ĐKXĐ: \(-1\le x\le8\) Đặt \(t=\sqrt{x+1}+\sqrt{8-x}\) ( Với \(t\ge0\))

\(\Rightarrow t^2=9+2\sqrt{\left(x+1\right)\left(8-x\right)}\)\(\Rightarrow\sqrt{\left(x+1\right)\left(8-x\right)}=\dfrac{t^2-9}{2}\)

\(\Rightarrow t+\dfrac{t^2-9}{2}=3\Rightarrow t^2+2t-15=0\)\(\Rightarrow\left(t+5\right)\left(t-3\right)=0\)

\(\left[{}\begin{matrix}t=-5\left(Loai\right)\\t=3\end{matrix}\right.\Rightarrow t=3\)

\(\Rightarrow3+\sqrt{\left(x+1\right)\left(8-x\right)}=3\) \(\Rightarrow\sqrt{\left(x+1\right)\left(8-x\right)}=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\) Thỏa mãn điều kiện .

Bài 2: 

a: Ta có: \(\sqrt{\sqrt{5}-x\sqrt{3}}=\sqrt{8+2\sqrt{15}}\)

\(\Leftrightarrow\sqrt{5}-x\sqrt{3}=8+2\sqrt{15}\)

\(\Leftrightarrow x\sqrt{3}=\sqrt{5}-8-2\sqrt{15}\)

\(\Leftrightarrow x=\dfrac{\sqrt{15}-8\sqrt{3}-6\sqrt{5}}{3}\)

b: Ta có: \(\sqrt{2+\sqrt{\sqrt{x}+3}}=3\)

\(\Leftrightarrow\sqrt{\sqrt{x}+3}=7\)

\(\Leftrightarrow\sqrt{x}=46\)

hay x=2116

MN ƠI GIÚP MK NHA

Bài 1 bạn tìm quanh quanh đây, mình thấy có bài y hệt rồi nên ko làm nữa

Bài 2 như sau:

ĐKXĐ: \(x\ge\dfrac{-1}{16}\)

\(x^2-x-20-2\left(\sqrt{16x+1}-9\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)-2\dfrac{\left(\sqrt{16x+1}-9\right)\left(\sqrt{16x+1}+9\right)}{\sqrt{16x+1}+9}=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)-\dfrac{32\left(x-5\right)}{\sqrt{16x+1}+9}=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4-\dfrac{32}{\sqrt{16x+1}+9}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\Rightarrow x=5\\x+4-\dfrac{32}{\sqrt{16x+1}+9}=0\left(1\right)\end{matrix}\right.\)

Xét phương trình (1): ta có \(x+4\ge-\dfrac{1}{16}+4=\dfrac{63}{16}\) \(\forall x\ge-\dfrac{1}{16}\)

\(\sqrt{16x+1}+9\ge9\Rightarrow\dfrac{32}{\sqrt{16x+1}+9}\le\dfrac{32}{9}\) \(\forall x\ge-\dfrac{1}{16}\)

Mà \(\dfrac{63}{16}-\dfrac{32}{9}=\dfrac{55}{144}>0\) \(\Rightarrow x+4-\dfrac{32}{\sqrt{16x+1}+9}>0\) \(\forall x\ge-\dfrac{1}{16}\)

\(\Rightarrow\) pt (1) vô nghiệm

Vậy pt đã cho có nghiệm duy nhất \(x=5\)

\(=\sqrt{\left(\sqrt{x+7}+1\right)^2}+\sqrt{x+7-\sqrt{x+7}-6}=4\)ĐK:\(x\ge-7\)

Đặt \(t=\sqrt{x+7}\left(t\ge0\right)\)

\(\Rightarrow t+1-4=\sqrt{t^2-t-6}\)

\(\Leftrightarrow t^2-6t+9=t^2-t-6\left(t\ge3\right)\)

\(\Leftrightarrow5t=15\)

\(\Leftrightarrow t=3\left(TM\right)\)\(\Rightarrow x=2\left(tm\right)\)

S={2}

b)ĐK:\(x\ge2\)

pt\(\Leftrightarrow\sqrt{x-2+2\sqrt{x-2}+2}-\sqrt{x-2-2\sqrt{x-2}+2}=-2\)

Đặt t= can(x-2)(t>=0)

Đến đây bạn giải tiếp nhé!

#Walker

b) ĐKXĐ:    \(x\ne1\)

Ta có:

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

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

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

Đặt \(\frac{x^2}{x-1}=a\)

Khi đó pt đã cho trở thành:

\(a^3-3a^2+3a-2=0\)

\(\Leftrightarrow\left(a-1\right)^3=1\Rightarrow a-1=1\Leftrightarrow a=2\)

Theo cách đặt:   \(\frac{x^2}{x-1}=2\Rightarrow x^2=2x-2\Leftrightarrow x^2-2x+1=-1\Leftrightarrow\left(x-1\right)^2=-1\left(ptvn\right)\)

a) ĐKXĐ:   \(x\ge8\)

Ta có:

\(x-\sqrt{x-8}-3\sqrt{x}+1=0\)

\(\Leftrightarrow x-9-\left(\sqrt{x-8}-1\right)-3\left(\sqrt{x}-3\right)=0\)

\(\Leftrightarrow x-9-\frac{x-9}{\sqrt{x-8}+1}-3.\frac{x-9}{\sqrt{x}+3}=0\)

\(\Leftrightarrow\left(x-9\right)\left(\frac{3}{\sqrt{x}+3}+\frac{1}{\sqrt{x-8}+1}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-9=0\\\frac{3}{\sqrt{x}+3}+\frac{1}{\sqrt{x-8}+1}-1=0\end{cases}}\)

+)  \(x-9=0\Leftrightarrow x=9\left(TMĐKXĐ\right)\)

+)  \(\frac{3}{\sqrt{x}+3}=\frac{\sqrt{x-8}}{\sqrt{x-8}+1}\Rightarrow\sqrt{x\left(x-8\right)}=3\)

\(\Leftrightarrow x^2-8x-9=0\Leftrightarrow\orbr{\begin{cases}x=9TMĐKXĐ\\x=-1\left(KTMĐKXĐ\right)\end{cases}}\)

Vaayh pt có 1 nghiệm là x=9