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a) \(x\left(5+3x\right)-\left(x+1\right)\left(3x-2\right)=12\)
\(5x+3x^2-3x^2+2x-3x+2=12\)
\(4x=10\)
\(x=\frac{5}{2}\)
vậy \(x=\frac{5}{2}\)
\(13x\left(x-8\right)-x+8=0\)
\(13x\left(x-8\right)-\left(x-8\right)=0\)
\(\left(13x-1\right)\left(x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}13x-1=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{13}\\x=8\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{1}{13}\\x=8\end{cases}}\)
(3x-5)2-(3x+1)2=8
<=>(9x2-30x+25)-(9x2+6x+1)=8
<=>9x2-30x+25-9x2-6x-1=8
<=>-36x+24=8
<=>-36x=-16
<=>x=4/9
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
a) 3x( 2x + 3) -(2x+5)(3x-2)=8
<=> 6x^2+9x-6x^2+4x-15x+10=8
<=> -2x+10=8
<=> -2x= 8-10 = -2
<=> x=1
b) (3x-4)(2x+1)-(6x+5)(x-3)=3
<=> 6x^2+3x-8x-4-6x^2+18x-5x+15=3
<=> -8x+11=3
<=> -8x= -8
<=> x=1
c, 2(3x-1)(2x+5)-6(2x-1)(x+2)=-6
<=> 2(6x^2+15x-2x-5)-6(2x^2+4x-x-2)=6
<=> 2(6x^2+13x-5)-6(2x^2+3x-2)=6
<=> 12x^2+ 26x-10-12x^2-18x+12=6
<=> 8x+2=6
<=> 8x=4
<=> x= 1/2
d, 3xy(x+y)-(x+y)(x^2 +y^2+2xy)+y^3=27
<=> 3x2y+3xy2-(x+y)(x+y)2+y3=27
<=> 3x2y+3xy2-(x+y)3+y3=27
<=> 3x2y +3xy2 -x3-3x2y-3xy2-y3+y3=27
<=> -x3=27
<=> x= \(-\sqrt[3]{27}\)= -3
a) \(3x\left(2x+3\right)-\left(2x+5x\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-\left(6x^2-4x+15x^2-10x\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x^2+10x=8\)
\(\Leftrightarrow23x-15x^2-8=0\)
\(\Leftrightarrow-15x\left(x-1\right)+8\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(8-15x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\8-15x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{8}{15}\end{cases}}}\)
Vậy ....
\(3x\left(2x+3\right)-\left(2x+5x\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-\left(6x^2-4x+15x^2-10x\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x^2+10x=8\)
\(\Leftrightarrow-15x^2+23x-8=0\)
\(\Leftrightarrow-\left(15x^2-23x+8\right)=0\)
\(\Leftrightarrow-\left(15x^2-15x-8x+8\right)=0\)
\(\Leftrightarrow-\left(15x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(15x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}1-x=0\\15x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{8}{15}\end{cases}}}\)
\(x^3-3x^2+3x-1=-8\)
\(\Leftrightarrow x-1=-2\)
hay x=-1
là x^2 cơ bạn ơi