1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

a: \(\dfrac{x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)

\(=\dfrac{x+10}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-\left(x+10\right)}{2\left(x+2\right)}\)

b: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)

\(=\dfrac{\left(2x-1\right)\left(2x+1\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(x-2\right)}\)

\(=\dfrac{3\left(2x-1\right)\left(2x+1\right)}{2\left(x-2\right)\left(x+4\right)}\)

c: \(=\dfrac{4y^2}{7x^4}\cdot\dfrac{35x^2}{-8y}=\dfrac{5}{x^2}\cdot\dfrac{-1}{2}\cdot y=\dfrac{-5y}{2x^2}\)

d: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)

a.(x+10) /(4*x)-8* 4 -(2*x)/x+2

-(127*x-10)/(4*x)

(5/2-127*x/4)/x

Câu a

a) \(\left(5x-2\right)\left(5x+2\right)-\left(5x+3\right)\left(5x-4\right)=0\)

\(\Leftrightarrow5x+8=8\)

\(\Leftrightarrow5x=8-8\)

\(\Leftrightarrow x=5.0\)

\(\Leftrightarrow x=0\)

b) 

câu b sao bạn

\(\left(4x-3\right)\left(4x+2\right)+\left(4x+5\right)\left(1-4x\right)=2.5^2\)

\(16x^2+8x-12x-6+4x-16x^2+5-20x=50\)

\(-20x-1=50\)

\(-20x=51\)

\(x=\frac{-51}{20}\)

Vậy \(x=\frac{-51}{20}\)

a ) \(\left(5x-2\right)\left(5x+2\right)-\left(5x+3\right)\left(5x-4\right)=8\)

\(\Leftrightarrow\left(5x\right)^2-4-\left(25x^2+15x-20x-12\right)=8\)

\(\Leftrightarrow25x^2-4-25x^2-15x+20x+12=8\)

\(\Leftrightarrow5x+8=8\)

\(\Leftrightarrow5x=0\)

\(\Leftrightarrow x=0\)

Vậy \(x=0\)

b ) \(\left(4x-3\right)\left(4x+2\right)+\left(4x+5\right)\left(1-4x\right)=2.5^2\)

\(\Leftrightarrow16x^2-12x+8x-6+4x+5-16x^2-20x=50\)

\(\Leftrightarrow-20x-1=50\)

\(\Leftrightarrow-20x=51\)

\(\Leftrightarrow x=-\dfrac{51}{20}\)

Vậy \(x=-\dfrac{51}{20}\)

ĐKXĐ:x∈R

\(\sqrt{2x^2-4x+12}=4x+8-2x^2\)

\(\Leftrightarrow2x^2-4x+12=\left(4x+8-2x^2\right)^2\)

\(\Leftrightarrow2x^2-4x+12=4x^4-16x^3-16x^2+64x+64\)

\(\Leftrightarrow4x^4-16x^3-18x^2+68x+52=0\)

\(\Leftrightarrow2x^4-8x^3-9x^2+34x+26=0\)

\(\Leftrightarrow2x^4-8x^3-9x^2+34x+26=0\)

rồi bạn phân tích đa thức thành nhân tử với nhân tử \(x^2-2x-2\) nhé

ta có

\(5x=-3y=4z\)

\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{3z}{45}=\frac{x-y+3z}{12+20+45}=\frac{7}{77}=\frac{1}{11}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{11}.12=\frac{12}{11}\\-y=\frac{1}{11}.20=\frac{20}{11}\\3z=\frac{1}{11}.45=\frac{45}{11}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{12}{11}\\y=-\frac{20}{11}\\z=\frac{45}{11}:3=\frac{15}{11}\end{cases}}\)

Vậy \(\hept{\begin{cases}x=\frac{12}{11}\\y=\frac{-20}{11}\\z=\frac{15}{11}\end{cases}}\)

c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)

\(\Leftrightarrow2\sqrt{x-1}=4\)

\(\Leftrightarrow x-1=4\)

hay x=5

e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)

\(\Leftrightarrow\left|2x-7\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)

a. ĐKXĐ: $x\in\mathbb{R}$

PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$

$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$

$\Leftrightarrow x\leq 2$

b. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$

$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$

$\Leftrightarrow 1=2\sqrt{x-2}$

$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$

$\Leftrightarrow \frac{1}{4}=x-2$

$\Leftrightarrow x=\frac{9}{4}$ (tm)

c. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \sqrt{x-1}+\sqrt{9}.\sqrt{x-1}-\sqrt{4}.\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x-1}+3\sqrt{x-1}-2\sqrt{x-1}=4$

$\Leftrightarrow 2\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x-1}=2$

$\Leftrightarrow x-1=4$

$\Leftrightarrow x=5$ (tm)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-2}-4\sqrt{\frac{4}{9}}\sqrt{x-2}+\sqrt{9}.\sqrt{x-2}-5=0$

$\Leftrightarrow \frac{1}{2}\sqrt{x-2}-\frac{8}{3}\sqrt{x-2}+3\sqrt{x-2}-5=0$

$\Leftrightarrow \frac{5}{6}\sqrt{x-2}-5=0$

$\Leftrightarrow \sqrt{x-2}=6$

$\Leftrightarrow x-2=36$

$\Leftrightarrow x=38$ (tm)

Đặt \(x^2+4x+8=a\)(a > 0 vì \(x^2+4x+8=\left(x+2\right)^2+4>0\)) , \(x^2+4x+4=b\left(b\ge0\right)\)
\(\Rightarrow a+b=x^2+4x+8+x^2+4x+4=2\left(x^2+4x+6\right)\)
Khi đó phương trình đã cho trở thành:
\(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\\ \Leftrightarrow a+b+2\sqrt{ab}=a+b\\ \Leftrightarrow2\sqrt{ab}=0\\\Leftrightarrow\sqrt{ab}=0\\ \Leftrightarrow ab=0\\ \Leftrightarrow\left[{}\begin{matrix}a=0\left(kh\text{ô}ngtm\right)\\b=0\end{matrix}\right.\)
\(\Rightarrow x^2+4x+4=0\\ \Leftrightarrow\left(x+2\right)^2=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2.\)

Kl: