a) ĐKXĐ: \(x\ge0\)

Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)

\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)

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

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)

1. ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow \sqrt{x-1}=5-\sqrt{x-4}$

$\Rightarrow x-1=25+x-4-10\sqrt{x-4}$

$\Leftrightarrow 22=10\sqrt{x-4}$

$\Leftrightarrow 2,2=\sqrt{x-4}$

$\Leftrightarrow 4,84=x-4\Leftrightarrow x=8,84$

(thỏa mãn)

2. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow (2x-2\sqrt{x})-(5\sqrt{x}-5)=0$

$\Leftrightarrow 2\sqrt{x}(\sqrt{x}-1)-5(\sqrt{x}-1)=0$

$\Leftrightarrow (\sqrt{x}-1)(2\sqrt{x}-5)=0$

$\Leftrightarrow \sqrt{x}-1=0$ hoặc $2\sqrt{x}-5=0$

$\Leftrightarrow x=1$ hoặc $x=\frac{25}{4}$ (tm)

3. ĐKXĐ: $x\geq 3$

Bình phương 2 vế thu được:

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

$\Leftrightarrow 4(2x+1)(x-3)=(x+2)^2$

$\Leftrightarrow 4(2x^2-5x-3)=x^2+4x+4$
$\Leftrightarrow 7x^2-24x-16=0$

$\Leftrightarrow (x-4)(7x+4)=0$

Do $x\geq 3$ nên $x=4$

Thử lại thấy thỏa mãn

Vậy $x=4$

4. ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow (x-4\sqrt{x}+4)+2021\sqrt{x-4}=0$

$\Leftrightarrow (\sqrt{x}-2)^2+2021\sqrt{x-4}=0$

Ta thấy, với mọi $x\geq 4$ thì:

$(\sqrt{x}-2)^2\ge 0$

$2021\sqrt{x-4}\geq 0$ 

Do đó để tổng của chúng bằng $0$ thì:
$\sqrt{x}-2=\sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)

\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)

\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)

a/ ĐKXĐ: \(x\ge-1\)

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

\(\Leftrightarrow2\left(\sqrt{x+1}+1\right)-\sqrt{x+1}=4\)

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

\(\Rightarrow x=3\)

b/ ĐKXĐ: \(x\ge1\)

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

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

Ta có \(VT\ge\left|\sqrt{x-1}-1+2-\sqrt{x-1}\right|=1\)

Nên dấu "=" xảy ra khi và chỉ khi:

\(1\le\sqrt{x-1}\le2\Rightarrow2\le x\le5\)

Vậy nghiệm của pt là \(2\le x\le5\)

c/ ĐKXĐ: \(x\ge1\)

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

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

- Với \(\sqrt{x-1}\ge1\Rightarrow x\ge2\) ta có:

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

\(\Leftrightarrow2=2\) (luôn đúng)

- Với \(1\le x< 2\) ta có:

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

\(\Leftrightarrow\sqrt{x-1}=1\Rightarrow x=2\left(l\right)\)

Vậy nghiệm của pt là \(x\ge2\)

d/ ĐKXĐ: \(-\le x\le1\)

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

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

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

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

TH1: \(\sqrt{4-4x^2}\ge1\Rightarrow-\frac{\sqrt{3}}{2}\le x\le\frac{\sqrt{3}}{2}\) ta có:

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

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

\(\Leftrightarrow4-4x^2=x^2+2x+1\)

\(\Leftrightarrow5x^2+2x-3=0\Rightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=\frac{3}{5}\end{matrix}\right.\)

TH2: \(\left[{}\begin{matrix}-1\le x< -\frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}< x\le1\end{matrix}\right.\) ta có:

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

\(\Leftrightarrow2x=0\Rightarrow x=0\left(l\right)\)

Vậy pt có nghiệm duy nhất \(x=\frac{3}{5}\)

a: \(\dfrac{6}{2-\sqrt{10}}-\dfrac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\sqrt{49+4\sqrt{10}}\)

\(=\dfrac{6\left(2+\sqrt{10}\right)}{4-10}-\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}+\sqrt{49+2\cdot2\sqrt{10}}\)

\(=\dfrac{6\left(2+\sqrt{10}\right)}{-6}-\sqrt{10}+\sqrt{49+2\cdot\sqrt{40}}\)

\(=-2-\sqrt{10}-\sqrt{10}+\sqrt{49+4\sqrt{10}}\)

\(=-2-2\sqrt{10}+\sqrt{49+4\sqrt{10}}\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)

\(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)

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

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

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

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

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

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

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

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

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

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

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

Trả lời nhanh giúp mình với mình cần gấp lắm

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

c) \(\sqrt[]{8+\sqrt[]{x}}+\sqrt{5-\sqrt[]{x}}=5\)

\(\Leftrightarrow\left(\sqrt[]{8+\sqrt[]{x}}+\sqrt{5-\sqrt[]{x}}\right)^2=25\left(1\right)\left(đkxđ:0\le x\le25\right)\)

Áp dụng Bất đẳng thức Bunhiacopxki cho 2 cặp số dương \(\left(1;\sqrt[]{8+\sqrt[]{x}}\right);\left(1;\sqrt{5-\sqrt[]{x}}\right)\)

\(\left(1.\sqrt[]{8+\sqrt[]{x}}+1.\sqrt{5-\sqrt[]{x}}\right)^2\le\left(1^2+1^2\right)\left(8+\sqrt[]{x}+5-\sqrt[]{x}\right)=26\)

\(\left(1\right)\Leftrightarrow26=25\left(vô.lý\right)\)

Vậy phương trình đã cho vô nghiệm

b) \(\sqrt[]{1+4x}+2\sqrt[]{2-x}+2\sqrt[]{\left(1+4x\right)\left(2-x\right)}=3\)  \(\left(đkxđ:-\dfrac{1}{4}\le x\le2\right)\)

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

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

Áp dụng Bất đẳng thức Bunhiacopxki :

\(\left(1.\sqrt[]{1+4x}+2\sqrt[]{2-x}\right)^2\le\left(1^2+2^2\right)\left(1+4x+2-x\right)=5\left(3x+3\right)\)

Áp dụng Bất đẳng thức Cauchy :

\(2\sqrt[]{\left(1+4x\right)\left(2-x\right)}\le1+4x+2-x=3x+3\)

Dấu "=" xảy ra khi và chỉ khi

\(1+4x=2-x\)

\(\Leftrightarrow x=\dfrac{1}{5}\left(thỏa.đk\right)\)

\(pt\left(1\right)\Leftrightarrow5\left(4x+3\right)=4x+3\)

\(\Leftrightarrow4\left(4x+3\right)=0\)

\(\Leftrightarrow x=-\dfrac{3}{4}\left(k.thỏa.x=\dfrac{1}{5}.vô.lý\right)\)

Vậy phương trình đã cho vô nghiệm

1)\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{26^2}=\sqrt{5}-2+26=24-\sqrt{5}\)

2) \(=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)

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

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

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

3. \(\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{\sqrt{x^2-2.x.1+1^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{|x-1|}{x-1}=\left[{}\begin{matrix}x-1>0\left(x>1\right)\\x-1< 0\left(x< 1\right)\end{matrix}\right.=\left[{}\begin{matrix}=1\\=\dfrac{x+1}{x-1}\end{matrix}\right.\)