=3

e) x=1

3 câu trả lời 732

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