a/ Giải rồi

b/ ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)

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

Pt trở thành:

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

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

\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)

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

\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)

\(\Leftrightarrow...\)

c/ Vế phải là \(181-4x\) hay \(18-14x\)?

d/ ĐKXĐ: \(x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=t>0\)

\(\Rightarrow t^2=2x+2\sqrt{x^2-16}\)

Pt trở thành:

\(\frac{t}{2}=\frac{t^2}{2}-6\)

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

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

\(\Leftrightarrow2x+2\sqrt{x^2-16}=16\)

\(\Leftrightarrow\sqrt{x^2-16}=8-x\left(x\le8\right)\)

\(\Leftrightarrow x^2-16=64-16x+x^2\)

\(\Rightarrow x=...\)

e/ ĐKXD: \(x>0\)

\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)

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

Pt trở thành:

\(5t=2\left(t^2-1\right)+4\)

\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow2x-4\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)

\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)

a/ ĐKXĐ: ...

\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)

Đặt \(\sqrt{x^2-5x-6}=a\ge0\)

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

\(\Rightarrow\sqrt{x^2-5x-6}=1\Leftrightarrow x^2-5x-7=0\)

b/ ĐKXĐ: ...

\(\Leftrightarrow5\sqrt{3x^2-4x-2}-2\left(3x^2-4x-2\right)+3=0\)

Đặt \(\sqrt{3x^2-4x-2}=a\ge0\)

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

\(\Rightarrow\sqrt{3x^2-4x-2}=3\Leftrightarrow3x^2-4x-11=0\)

c/ \(\Leftrightarrow x^2+2x-6+\sqrt{2x^2+4x+3}=0\)

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

\(\frac{a^2-3}{2}-6+a=0\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x-6=0\)

d/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{3x-1}{x}}=a>0\)

\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\)

\(\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)

\(\Rightarrow a=1\Rightarrow\sqrt{\frac{3x-1}{x}}=1\Leftrightarrow3x-1=x\)

e/ĐKXĐ: ...

\(\Leftrightarrow2\sqrt{\frac{6x-1}{x}}=\frac{x}{6x-1}+1\)

Đặt \(\sqrt{\frac{6x-1}{x}}=a>0\)

\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)

\(\Rightarrow a=1\Rightarrow\sqrt{\frac{6x-1}{x}}=1\Rightarrow6x-1=x\)

f/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{x}{2x-1}}=a>0\)

\(\frac{1}{a}+1+a=3a^2\)

\(\Leftrightarrow3a^3-a^2-a-1=0\)

\(\Leftrightarrow\left(a-1\right)\left(3a^2+2a+1\right)=0\)

\(\Leftrightarrow a=1\Rightarrow\sqrt{\frac{x}{2x-1}}=1\Rightarrow x=2x-1\)

A = \(x^2+3x-7=x^2+2x\frac{3}{2}+\frac{9}{4}-\frac{37}{4}\)

\(=\left(x+\frac{3}{2}\right)^2-\frac{37}{4}\ge-\frac{37}{4}\)

\(\Rightarrow\)min A = \(-\frac{37}{4}\Leftrightarrow x=-\frac{3}{2}\)

B = \(x-5\sqrt{x}-1\) ĐKXĐ: \(x\ge0\)

\(=x-2\sqrt{x}\frac{5}{2}+\frac{25}{4}-\frac{29}{4}=\left(\sqrt{x}-\frac{5}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)

\(\Rightarrow\)min B = \(-\frac{29}{4}\Leftrightarrow x=\frac{25}{4}\)( thỏa mãn)

C = \(\frac{-4}{\sqrt{x}+7}\) ĐKXĐ:\(x\ge0\)

Ta có: \(\sqrt{x}+7\ge7\Rightarrow\frac{4}{\sqrt{x}+7}\le\frac{4}{7}\)\(\Leftrightarrow\frac{-4}{\sqrt{x}+7}\ge-\frac{4}{7}\)

\(\Rightarrow\)min C = \(-\frac{4}{7}\Leftrightarrow x=0\)

D = \(\frac{\sqrt{x}+1}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)

\(=1-\frac{2}{\sqrt{x}+3}\ge1-\frac{2}{3}=\frac{1}{3}\)

\(\Rightarrow\)min D = \(\frac{1}{3}\Leftrightarrow x=0\)

E = \(\frac{x+7}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)

\(=\frac{x-9+16}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-6\ge2\sqrt{16}-6=2\)

\(\Rightarrow\)min E = \(2\Leftrightarrow x=1\)(thỏa mãn)

F = \(\frac{x^2+3x+5}{x^2}\) ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow\)\(x^2\left(F-1\right)-3x-5=0\)

△ = \(3^2+20\left(F-1\right)\ge0\)\(\Leftrightarrow F\ge\frac{11}{20}\)

\(\Rightarrow\)min F = \(\frac{11}{20}\Leftrightarrow x=-\frac{10}{3}\)( thỏa mãn)

a)\(ĐK:-3\le x\le6\)

\(PT\Leftrightarrow\sqrt{x+3}+\sqrt{6-x}=3\)

\(\Leftrightarrow x+3+6-x+2\sqrt{\left(x+3\right)\left(6-x\right)}=9\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\left(tm\right)\)

b/ ĐKXĐ: \(x\ge7\)

\(\sqrt{3x-2}=1+\sqrt{x-7}\)

\(\Leftrightarrow3x-2=x-6+2\sqrt{x-7}\)

\(\Leftrightarrow x+2=\sqrt{x-7}\)

\(\Leftrightarrow x^2+4x+4=x-7\)

\(\Leftrightarrow x^2+3x+11=0\) (vô nghiệm)

c/ ĐKXĐ: \(x\ge1;x\ne50\)

\(1-\sqrt{3x+1}=\sqrt{x-1}-7\)

\(\Leftrightarrow\sqrt{x-1}+\sqrt{3x+1}=8\)

\(\Leftrightarrow4x+2\sqrt{3x^2-2x-1}=64\)

\(\Leftrightarrow\sqrt{3x^2-2x-1}=32-2x\) (\(x\le16\))

\(\Leftrightarrow3x^2-2x-1=\left(32-2x\right)^2\)

d/ ĐKXĐ: \(x\ge\frac{4}{7};x\ne\frac{13}{7}\)

\(\Leftrightarrow\sqrt{x+1}=\sqrt{7x-4}-3\)

\(\Leftrightarrow\sqrt{x+1}+3=\sqrt{7x-4}\)

\(\Leftrightarrow x+10+6\sqrt{x+1}=7x-4\)

\(\Leftrightarrow3\sqrt{x+1}=3x-7\) (\(x\ge\frac{7}{3}\))

\(\Leftrightarrow9\left(x+1\right)=\left(3x-7\right)^2\)

\(\Leftrightarrow...\)

e/ Giống câu b

f/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le-\frac{1}{3}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\sqrt{\frac{3x+1}{2x-1}}=a\ge0\\\sqrt{\frac{x-1}{2x-1}}=b\ge0\end{matrix}\right.\) ta được hệ:

\(\left\{{}\begin{matrix}2a-b=2\\a^2+5b^2=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=2a-2\\a^2+5b^2=4\end{matrix}\right.\)

\(\Rightarrow a^2+5\left(2a-2\right)^2=4\)

\(\Leftrightarrow a^2+20\left(a^2-2a+1\right)-4=0\)

\(\Leftrightarrow21a^2-40a+16=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{4}{3}\\a=\frac{4}{7}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{\frac{3x+1}{2x-1}}=\frac{4}{3}\\\sqrt{\frac{3x+1}{2x-1}}=\frac{4}{7}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{3x+1}{2x-1}=\frac{16}{9}\\\frac{3x+1}{2x-1}=\frac{16}{49}\end{matrix}\right.\) \(\Leftrightarrow...\)

a) \(\left\{{}\begin{matrix}x\ge0\\-\sqrt{x+7}< 0\\-5x-4\ne0\\-3x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x+7>0\\-5x\ne4\\-3x\ne-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>-7\\x\ne\frac{-4}{5}\\x\ne\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne\frac{2}{3}\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}x\ge0\\x+4\ne0\\x-2\ge0\\-2x-10\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-4\\x\ge2\\-2x\ne10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne-5\end{matrix}\right.\Leftrightarrow x\ge2\)

c) \(\left\{{}\begin{matrix}x\ge0\\-x-3\ne0\\2x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-3\\x\ne-\frac{3}{2}\end{matrix}\right.\Leftrightarrow x\ge0\)

d) \(\left\{{}\begin{matrix}2x-7\ge0\\x\ge0\\3x-4\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{7}{2}\\x\ge0\\x\ne\frac{4}{3}\\x\ne3\end{matrix}\right.\Leftrightarrow x\ge\frac{7}{2}\)

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 \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)

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

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