a: \(\Leftrightarrow\sqrt{6}\left(x+1\right)=5\sqrt{6}\)

=>x+1=5

=>x=4

b: =>x^2/10=1,1

=>x^2=11

=>x=căn 11 hoặc x=-căn 11

c: =>(4x+3)/(x+1)=9 và (4x+3)/(x+1)>=0

=>4x+3=9x+9

=>-5x=6

=>x=-6/5

d: =>(2x-3)/(x-1)=4 và x-1>0 và 2x-3>=0

=>2x-3=4x-4 và x>=3/2

=->-2x=-1 và x>=3/2

=>x=1/2 và x>=3/2

=>Ko có x thỏa mãn

e: Đặt căn x=a(a>=0)

PT sẽ là a^2-a-5=0

=>\(\left[{}\begin{matrix}a=\dfrac{1+\sqrt{21}}{2}\left(nhận\right)\\a=\dfrac{1-\sqrt{21}}{2}\left(loại\right)\end{matrix}\right.\)

=>x=(1+căn 21)^2/4=(11+căn 21)/2

Lời giải:
ĐKXĐ: \(1\le x\leq 2\)

Ta có: \((\sqrt{2-x}+1)(\sqrt{x+3}-\sqrt{x-1})=4\)

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

\(\Leftrightarrow (\sqrt{2-x}+1).\frac{4}{\sqrt{x+3}+\sqrt{x-1}}=4\Rightarrow \sqrt{2-x}+1=\sqrt{x+3}+\sqrt{x-1}\)

\(\Leftrightarrow (\sqrt{x+3}-2)+\sqrt{x-1}-(\sqrt{2-x}-1)=0\)

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

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

Hiển nhiên biểu thức trong ngoặc lớn luôn lớn hơnm $0$

Do đó \(\sqrt{x-1}=0\Leftrightarrow x=1\) (thỏa mãn)

đằng giữa 2 căn là dấu cộng nha ~

a.

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

\(\Rightarrow\sqrt{\left(\sqrt{x}\right)^2+2.2.\sqrt{x}+2^2}=5x+2\)

\(\Rightarrow\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

\(\Rightarrow\sqrt{x}+2=5x+2\)

\(\Rightarrow\sqrt{x}=5x\)

\(\Rightarrow x=25x^2\)

\(\Rightarrow x=0\)
Vậy nghiệm của phương trình là x = 0

b)

\(\sqrt{x-2\sqrt{x}+1}-\sqrt{x-4\sqrt{x}+4}=10\)

\(\Rightarrow\sqrt{\left(\sqrt{x}-1\right)^2}-\sqrt{\left(\sqrt{x}-2\right)^2=10}\)

\(\Rightarrow\sqrt{x}-1-\sqrt{x}+2=10\)

\(\Rightarrow1=10\) (Vô lí)

Vậy phương trình đã cho vô nghiệm

ĐK:\(-\frac{1}{2}\le x\le4\)

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

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

\(\Leftrightarrow\frac{4-x-\left(\frac{1}{2}x-2\right)^2}{\sqrt{4-x}+\frac{1}{2}x-2}+\frac{2x+1-\left(-\frac{1}{2}x-1\right)^2}{\sqrt{2x+1}+\frac{1}{2}x-1}=0\)

\(\Leftrightarrow\frac{\frac{-\left(x^2-4x\right)}{4}}{\sqrt{4-x}+\frac{1}{2}x-2}+\frac{\frac{-\left(x^2-4x\right)}{4}}{\sqrt{2x+1}+\frac{1}{2}x-1}=0\)

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

Thấy: \(\frac{1}{\sqrt{4-x}+\frac{1}{2}x-2}+\frac{1}{\sqrt{2x+1}+\frac{1}{2}x-1}>0\)

\(\Rightarrow\frac{-x\left(x-4\right)}{4}=0\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

4) Ta có: \(\left(x+3\right)\cdot\sqrt{10-x^2}=x^2-x-12\)

\(\Leftrightarrow\left(x+3\right)\cdot\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\10-x^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-8x+16-10+x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x^2-8x+6=0\end{matrix}\right.\)

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

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

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge2\)

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

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

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

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

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

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

\(\Rightarrow x=2\)

3.

ĐKXĐ: \(x\ge-1\)

\(x^2+x-12+12\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\dfrac{12\left(x-3\right)}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4+\dfrac{12}{\sqrt{x+1}+2}\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)