5(x^2-1)+x(1=5x)=x-2 giúp mk vs mn
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\(x(x-5)(x+5)-(x+2)(x^2-2x+4)\)
\(\Leftrightarrow x(x^2-25)-(x^3+8)\)
\(\Leftrightarrow x^3-25x-x^3-8\)
\(\Leftrightarrow-25x=11\Leftrightarrow x=-\frac{11}{25}\)
\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)\)
\(\rightarrow\)\(x\left(x^2-25\right)-\left(x^3+8\right)\)
\(\rightarrow\)\(x^3-25x-x^3-8\)
\(\rightarrow\)\(-25x=11\Leftrightarrow x=-\frac{11}{25}\)
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x ( x-5) (x+5) -(x+2)(x2 -2x +4 ) =17
x. (x2 -25) - ( x3 + 8 ) =17
x3 -25x -x3 -8 =17
-25x -8 =17
-25x = 25
x=-1
vậy x =-1
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\(=\left(2x\right)^2+\left(3y\right)^2-\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]\)
\(=4x^2-9y^2-\left(2x^3-3y^3\right)\)
\(=4x^2-9y^2-2x^3+3y^3\)
áp dụng hằng đẳng thức số 4 và số 6
Giải
(2x+3y)(2x-3y)-(2x+3y)(4x2-6xy+9y2)
=4x2-9y2-(8x3+27y3)
=4x2-9y3-8x3-27y3
=-4x2-36y3
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\(6x^2-\left(2x-3\right)\left(3x-2\right)=1\)
\(\Rightarrow6x^2-\left(6x^2-4x-9x+4\right)=1\)
\(\Rightarrow6x^2-6x^2-13x+6=1\)
\(\Rightarrow-13x+6=1\)
\(\Rightarrow-13x=-7\)
\(\Rightarrow x=\frac{7}{13}\)
\(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-2=0\)
\(\Rightarrow x^3+3x^2+3x+1-x^3+1-2=0\)
\(\Rightarrow3x^2+3x=0\)
\(\Rightarrow3x\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
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\(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)
`5x (x-3)-x+3=0`
`->5x (x-3)-(x-3)=0`
`-> (x-3)(5x-1)=0`
TH1 : `x-3=0 ->x=3`
TH2 : `5x-1=0 ->x=1/5`
Vậy `x=3,x=1/5`
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`(2x^2 +1/3)^3`
`= (2x^2)^3 + 3 . (2x^2)^2 . 1/3 + 3 . 2x^2 . (1/3)^2 + (1/3)^3`
`= 8x^6 + 4x^4 + 6/9 x^2 + 1/27`
\(5x^2\)\(-15x+5+x-5x^2\)\(=x-2\)
\(-15x+5=-2\)
\(-15x=-2-5=-7\)
\(x=-7:-15=\frac{-7}{-15}\)\(=\frac{7}{15}\)
\(x=\frac{7}{15}\)