a.

\(\sqrt{4x^2+4x+1}-\sqrt{25x^2+10x+1}=0\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}-\sqrt{\left(5x+1\right)^2}=0\)

\(\Leftrightarrow2x+1-\left(5x+1\right)=0\)

\(\Leftrightarrow-3x=0\Leftrightarrow x=0\)

b.

\(\sqrt{x^4-16x^2+64}=\sqrt{25x^2+10x+1}\)

\(\Leftrightarrow\sqrt{\left(x^2-8\right)^2}=\sqrt{\left(5x+1\right)^2}\)

\(\Leftrightarrow x^2-8=5x+1\)

\(\Leftrightarrow x^2-5x+\dfrac{25}{4}=\dfrac{61}{4}\)

\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{61}{4}\)

............................

tương tự ..

c: \(\Leftrightarrow\sqrt{x-5}\left(\sqrt{x+5}-1\right)=0\)

=>x-5=0 hoặc x+5=1

=>x=-4 hoặc x=5

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

=>2x+3=0 hoặc 2x-3=4

=>x=7/2 hoặc x=-3/2

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

=>x-2=0 hoặc 3 căn x+2=1

=>x=2 hoặc x+2=1/9

=>x=-17/9 hoặc x=2

\(\sqrt{25x^2+80x+64}+\sqrt{9x^2-6x+1}=\sqrt{4x^2+36x+81}\)

\(pt\Leftrightarrow\sqrt{\left(5x+8\right)^2}+\sqrt{\left(3x-1\right)^2}=\sqrt{\left(2x+9\right)^2}\)

\(\Leftrightarrow\left|5x+8\right|+\left|3x-1\right|=\left|2x+9\right|\)

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(VT=\left|5x+8\right|+\left|-\left(3x-1\right)\right|\)

\(=\left|5x+8\right|+\left|-3x+1\right|\)

\(\ge\left|5x+8-3x+1\right|=\left|2x+9\right|=VP\)

Đẳng thức xảy ra khi \(-\frac{8}{5}\le x\le\frac{1}{3}\)

P.s:thực ra thì áp dụng căn a+căn b>= căn a+b ngay từ đầu luôn cx dc tùy

Sửa đề: \(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25x-125}+6\)

\(\Leftrightarrow\sqrt{x-5}+\dfrac{1}{3}\cdot3\cdot\sqrt{x-5}-\dfrac{1}{5}\cdot5\sqrt{x-5}=6\)

\(\Leftrightarrow\sqrt{x-5}=6\)

=>x-5=36

hay x=41

\(ĐK:x\ge2\)

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

<=>\(x^2-5x+4=2\sqrt{2\left(x-2\right)}\)

<=>\(x^2-5x+4+x-2+2=\left(x-2\right)+2\sqrt{2\left(x-2\right)}+2\)

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

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

<=> \(\left(x-2-\sqrt{x-2}-2\right)\left(x-2+\sqrt{x-2}+2\right)=0\)

<=>\(\left(x-\sqrt{x-2}-4\right)\left(x+\sqrt{x-2}\right)=0\)

Xét \(x-\sqrt{x-2}-4=0\)

<=>\(x^2-8x+16=x-2\)

<=>\(x^2-9x+18=0\)

=> x=6;3(nhận)

Xet1\(x+\sqrt{x-2}=0\)

Do x\(\ge2\)=> pt vô nghiệm

Vậy ...

pt <=> \(2x^2-20x+54-2\sqrt{x-4}-2\sqrt{6-x}=0\)

<=> \(\left(2x^2-20x+50\right)+\left(x-4-2\sqrt{x-4}+1\right)+\left(6-x-2\sqrt{6-x}+1\right)=0\)

<=> \(2\left(x-5\right)^2+\left(\sqrt{x-4}-1\right)^2+\left(\sqrt{6-x}-1\right)^2=0\)

<=> x = 5

ĐKXĐ :  \(-4\le x\le4\)

TA CÓ : \(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=2x\)

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

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

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

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

\(\Leftrightarrow x=0\)HOẶC  \(\sqrt{4-x}-2\sqrt{x+4}-2=0\)

VỚI \(\sqrt{4-x}-2\sqrt{x+4}-2=0\)

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

\(\Leftrightarrow4-x+4-4\sqrt{4-x}=4x+16\)

\(\Leftrightarrow8-x-4x-16=4\sqrt{4-x}\)

\(\Leftrightarrow-5x-8=4\sqrt{4-x}\)ĐK : \(-4\le x\le\frac{-8}{5}\)

\(\Leftrightarrow\left[-\left(5x+8\right)\right]^2=16\left(4-x\right)\)

\(\Leftrightarrow25x^2+64+80x=64-16x\)

\(\Leftrightarrow25x^2+96x=0\Leftrightarrow x\left(25x+96\right)=0\)

\(\Leftrightarrow x=0\)HOẶC \(x=\frac{-96}{25}\)(THỎA MÃN ĐK )                                                                               

                                                                                               VẬY PT CÓ 2 NGHIỆM \(x\in\left[0;\frac{-96}{25}\right]\)

P/S : CÁCH CỦA MÌNH KHÁ DÀI VÀ CHI TIẾT QUÁ . BẠN CÓ THỂ THAM KHẢO CÁCH KHÁC NHANH HƠN :>