b,\(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}-16\sqrt{x+1}=0\) (dk \(x\ge-1\)

\(\Leftrightarrow\sqrt{x+1}\left(4-3+2-16\right)=0\)

\(\Leftrightarrow\sqrt{x+1}.-13=0\)

\(\Leftrightarrow x=-1\)

h) 

ĐKXĐ: $x\geq -5$

PT $\Leftrightarrow \sqrt{x+5}=6$

$\Rightarrow x+5=36\Rightarrow x=31$ (thỏa mãn)

i) ĐKXĐ: $x\geq 5$

PT \(\Leftrightarrow \sqrt{x-5}+4\sqrt{x-5}-\sqrt{x-5}=12\)

\(\Leftrightarrow 4\sqrt{x-5}=12\Leftrightarrow \sqrt{x-5}=3\Rightarrow x-5=9\Rightarrow x=14\) (thỏa mãn)

j) 

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 3\sqrt{2x}+\sqrt{2x}-6\sqrt{2x}+4=0$

$\Leftrightarrow -2\sqrt{2x}+4=0$

$\Leftrightarrow \sqrt{2x}=2$

$\Rightarrow x=2$ (thỏa mãn)

k) ĐK: $x^2\geq 5$

PT $\Leftrightarrow 2\sqrt{x^2-5}-\frac{1}{3}\sqrt{x^2-5}+\frac{3}{4}\sqrt{x^2-5}-\frac{5}{12}\sqrt{x^2-5}=4$

$\Leftrightarrow 2\sqrt{x^2-5}=4$

$\Leftrightarrow \sqrt{x^2-5}=2$

$\Rightarrow x^2-5=4$

$\Leftrightarrow x^2=9\Rightarrow x=\pm 3$ (đều thỏa mãn)

l) ĐKXĐ: $x\geq -1$

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

$\Leftrightarrow 4\sqrt{x+1}=4$

$\Leftrightarrow \sqrt{x+1}=1$

$\Rightarrow x+1=1$

$\Rightarrow x=0$

m) 

ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow 4\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}+3\sqrt{x+1}$

$\Leftrightarrow 6\sqrt{x+1}=16+2\sqrt{x+1}$

$\Leftrightarrow 4\sqrt{x+1}=16$

$\Leftrightarrow \sqrt{x+1}=4$

$\Rightarrow x=15$ (thỏa mãn)

Điều kiện xác định phương trình \(-2\le x\le2.\)

Phương trình tương đương với \(3x-2=0\)  hoặc

\(\frac{1}{\sqrt{2x+4}+2\sqrt{2-x}}=\frac{2}{\sqrt{9x^2+16}}\leftrightarrow\sqrt{9x^2+16}=2\sqrt{2x+4}+4\sqrt{2-x}\)

Trường hợp 1. \(3x-2=0\leftrightarrow x=\frac{3}{2}.\)

Trường hợp 2. \(\sqrt{9x^2+16}=2\sqrt{2x+4}+4\sqrt{2-x}\).

Ta đánh giá vế trái như sau: theo bất đẳng thức Bunhia \(\sqrt{9x^2+16}\ge\sqrt{6}x+\frac{4}{\sqrt{3}}\).

Mặt khác vế phải không vượt quá \(\sqrt{3+2\sqrt{2}}\cdot\sqrt{\frac{8x+16}{3+2\sqrt{2}}}+\sqrt[4]{2}\cdot\sqrt{\frac{32-16x}{\sqrt{2}}}\le\sqrt{6}x+\frac{4}{\sqrt{3}}\)

Vì vậy ta có dấu bằng xảy ra, hay \(x=\frac{4\sqrt{2}}{3}.\)

Trần Thị Diễm Quỳnh ảo tưởng sức manh ak

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)

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

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

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

3: Ta có: \(\sqrt{4x+1}=x+1\)

\(\Leftrightarrow x^2+2x+1=4x+1\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)

\(\Leftrightarrow3\sqrt{x-1}=15\)

\(\Leftrightarrow x-1=25\)

hay x=26

5: Ta có: \(\sqrt{4x^2-12x+9}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)

ĐK : x ≥ 0

⇔ \(\frac{3}{4}\sqrt{x}-\sqrt{3^2x}-\frac{1}{4}\sqrt{3^2x}=-5\)

⇔ \(\frac{3}{4}\sqrt{x}-3\sqrt{x}-\frac{1}{4}\cdot3\sqrt{x}=-5\)

⇔ \(-\frac{9}{4}\sqrt{x}-\frac{3}{4}\sqrt{x}=-5\)

⇔ \(-3\sqrt{x}=-5\)

⇔ \(\sqrt{x}=15\)

⇔ \(x=225\)( tm )

b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)

ĐK : x ≤ 3

⇔ \(\sqrt{3-x}-\sqrt{3^2\left(3-x\right)}+\frac{5}{4}\sqrt{4^2\left(3-x\right)}=6\)

⇔ \(\sqrt{3-x}-3\sqrt{3-x}+\frac{5}{4}\cdot4\sqrt{3-x}=6\)

⇔ \(-2\sqrt{3-x}+5\sqrt{3-x}=6\)

⇔ \(3\sqrt{3-x}=6\)

⇔ \(\sqrt{3-x}=2\)

⇔ \(3-x=4\)

⇔ \(x=-1\)( tm )

c) \(\sqrt{9x^2+12x+4}=4\)

⇔ \(\sqrt{\left(3x+2\right)^2}=4\)

⇔ \(\left|3x+2\right|=4\)

⇔ \(\orbr{\begin{cases}3x+2=4\\3x+2=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-2\end{cases}}\)

d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)

ĐK : x ≥ 1

⇔  \(\frac{1}{3}\sqrt{x-1}+2\sqrt{2^2\left(x-1\right)}-12\sqrt{\left(\frac{1}{5}\right)^2\cdot\left(x-1\right)}=\frac{29}{15}\)

⇔  \(\frac{1}{3}\sqrt{x-1}+2\cdot2\sqrt{x-1}-12\cdot\frac{1}{5}\sqrt{x-1}=\frac{29}{15}\)

⇔  \(\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}\)

⇔ \(\frac{29}{15}\sqrt{x-1}=\frac{29}{15}\)

⇔ \(\sqrt{x-1}=1\)

⇔ \(x-1=1\)

⇔ \(x=2\)( tm )