Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\Leftrightarrow\left(x^2-3\right)^2=0\\ \Leftrightarrow x^2-3=0\\ \Leftrightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\\ b,\Leftrightarrow8x^3+12x^2+6x+1-64=0\\ \Leftrightarrow\left(2x+1\right)^3-4^3=0\\ \Leftrightarrow\left(2x+1-4\right)\left[\left(2x+1\right)^2+4\left(2x+1\right)+16\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=3\\4x^2+4x+1+8x+4+16=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\4x^2+12x+17=0\left(1\right)\end{matrix}\right.\)
Xét \(\left(1\right)\Leftrightarrow\left(2x+3\right)^2+8=0\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
Vậy pt có nghiệm \(x=\dfrac{3}{2}\)
\(c,\Leftrightarrow\left(3-2x-5\right)\left(3-2x+5\right)=0\\ \Leftrightarrow\left(-2-2x\right)\left(8-2x\right)=0\\ \Leftrightarrow-2\left(x+1\right)\cdot2\left(4-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
\(6,\Rightarrow6x^2+12x+6-2x^3-6x^2-6x-2+2x^3-2=1\\ \Rightarrow6x=-1\Rightarrow x=-\dfrac{1}{6}\\ 7,\Rightarrow\left(x-2\right)\left(x-2-x-2\right)=0\\ \Rightarrow-4\left(x-2\right)=0\\ \Rightarrow x-2=0\Rightarrow x=2\\ 8,\Rightarrow\left(x-2\right)^2-25=0\\ \Rightarrow\left(x-2-5\right)\left(x-2+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)
d: \(\Leftrightarrow3x^2-6x-2x+4=0\)
=>(x-2)(3x-2)=0
=>x=2 hoặc x=2/3
e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)
=>x(x-3)(x+1)=0
hay \(x\in\left\{0;3;-1\right\}\)
f: \(\Leftrightarrow x^2-5x-2+x=0\)
\(\Leftrightarrow x^2-4x-2=0\)
\(\Leftrightarrow\left(x-2\right)^2=6\)
hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
a.
$x^4-6x^2+9=0$
$\Leftrightarrow (x^2-3)^2=0$
$\Leftrightarrow x^2-3=0$
$\Leftrightarrow x^2=3$
$\Leftrightarrow x=\pm \sqrt{3}$
b.
$8x^3+12x^2+6x-63=0$
$\Leftrightarrow (8x^2+12x^2+6x+1)-64=0$
$\Leftrightarrow (2x+1)^3=64=4^3$
$\Leftrightarrow 2x+1=4$
$\Leftrightarrow x=\frac{3}{2}$
c. $(3-2x)^2-25=0$
$\Leftrightarrow (3-2x)^2-5^2=0$
$\Leftrightarrow (3-2x-5)(3-2x+5)=0$
$\Leftrightarrow (-2-2x)(8-2x)=0$
$\Leftrightarrow -2-2x=0$ hoặc $8-2x=0$
$\Leftrightarrow x=-1$ hoặc $x=4$
d.
$6(x+1)^2-2(x+1)^3+2(x-1)(x^2+x+1)=1$
$\Leftrightarrow (x+1)^2[6-2(x+1)]+2(x^3-1)=1$
$\Leftrightarrow (x+1)^2(4-2x)+2x^3-3=0$
$\Leftrightarrow 6x+1=0$
$\Leftrightarrow x=\frac{-1}{6}$
e. $(x-2)^2-(x-2)(x+2)=0$
$\Leftrightarrow (x-2)[(x-2)-(x+2)]=0$
$\Leftrightarrow (x-2)(-4)=0$
$\Leftrightarrow x-2=0$
$\Leftrightarrow x=2$
f. $x^2-4x+4=25$
$\Leftrightarrow (x-2)^2=5^2=(-5)^2$
$\Leftrightarrow x-2=5$ hoặc $x-2=-5$
$\Leftrightarrow x=7$ hoặc $x=-3$