Tìm x :
a) \(x^2-4=8.\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)=8.\left(x-2\right)\)
\(\Leftrightarrow x+2=8\)
\(\Leftrightarrow x=6\)
Vậy : \(x=6\)
b) \(4x^2-12x+9=\left(5-x\right)^2\)
\(\Leftrightarrow\left(2x\right)^2-2.2x.3+3^2=\left(5-x\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(5-x\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=5-x\\2x-3=x-5\end{matrix}\right.\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}3x=8\Rightarrow x=\frac{8}{3}\\x=-2\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{8}{3},-2\right\}\)
a)(x-2)(x+2)(x\(^2\)+4)-(x\(^2\)-3)(x\(^2\)+3)
= (x\(^2\)-4)(x\(^2\)+4)-(x\(^4\)-9)
= x\(^4\)-16-x\(^4\)+9= -7
b)(x-2)\(^2\)-(x+1)(x\(^2\)-x+1)+6(x-1)\(^2\)
= (x-2)\(^2\)-(x\(^3\)+1)+6(x\(^2\)-2x+1)
= x\(^2\)-4x +4-x\(^3\)-1+6x\(^2\)-12x+6
= -x\(^3\)+7x\(^2\)-16x+10
a)x\(^2\)-4=8(x-2)
<=>x\(^2\)-8x+16=4
<=>(x-4)\(^2\)=4
=>x-4=4 hoặc x-4=-4
=> x=8 hoặc x=0
b) 4x\(^2\)-12x+9=(5-x)\(^2\)
<=>(2x-3)\(^2\)-(5-x)\(^2\)=0
<=>(2x-3+5-x)(2x-3-5+x)=0
<=>(x+2)(3x-8)=0
=>x+2=0 hoặc 3x-8=0
=>x=-2 hoặc x=\(\frac{8}{3}\)
Sửa lại:
a)x\(^2\)-4=8(x-2)
<=>x\(^2\)-8x+16=4
<=>(x-4)\(^2\)=4
=>x-4=2 hoặc x-4=-2
=>x=6 hoặc x=2
