a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le2\end{matrix}\right.\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x>\dfrac{2\sqrt{14}}{7}\\x< -\dfrac{2\sqrt{14}}{7}\end{matrix}\right.\)

c: ĐKXĐ: \(x=\dfrac{1}{3}\)

d: ĐKXĐ: \(-\dfrac{2}{3}< x\le\sqrt{3}\)

a.

ĐKXĐ: \(x\ge1\)

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

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

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

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

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

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-1\)

\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)

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

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

\(\Leftrightarrow x=3\)

c.

ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)

\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)

\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)

Dấu "=" xảy ra khi và chỉ khi \(x=-1\)

d.

ĐKXĐ: \(x>1\)

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x-1}=b>0\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^2-1}{a}=\dfrac{1-b^2}{b}\)

\(\Leftrightarrow a-\dfrac{1}{a}=\dfrac{1}{b}-b\)

\(\Leftrightarrow a+b-\dfrac{a+b}{ab}=0\)

\(\Leftrightarrow\left(a+b\right)\left(1-\dfrac{1}{ab}\right)=0\)

\(\Leftrightarrow1-\dfrac{1}{ab}=0\)

\(\Leftrightarrow ab=1\)

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

\(\Leftrightarrow x^3-1=1\)

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

Bài 1:
a. ĐKXĐ: $x\geq \frac{2}{5}$

PT $\Leftrightarrow 5x-2=7^2=49$

$\Leftrightarrow 5x=51$

$\Leftrightarrow x=\frac{51}{5}=10,2$

b. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{25(x-3)}=24$

$\Leftrightarrow 3\sqrt{x-3}+5\sqrt{x-3}=24$

$\Leftrightarrow 8\sqrt{x-3}=24$

$\Leftrightarrow \sqrt{x-3}=3$

$\Leftrightarrow x-3=9$

$\Leftrightarrow x=12$ (tm)

Bài 1:

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow x^2-5x+6-2(\sqrt{x-2}-1)=0$

$\Leftrightarrow (x-2)(x-3)-2.\frac{x-3}{\sqrt{x-2}+1}=0$

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

$x-3=0$ hoặc $x-2=\frac{2}{\sqrt{x-2}+1}$

Nếu $x-3=0$

$\Leftrightarrow x=3$ (tm) 

Nếu $x-2=\frac{2}{\sqrt{x-2}+1}$

$\Leftrightarrow a^2=\frac{2}{a+1}$ (đặt $\sqrt{x-2}=a$)

$\Leftrightarrow a^3+a^2-2=0$

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

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

Hiển nhiên $a^2+2a+2=(a+1)^2+1>0$ với mọi $a$ nên $a-1=0$

$\Leftrightarrow a=1\Leftrightarrow \sqrt{x-2}=1\Leftrightarrow x=3$ (tm)

Vậy pt có nghiệm duy nhất $x=3$.

Bài 2:

ĐKXĐ: $x\geq 0; x\neq 4$

\(A=\frac{\sqrt{x}(\sqrt{x}-2)-\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}+2)\sqrt{x}-2)}.\frac{\sqrt{x}+2}{2}\\ =\frac{-4\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}+2}{2}\\ =\frac{-2\sqrt{x}}{\sqrt{x}-2}=\frac{2\sqrt{x}}{2-\sqrt{x}}\)

a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$

$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$

$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$

Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}=\frac{12}{5}$

$\Leftrightarrow x=5,76$ (thỏa mãn)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 2\sqrt{x-2}=8$

$\Leftrightarrow \sqrt{x-2}=4$

$\Leftrightarrow x=4^2+2=18$ (tm)

b. ĐKXĐ: $x^2\geq 5$

PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$

$\Leftrightarrow \sqrt{x^2-5}=0$

$\Leftrightarrow x=\pm \sqrt{5}$

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

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

=>x-2=16

hay x=18

b: \(\Leftrightarrow\left|3x+2\right|=4x\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)

c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

\(\Leftrightarrow4\sqrt{x-2}=40\)

=>x-2=100

hay x=102

d: =>5x-6=9

hay x=3

\(a,\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\left(dk:x\ge2\right)\)

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

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

\(\Leftrightarrow x-2=16\)

\(\Leftrightarrow x=18\left(tmdk\right)\)

b,\(\sqrt{9x^2-12x+4=3x\left(dk:x\ge0\right)}\)

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

\(\Leftrightarrow\left|3x-2\right|=3x\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=3x\\3x-2=-3x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=\dfrac{1}{3}\left(tmdk\right)\end{matrix}\right.\)

Các câu còn lại làm tương tự nhé 

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

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

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

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)

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

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

\(\left|x-2\right|=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)

a.

\(y'=-\dfrac{3}{2}x^3+\dfrac{6}{5}x^2-x+5\)

b.

\(y'=\dfrac{\left(x^2+4x+5\right)'}{2\sqrt{x^2+4x+5}}=\dfrac{2x+4}{2\sqrt{x^2+4x+5}}=\dfrac{x+2}{\sqrt{x^2+4x+5}}\)

c.

\(y=\left(3x-2\right)^{\dfrac{1}{3}}\Rightarrow y'=\dfrac{1}{3}\left(3x-2\right)^{-\dfrac{2}{3}}=\dfrac{1}{3\sqrt[3]{\left(3x-2\right)^2}}\)

d.

\(y'=2\sqrt{x+2}+\dfrac{2x-1}{2\sqrt{x+2}}=\dfrac{6x+7}{2\sqrt{x+2}}\)

e.

\(y'=3sin^2\left(\dfrac{\pi}{3}-5x\right).\left[sin\left(\dfrac{\pi}{3}-5x\right)\right]'=-15sin^2\left(\dfrac{\pi}{3}-5x\right).cos\left(\dfrac{\pi}{3}-5x\right)\)

g.

\(y'=4cot^3\left(\dfrac{\pi}{6}-3x\right)\left[cot\left(\dfrac{\pi}{3}-3x\right)\right]'=12cot^3\left(\dfrac{\pi}{6}-3x\right).\dfrac{1}{sin^2\left(\dfrac{\pi}{3}-3x\right)}\)

Câu a bạn coi lại đề

b. ĐKXĐ: \(x\ge0;x\ne1\)

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

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

\(\Leftrightarrow5x+1+2\sqrt{3x\left(2x+1\right)}=3x+2\)

\(\Leftrightarrow2\sqrt{6x^2+3x}=1-2x\) (\(x\le\dfrac{1}{2}\) )

\(\Leftrightarrow4\left(6x^2+3x\right)=4x^2-4x+1\)

\(\Leftrightarrow20x^2+16x-1=0\)

\(\Rightarrow x=\dfrac{-4+\sqrt{21}}{10}\)

Bạn xem lại đề câu a.