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) ĐK x>2

a) Để giá trị của biểu thức \(\frac{x}{x^2-4}+\sqrt{x-2}\)xác định được thì

\(\left\{{}\begin{matrix}x^2-4\ne0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\notin\left\{2;-2\right\}\\x\ge2\end{matrix}\right.\Leftrightarrow x>2\)

b) Để giá trị của biểu thức \(\frac{\sqrt{x}}{\left|x\right|-1}\) xác định được thì

\(\left\{{}\begin{matrix}x\ge0\\\left|x\right|-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left|x\right|\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\notin\left\{1;-1\right\}\end{matrix}\right.\Leftrightarrow0\le x\ne1\)

c/ \(C=\sqrt{x^2-6x+9}+\sqrt{x^2+10x+25}\)

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

\(=|3-x|+|x+5|\ge|3-x+x+5|=8\)

d/ \(D=\sqrt{x^2-6x+9}+\sqrt{4x^2+24x+36}\)

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

\(=|3-x|+|x+3|+|x+3|\ge|3-x+x+3|+0=6\)

e/ \(2E=\sqrt{x^2}+2\sqrt{x^2-2x+1}\)

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

\(=|x|+|1-x|+|x-1|\ge|x+1-x|+0=1\)

\(\Rightarrow E\ge\frac{1}{2}\)

a) ĐK: \(x\geq \frac{1}{2}\)

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

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

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

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

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

Đối với $(*)$:

Vì \(x\geq \frac{1}{2}\Rightarrow \sqrt{2x-1}+\sqrt{x+1}\geq \sqrt{\frac{1}{2}+1}>1\)

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

Do đó $(*)$ vô nghiệm

Vậy pt có nghiệm duy nhất $x=2$

b) ĐK:.....

\(\sqrt{2x^2-3x+10}+\sqrt{2x^2-5x+4}=x+3\)

TH1:

\(\sqrt{2x^2-3x+10}=\sqrt{2x^2-5x+4}\)

\(\Rightarrow 2x^2-3x+10=2x^2-5x+4\)

\(\Rightarrow 2x+6=0\Rightarrow x=-3\) (thử lại thấy không thỏa mãn)

TH2: \(\sqrt{2x^2-3x+10}\neq \sqrt{2x^2-5x+4}\), tức là \(x\neq -3\)

PT ban đầu tương đương với:

\(\frac{(2x^2-3x+10)-(2x^2-5x+4)}{\sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}}=x+3\)

\(\Leftrightarrow \frac{2(x+3)}{\sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}}=x+3\)

\(\Leftrightarrow \frac{2}{\sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}}=1\) (do \(x\neq -3\) )

\(\Rightarrow \sqrt{2x^2-3x+10}-\sqrt{2x^2-5x+4}=2\)

\(\Rightarrow \sqrt{2x^2-3x+10}=2+\sqrt{2x^2-5x+4}\)

Bình phương 2 vế:

\(2x^2-3x+10=4+2x^2-5x+4+4\sqrt{2x^2-5x+4}\)

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

\(\Rightarrow (x+1)^2=4(2x^2-5x+4)\)

\(\Rightarrow 7x^2-22x+15=0\Rightarrow \left[\begin{matrix} x=\frac{15}{7}\\ x=1\end{matrix}\right.\) (thử đều thấy t/m)

Vậy...........

c) ĐK: \(-2\leq x\leq 3\)

Ta có: \(\sqrt{x+2}-\sqrt{3-x}=x^2-6x+9\)

\(\Leftrightarrow (\sqrt{x+2}-2)-(\sqrt{3-x}-1)=x^2-6x+8\)

\(\Leftrightarrow \frac{x+2-4}{\sqrt{x+2}+2}-\frac{(3-x)-1}{\sqrt{3-x}+1}=(x-2)(x-4)\)

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

\(\Rightarrow \left[\begin{matrix} x-2=0\rightarrow x=2\\ \frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}=x-4(*)\end{matrix}\right.\)

Đối với $(*)$. Ta thấy vế trái luôn lớn hơn $0$, vế phải nhỏ hơn $0$ do $x\leq 3$ nên $(*)$ vô nghiệm

Vậy pt có nghiệm duy nhất $x=2$

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

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

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

a) ĐK: \(\left\{{}\begin{matrix}x\ne-1\\\frac{4-x}{x+1}\ge0\end{matrix}\right.\). Lập bảng xét dấu sẽ được \(-1< x\le4\)

b) Tương tự

c)(em ko chắc) ĐK: \(\left\{{}\begin{matrix}x^2-4\ge0\left(1\right)\\\frac{x-2}{x+1}\ge0\left(2\right)\\x\ne-1\end{matrix}\right.\). Giải (1) ta được \(x\le-2\text{hoặc }x\ge2\)

Giải (2) được \(x\le-1\text{ hoặc }x\ge2\)

Kết hợp lại ta được: \(x\le-2\text{hoặc }x\ge2\)