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)

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

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

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

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

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

\(ĐK:x\ge5\)

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

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

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

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

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 \(\dfrac{5}{3}+\sqrt{x+3}>0\))

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

\(dat:\sqrt{x-5}=a\Rightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\frac{1}{3}=\sqrt{9\left(x-5\right)}\Rightarrow\sqrt{4}.a+a-\frac{1}{3}=\sqrt{9}.a\Rightarrow3a-\frac{1}{3}=3a\left(voli\right)\Rightarrow vonghiem\)

câu a chắc đề như zầy pk bạn???

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

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}+3\sqrt{x-5}=\frac{13}{3}\)

\(\Leftrightarrow6\sqrt{x-5}=\frac{13}{3}\Rightarrow\sqrt{x-5}=\frac{13}{18}\Leftrightarrow x=\frac{1789}{324}\)

b)đề như này đúng ko bạn??

\(\sqrt{16-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}\)

\(\Leftrightarrow4\sqrt{1-2x}-2\sqrt{3x}=\sqrt{3x}+3\sqrt{1-2x}\)

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

\(\Leftrightarrow1-2x=27x\Leftrightarrow x=\frac{1}{29}\)

câu c\(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)

Xét điều kiện \(\left\{{}\begin{matrix}x\le1\\x\ge5\end{matrix}\right.\)không tồn tại số nào nằm trong khoảng này

Vậy pt trên vô nghiệm

ĐK: \(x\ge0\)\(4\sqrt{x}-2\sqrt{9x}+16\sqrt{x}=5\)  5  (=) \(\sqrt{x}\left(4-2\sqrt{9}+16\right)=5\) (=) \(\sqrt{x}.14=5\)(=) x=\(\frac{25}{196}\)

ĐK: \(x\ge-5\)PT(=) \(\sqrt{5+x}\left(\sqrt{4}-3+\frac{4}{3}.3\right)=6\) (=) \(\sqrt{5+x}.3=6\) (=)\(\sqrt{5+x}=2\)(=) X = -1 (nhận)

Giải:

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

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

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

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

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

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

\(\Leftrightarrow\)\(\sqrt{x-5}\)= 2 . Đk : x \(\ge\)5

\(\Rightarrow\)x - 5 = 4

\(\Leftrightarrow\)x = 9 ( thỏa mãn )

Vậy phương trình đã cho có tập nghiệm S = \(\left\{9\right\}\)

a: ĐKXĐ: x-5>=0

=>x>=5

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

=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

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

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x-1>=0

=>x>=1

\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)

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

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

=>\(\sqrt{x-1}=-2\)(vô lý)

Vậy: Phương trình vô nghiệm

c: ĐKXĐ: x-2>=0

=>x>=2

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot\sqrt{9x-18}+6\cdot\sqrt{\dfrac{x-2}{81}}=-4\)

=>\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)

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

=>\(-\sqrt{x-2}=-4\)

=>x-2=16

=>x=18(nhận)

d: ĐKXĐ: x+3>=0

=>x>=-3

\(\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\cdot\sqrt{16x+48}=0\)

=>\(3\sqrt{x+3}+4\sqrt{x+3}-\dfrac{3}{4}\cdot4\sqrt{x+3}=0\)

=>\(4\sqrt{x+3}=0\)

=>x+3=0

=>x=-3(nhận)

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

= \(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)

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

= \(2\sqrt{x-5}=4\)

= \(\sqrt{x-5}=2\)

= \(\left|x-5\right|=4\)

=> \(x-5=\pm4\)

\(x=\pm4+5\)

\(x=9;x=1\)

Vậy x=9; x=1

b) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)

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

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

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

=>\(\left|x-1\right|=-2\)

\(x-1=\mp2\)

\(x=-3;x=1\)

Vậy x=-3; x=1