a/ ĐKXĐ: ...

\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)

\(\Rightarrow x+\frac{1}{4x}=a^2-1\)

Pt trở thành:

\(3a=2\left(a^2-1\right)-7\)

\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)

b/ ĐKXĐ:

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)

\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

d/ ĐKXĐ: ...

\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)

\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)

\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)

\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

e/ ĐKXĐ: ...

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

Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

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

Đặt \(\sqrt{\frac{x^2-1}{x}}=a\ge0\)

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

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

f/ ĐKXĐ: ...

\(\Leftrightarrow x^2-6+x\sqrt{\frac{x^2-6}{x}}-6x=0\)

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{x^2-6}{x}+\sqrt{\frac{x^2-6}{x}}-6=0\)

Đặt \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)

\(a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{\frac{x^2-6}{x}}=2\Leftrightarrow x^2-4x-6=0\)

a/ ĐKXĐ: ...

\(\Leftrightarrow x+8+\sqrt{x+8}-\left(x+8\right)=\sqrt{x}+\sqrt{x+3}\)

\(\Leftrightarrow\sqrt{x+8}=\sqrt{x}+\sqrt{x+3}\)

\(\Leftrightarrow x+8=2x+3+2\sqrt{x^2+3x}\)

\(\Leftrightarrow5-x=2\sqrt{x^2+3x}\) (\(x\le5\))

\(\Leftrightarrow x^2-10x+25=4\left(x^2+3x\right)\)

\(\Leftrightarrow...\)

b/ ĐKXĐ: \(2\le x\le5\)

\(\Leftrightarrow2\left(x-2\right)+\sqrt{2\left(x-2\right)}\left(\sqrt{5-x}-\sqrt{3x-3}\right)=0\)

\(\Leftrightarrow\sqrt{2\left(x-2\right)}\left(\sqrt{2x-4}+\sqrt{5-x}-\sqrt{3x-3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\sqrt{2x-4}+\sqrt{5-x}=\sqrt{3x-3}\left(1\right)\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left(2x-4\right)\left(5-x\right)=\left(x-2\right)^2\)

\(\Leftrightarrow...\)

c/ ĐKXĐ: \(x\le12\)

\(\Leftrightarrow\sqrt[3]{24+x}\sqrt{12-x}-6\sqrt{12-x}+12-x=0\)

\(\Leftrightarrow\sqrt{12-x}\left(\sqrt[3]{24+x}-6+\sqrt{12-x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\\sqrt[3]{24+x}+\sqrt{12-x}=6\left(1\right)\end{matrix}\right.\)

Xét (1):

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{24+x}=a\\\sqrt{12-x}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=6\\a^3+b^2=36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^3+b^2=36\end{matrix}\right.\)

\(\Leftrightarrow a^3+\left(6-a\right)^2=36\)

\(\Leftrightarrow a^3+a^2-12a=0\)

\(\Leftrightarrow a\left(a^2+a-12\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=3\\a=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{24+x}=0\\\sqrt[3]{24+x}=3\\\sqrt[3]{24+x}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}24+x=0\\24+x=27\\24+x=-64\end{matrix}\right.\)

d/ ĐKXĐ: \(x\le\frac{3}{2}\) ; \(x\ne\frac{3}{8};x\ne-\frac{13}{24}\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2\sqrt{3-2x}-3}-\frac{1}{3-2\sqrt[3]{5+3x}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\frac{1}{2\sqrt{3-2x}-3}=\frac{1}{3-2\sqrt[3]{5+3x}}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\sqrt{3-2x}-3=3-2\sqrt[3]{5+3x}\)

\(\Leftrightarrow\sqrt[3]{5+3x}+\sqrt{3-2x}=3\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{5+3x}=a\\\sqrt{3-2x}=b\ge0\end{matrix}\right.\) ta được:

\(\left\{{}\begin{matrix}a+b=3\\2a^3+3b^2=19\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\2a^3+3b^2=19\end{matrix}\right.\)

\(\Leftrightarrow2a^3+3\left(3-a\right)^2=19\)

\(\Leftrightarrow2a^3+3a^2-18a+8=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-4\\a=\frac{1}{2}\\a=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{5+3x}=-4\\\sqrt[3]{5+3x}=\frac{1}{2}\\\sqrt[3]{5+3x}=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5+3x=-64\\5+3x=\frac{1}{8}\\5+3x=8\end{matrix}\right.\)

từ dòng cuối là sai rồi bạn à

Bạn bỏ dòng cuối đi còn lại đúng rồi

Ở tử đặt nhân tử chung căn x chung  rồi lại đặt căn x +1 chung

Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra 

rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)

cảm ơn bạn nha 

a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)

\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)

\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)

Đặt \(\sqrt{2x+3}+\sqrt{4-x}=a>0\Rightarrow a^2=x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\)

Phương trình trở thành:

\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)

\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)

Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(

b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)

Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)

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

Pt trở thành:

\(a+10\left(\frac{a^2-5}{4}\right)=13\)

\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)

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

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

c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)

\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)

Đặt \(x\sqrt{2x^2+4}=a\) ta được:

\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)

d/ ĐKXĐ: \(x\ge1\)

Nhân cả tử và mẫu của vế phải với liên hợp của nó ta được:

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

Đặt \(\sqrt{x+1}+\sqrt{x-1}=a>0\)

\(\Rightarrow a^2-3=\frac{a}{2}\Rightarrow2a^2-a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

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

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

\(\Leftrightarrow\sqrt{x^2-1}=2-x\) (\(x\le2\))

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

\(\Rightarrow x=\frac{5}{4}\)

a, \(5\sqrt{2x^2+3x+9}=2x^2+3x+3\) (*)

Đặt \(2x^2+3x=a\left(a\ge-9\right)\)

=> \(5\sqrt{a+9}=a+3\)

<=> \(25\left(a+9\right)=a^2+6a+9\)

<=> \(25a+225=a^2+6a+9\)

<=> \(0=a^2+6a+9-25a-225=a^2-19a-216\)

<=> 0= \(a^2-27a+8a-216\)

<=> \(\left(a-27\right)\left(a+8\right)=0\)

=> \(\left[{}\begin{matrix}a=27\\a=-8\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}2x^2+3x=27\\2x^2+3x=-8\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}2x^2+3x-27=0\\2x^2+3x+8=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\left(x-3\right)\left(2x+9\right)=0\\2\left(x^2+2.\frac{3}{4}+\frac{9}{16}\right)+\frac{55}{8}=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\frac{9}{2}\left(tm\right)\\2\left(x+\frac{3}{4}\right)^2=-\frac{55}{8}\left(ktm\right)\end{matrix}\right.\)

Vậy pt (*) có tập nghiệm \(S=\left\{3,-\frac{9}{2}\right\}\)

b, \(9-\sqrt{81-7x^3}=\frac{x^3}{2}\left(đk:x\le\sqrt[3]{\frac{81}{7}}\right)\)(*)

<=> \(\sqrt{81-7x^3}=9-\frac{x^3}{2}\)

<=>\(81-7x^3=\left(9-\frac{x^3}{2}\right)^2=81-9x^3+\frac{x^6}{4}\)

<=> \(-7x^3+9x^3-\frac{x^6}{4}=0\) <=> \(2x^3-\frac{x^6}{4}=0\)<=> \(8x^3-x^6=0\)

<=> \(x^3\left(8-x^2\right)=0\)

=> \(\left[{}\begin{matrix}x=0\\8=x^2\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=\pm2\sqrt{2}\left(ktm\right)\end{matrix}\right.\)

Vậy pt (*) có nghiệm x=0

d,\(\sqrt{9x-2x^2}-9x+2x^2+6=0\) (*) (đk: \(0\le x\le\frac{1}{2}\))

<=> \(\sqrt{9x-2x^2}-\left(9x-2x^2\right)+6=0\)

Đặt \(\sqrt{9x-2x^2}=a\left(a\ge0\right)\)

Có \(a-a^2+6=0\)

<=> \(a^2-a-6=0\) <=> \(a^2-3x+2x-6=0\)

<=> \(\left(a-3\right)\left(a+2\right)=0\)

=> \(a-3=0\) (vì a+2>0 vs mọi \(a\ge0\))

<=> a=3 <=>\(\sqrt{9x-2x^2}=3\) <=> \(9x-2x^2=9\)

<=> 0=\(2x^2-9x+9\) <=> \(2x^2-6x-3x+9=0\) <=>\(\left(2x-3\right)\left(x-3\right)=0\)

=> \(\left[{}\begin{matrix}2x=3\\x=3\end{matrix}\right.< =>\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)(t/m)

Vậy pt (*) có tập nghiệm \(S=\left\{\frac{3}{2},3\right\}\)

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/ ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)

\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)

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

\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)

b/ ĐKXĐ: ....

Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)

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

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

\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)

a/ ĐK: \(x\ge0\)

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

Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:

\(a=x^2-\left(a^2-x\right)\)

\(\Leftrightarrow x^2-a^2+x-a=0\)

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

\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))

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

d/ ĐKXĐ: ...

\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)

\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)

\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)

\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)

\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))

e/ ĐKXĐ: \(x\ge0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+3x+5}=a>0\\\sqrt{x^2-2x+5}=b>0\\\sqrt{x}=c\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=5c^2\)

Ta được hệ: \(\left\{{}\begin{matrix}a^2-b^2=5c^2\\a+b=5c\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=5c^2\\a+b=5c\end{matrix}\right.\)

\(\Rightarrow5c\left(a-b\right)=5c^2\)

\(\Leftrightarrow\left[{}\begin{matrix}c=0\\a-b=c\end{matrix}\right.\)

f/ ĐKXĐ: \(x>0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x}=4x\\x+3=4x\end{matrix}\right.\)