Điều kiện xác định : \(x\ge2\)

Ta có : \(\sqrt{x+8+2\sqrt{x+7}}+\sqrt{x+1-\sqrt{x+7}}=4\)

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

\(\Leftrightarrow\sqrt{x+7}+\sqrt{\left(x+7\right)-\sqrt{x+7}-6}-3=0\)

Đặt \(t=\sqrt{x+7},t\ge0\) , pt trở thành \(t+\sqrt{t^2-t-6}-3=0\)

\(\Leftrightarrow\left(t-3\right)+\sqrt{\left(t-3\right)\left(t+2\right)}=0\)

\(\Leftrightarrow\sqrt{t-3}\left(\sqrt{t-3}+\sqrt{t+2}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{t-3}=0\\\sqrt{t-3}+\sqrt{t+2}=0\end{array}\right.\)

Vì \(\sqrt{t-3}\ge0,\sqrt{t+2}\ge0\Rightarrow\sqrt{t-3}+\sqrt{t+2}\ge0\) . Dấu "=" không đồng thời xảy ra nên pt vô nghiệm.

Vậy t = 3 => x = 2

pt có nghiệm x = 2

a.

ĐKXĐ: \(1\le x\le7\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Biến đổi pt đầu:

\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a^2b^2-b^4=b-a\)

\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)

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

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)

Thế vào pt dưới:

\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)

\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)

\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)

\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)

\(\Leftrightarrow...\)

a/ \(\Leftrightarrow\sqrt{x^2+x+3}-\sqrt{x^2+2}+\sqrt{x^2+x+8}-\sqrt{x^2+7}=0\)

\(\Leftrightarrow\frac{x+1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{x+1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}\right)=0\)

\(\Leftrightarrow x+1=0\) (ngoặc to phía sau luôn dương)

\(\Rightarrow x=-1\)

b/

\(\sqrt{7-x^2+x\sqrt{x+5}}=\sqrt{3-2x-x^2}\) (1)

\(\Rightarrow7-x^2+x\sqrt{x+5}=3-2x-x^2\)

\(\Leftrightarrow x\sqrt{x+5}=-2x-4\)

\(\Rightarrow x^2\left(x+5\right)=4x^2+16x+16\)

\(\Rightarrow x^3+x^2-16\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x^2-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)

Do các phép biến đổi ko tương đương nên cần thay nghiệm vào (1) để kiểm tra

c/ ĐKXĐ: \(x\ge\frac{5}{3}\)

\(\Leftrightarrow\sqrt{10x+1}-\sqrt{9x+4}+\sqrt{3x-5}-\sqrt{2x-2}=0\)

\(\Leftrightarrow\frac{x-3}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{x-3}{\sqrt{3x-5}+\sqrt{2x-2}}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{1}{\sqrt{3x-5}+\sqrt{2x-2}}\right)=0\)

\(\Leftrightarrow x-3=0\) (ngoặc phía sau luôn dương)

d/ Đề bài là \(2\sqrt{2x+3}\) hay \(2\sqrt{2x-3}\) bạn?

e/ ĐKXĐ: \(x\ge-3\)

\(\Leftrightarrow\sqrt{x+3+2\sqrt{x+3}+1}=x+4\)

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

\(\Leftrightarrow\sqrt{x+3}+1=x+4\)

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

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

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

d, ĐK: \(x\ge\frac{3}{2}\)

\(PT\Leftrightarrow\sqrt{5x-6}-\sqrt{x+7}+2\sqrt{2x-3}-\sqrt{4x+1}=0\)

\(\Leftrightarrow\frac{4x-13}{\sqrt{5x-6}+\sqrt{x-7}}+\frac{4x-13}{2\sqrt{2x-3}+\sqrt{4x+1}}=0\)

\(\Leftrightarrow\left(4x-13\right)\left(\frac{1}{\sqrt{5x-6}+\sqrt{x+7}}+\frac{1}{2\sqrt{2x-3}+\sqrt{4x+1}}\right)=0\)

Vì \(\frac{1}{\sqrt{5x-6}+\sqrt{x+7}}+\frac{1}{2\sqrt{2x-3}+\sqrt{4x+1}}>0\)

\(\Leftrightarrow x=\frac{13}{4}=3,25\)(tm)

\(\sqrt{\frac{-6}{1+x}}=5\)

\(\Leftrightarrow\sqrt{\frac{-6}{1+x}}^2=5^2\)

\(\Leftrightarrow\frac{-6}{1+x}=25\)

\(\Leftrightarrow x+1=\frac{-6}{25}\)

\(\Leftrightarrow x=\frac{-6}{25}-1=\frac{-31}{25}\)

\(\sqrt{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}=2\)

\(\Leftrightarrow\sqrt{x-49}=2\)

\(\Leftrightarrow x-49=4\Leftrightarrow x=53\)

\(\sqrt{\frac{1}{4}-2a}=3\)

\(\Leftrightarrow\frac{1}{4}-2a=9\)

\(\Leftrightarrow2a=\frac{1}{4}-\frac{36}{4}\)

\(\Leftrightarrow2a=\frac{-35}{4}\)

\(\Leftrightarrow a=\frac{-35}{8}\)

đội tuyển toán tự làm đi m 

a) ĐKXĐ: 1\(\le x\le7\)

phương trình <=> \(x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(7-x\right)\left(x-1\right)}=0\\ \Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\\ \Leftrightarrow\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}-\sqrt{7-x}\right)=0\\\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=7-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\left(thoả.mãn\right) \)

Vậy S={5,4} là tập nghiệm của phương trình

b) PT <=> \(2x^2-6x+4=\sqrt[2]{\left(x+2\right)\left(x^2-2x+4\right)}\)

Đặt \(\sqrt[2]{x+2}=y,\sqrt[2]{x^2-2x+4}=z\) (y,z>=0)

=> z^2-y^2=x^2-3x+2

pt<=> 2z^2-2y^2=3yz <=> (2z+y)(z-2y)=0

đến đó tự làm tự đặt dkxd

c) Đặt 2 cái căn là a,b => 2a+b=8

và 2a^3 -b^3=1

Thế b=8-2a. pt<=> 2a^3 -(8-2a)^3=1. Đến đó tự giải

b,

+ Với \(x=0\) \(\Rightarrow PTVN\)

+ Với \(x\ne0\), chia cả 2 vế cho \(x^2\) :

\(PT\Leftrightarrow x^2-16x+46+\frac{144}{x}+\frac{81}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\frac{81}{x^2}\right)-16\left(x-\frac{9}{x}\right)+46=0\)

Đặt \(x-\frac{9}{x}=t\Rightarrow t^2=x^2+\frac{81}{x^2}-18\)

\(\Leftrightarrow t^2+18-16t+46=0\)

\(\Leftrightarrow t^2-16t+64=0\Rightarrow t=8\)

\(\Leftrightarrow x-\frac{9}{x}=8\Leftrightarrow x^2-8x-9=0\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\) (t/m)