ĐKXĐ: bạn tự đặt nhé

đặt \(\sqrt{x^2-8}=a\) và \(\sqrt{x-2}=b\)==> \(a^2+b^2=x^2+x-10\)<=> \(x^2+x-9=a^2+b^2+1\)

khi đó pt đầu trở thành 

\(a^2+b^2+1=a+b+ab\)

<=> \(2a^2+2b^2+2-2a-2b-2ab=0\)

<=> \(\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)

<=>\(\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\)

<=> \(a=b=1\)

==> \(\sqrt{x-2}=1\)<=>x=3 

vậy x=3

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

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

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

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

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

\(\Leftrightarrow...\)

\(a,\left(\dfrac{1}{4}\right)^{x-2}=\sqrt{8}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^{2x-4}=\left(\dfrac{1}{2}\right)^{-\dfrac{3}{2}}\\ \Leftrightarrow2x-4=-\dfrac{3}{2}\\ \Leftrightarrow2x=\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{4}\)

\(b,9^{2x-1}=81\cdot27^x\\ \Leftrightarrow3^{4x-2}=3^{4+3x}\\ \Leftrightarrow4x-2=4+3x\\ \Leftrightarrow x=6\)

c, ĐK: \(x-2>0\Rightarrow x>2\)

\(2log_5\left(x-2\right)=log_59\\ \Leftrightarrow log_5\left(x-2\right)^2=log_59\\ \Leftrightarrow\left(x-2\right)^2=3^2\\ \Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = 5.

d, ĐK: \(x-1>0\Leftrightarrow x>1\)

\(log_2\left(3x+1\right)=2-log_2\left(x-1\right)\\ \Leftrightarrow log_2\left(3x+1\right)\left(x-1\right)=2\\ \Leftrightarrow3x^2-2x-1=4\\ \Leftrightarrow3x^2-2x-5=0\\ \Leftrightarrow\left(3x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

Vậy phương trình có nghiệm \(x=\dfrac{5}{3}\)

a. 

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$

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

$\Leftrightarrow 2x=9$

$\Leftrightarrow x=\frac{9}{2}$ (tm)

b.

ĐKXĐ: $x\geq -2$

PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$

$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$

$\Leftrightarrow 3\sqrt{x+2}=15$

$\Leftrightarrow \sqrt{x+2}=5$

$\Leftrightarrow x+2=25$

$\Leftrightarrow x=23$ (tm)

c.

$\sqrt{(x-2)^2}=12$

$\Leftrightarrow |x-2|=12$

$\Leftrightarrow x-2=12$ hoặc $x-2=-12$

$\Leftrightarrow x=14$ hoặc $x=-10$

e.

PT $\Leftrightarrow |2x-1|-x=3$

Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$

$\Leftrightarrow x=4$ (tm)

Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

ĐK: mọi x thuộc R

Ta có:\(x^2+5x+9=\left(x+5\right)\sqrt{x^2+9}\)

   \(\Leftrightarrow\left(x+5\right)\sqrt{x^2+9}-x^2-5x-9=0\)

   \(\Leftrightarrow\left(x+5\right)\left(\sqrt{x^2+9}-5\right)-\left(x^2-16\right)=0\)

   \(\Leftrightarrow\left(x+5\right).\dfrac{x^2-16}{\sqrt{x^2+9}+5}-\left(x-4\right)\left(x+4\right)=0\)

   \(\Leftrightarrow\left(x+5\right).\dfrac{\left(x-4\right)\left(x+4\right)}{\sqrt{x^2+9}+5}-\left(x-4\right)\left(x+4\right)=0\)

   \(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{x+5}{\text{​​}\sqrt{x^2+9}+5}-1\right)=0\)

   \(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\\dfrac{x+5}{\text{​​}\sqrt{x^2+9}+5}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\\dfrac{x+5}{\text{​​}\sqrt{x^2+9}+5}=1\left(1\right)\end{matrix}\right.\)

Giải (1) ta có:

\(\left(1\right)\Leftrightarrow x+5=\sqrt{x^2+9}+5\)

     \(\Leftrightarrow x=\sqrt{x^2+9}\)

     \(\Leftrightarrow x^2=x^2+9\)

     \(\Leftrightarrow0=9\) (vô lí)

Vậy phương trình có 2 nghiệm là ... 

a. Đề bài sai, phương trình không giải được

b.

ĐKXĐ: \(x\ge-\dfrac{2}{3}\)

\(\left(2x+10\right)\left(\dfrac{1-\left(3+2x\right)}{1+\sqrt{3+2x}}\right)^2=4\left(x+1\right)^2\)

\(\Leftrightarrow\dfrac{\left(2x+10\right)4.\left(x+1\right)^2}{\left(1+\sqrt{3+2x}\right)^2}=4\left(x+1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x+1\right)^2=0\Rightarrow x=-1\\2x+10=\left(1+\sqrt{3+2x}\right)^2\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow2x+10=2x+4+2\sqrt{2x+3}\)

\(\Leftrightarrow\sqrt{2x+3}=3\)

\(\Leftrightarrow x=3\)

cho em hỏi , em thấy câu a có nghiệm mà

a, ĐK: \(x\le-1,x\ge3\)

\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)

\(\Leftrightarrow\left(2\sqrt{x^2-2x-3}+3\right).\left(\sqrt{x^2-2x-3}-1\right)=0\)

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

\(\Leftrightarrow x^2-2x-3=1\)

\(\Leftrightarrow x^2-2x-4=0\)

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

b, ĐK: \(-2\le x\le2\)

Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)

Khi đó phương trình tương đương:

\(3t-t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)

Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm

Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)