1.

ĐK: \(-1\le x\le4\)

Đặt \(\sqrt{x+1}+\sqrt{4-x}=t\left(t\ge0\right)\)

\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{t^2-5}{2}\)

\(PT\Leftrightarrow t+\frac{t^2-5}{2}=5\Rightarrow t^2+2t-15=0\) \(\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)

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

2.

ĐK:\(x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t\ge0\right)\)

\(\Rightarrow2\sqrt{x^2-16}=t^2-2x\)

\(PT\Leftrightarrow t=2x-12+t^2-2x\)

\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\) Giải tiếp như trên.

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé

a/ ĐKXĐ: ...

\(\Leftrightarrow2\sqrt{\frac{x}{x-1}}-\sqrt{\frac{x-1}{x}}=\frac{2\left(x-1\right)}{x}+3\)

Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)

\(\frac{2}{a}-a=2a^2+3\Leftrightarrow2a^3+a^2+3a-2=0\)

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

\(\Rightarrow\sqrt{\frac{x-1}{x}}=\frac{1}{2}\Leftrightarrow4\left(x-1\right)=x\)

b/ ĐKXĐ: ...

\(\Leftrightarrow3\sqrt{\frac{2x}{x-1}}+4\sqrt{\frac{x-1}{2x}}=\frac{3\left(x-1\right)}{2x}+10\)

Đặt \(\sqrt{\frac{x-1}{2x}}=a>0\)

\(\frac{3}{a}+4a=3a^2+10\Leftrightarrow3a^3-4a^2+10a-3=0\)

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

\(\Leftrightarrow\sqrt{\frac{x-1}{2x}}=\frac{1}{3}\Leftrightarrow9\left(x-1\right)=2x\)

c/ ĐKXĐ: ...

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

Đặt \(\sqrt{\frac{3-2x}{x}}=a>0\)

\(\frac{1}{a}+5a=4a^2+5\Leftrightarrow4a^3-5a^2+5a-1=0\)

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

\(\Leftrightarrow\sqrt{\frac{3-2x}{x}}=\frac{1}{4}\Leftrightarrow16\left(3-2x\right)=x\)

d/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)

\(a^2-2a=3\Leftrightarrow a^2-2a-3=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=3\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{\frac{x-1}{x}}=3\Leftrightarrow x-1=9x\)

e/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{x}{x-1}}=a>0\)

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

\(\Rightarrow\left[{}\begin{matrix}a=\sqrt{2}\\a=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

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

f/ ĐKXĐ: ...

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

Bình phương 2 vế:

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

\(\Rightarrow x=1\)

a

Lời giải:

a) ĐK: \(x>0; x\neq 25; x\neq 36\)

PT \(\Rightarrow (\sqrt{x}-2)(\sqrt{x}-6)=(\sqrt{x}-5)(\sqrt{x}-4)\)

\(\Leftrightarrow x-8\sqrt{x}+12=x-9\sqrt{x}+20\)

\(\Leftrightarrow \sqrt{x}=8\Rightarrow x=64\) (thỏa mãn)

Vậy.......

b)

ĐK: \(x\geq \frac{-1}{2}\)

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

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

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

\(\Rightarrow x=\frac{3^2-1}{2}=4\) (thỏa mãn)

c)

ĐK: \(x\geq 2\)

PT \(\Leftrightarrow \sqrt{4(x-2)}-\frac{1}{2}\sqrt{x-2}+\sqrt{9(x-2)}=9\)

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

\(\Leftrightarrow \frac{9}{2}\sqrt{x-2}=9\Leftrightarrow \sqrt{x-2}=2\Rightarrow x=2^2+2=6\) (thỏa mãn)

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

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

\(=\dfrac{3}{\sqrt{x}-2}\)

Trả lời nhanh giúp mình với mình cần gấp lắm

1/ Đặt \(\sqrt{9-x^2}=a\ge0\)

\(\Rightarrow\frac{9-a^2}{3+a}+\frac{1}{12-4a}=1\)

\(\Leftrightarrow4a^2-20a+25=0\)

\(\Leftrightarrow a=\frac{5}{2}\)

\(\Rightarrow\sqrt{9-x^2}=\frac{5}{2}\)

\(\Leftrightarrow x^2=\frac{11}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{\sqrt{11}}{2}\\x=\frac{\sqrt{11}}{2}\end{cases}}\)

2/ \(\frac{9}{x^2}+\frac{2x}{\sqrt{2x^2+9}}-1=0\)

\(\Leftrightarrow\frac{9+2x^2}{x^2}+\frac{2x}{\sqrt{2x^2+9}}-3=0\)

Đặt \(\frac{x}{\sqrt{2x^2+9}}=a\)

\(\Rightarrow\frac{1}{a^2}+2a-3=0\)

\(\Leftrightarrow2a^3-3a^2+1=0\)

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

Làm nốt nhé

3/ \(x+\frac{x}{\sqrt{x^2-1}}=2\sqrt{2}\)

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

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

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

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