\(log_216=log_22^4=4\)

\(log_32187=log_33^7=7\)

\(log_{10}\dfrac{1}{100}=log_{10}10^{-2}=-2\)

\(log10000=log10^4=4\)

\(9^{log_312}=3^{2log_312}=3^{log_3144}=144\)

\(8^{log_25}=2^{3log_25}=2^{log_2125}=125\)

\(\left(\dfrac{1}{25}\right)^{log_5\dfrac{1}{3}}=5^{-2log_5\dfrac{1}{3}}=5^{log_59}=9\)

\(\left(\dfrac{1}{4}\right)^{log_23}=2^{-2log_23}=2^{log_2\dfrac{1}{9}}=\dfrac{1}{9}\)

a) Ta có: \(-\dfrac{3}{2}\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2\cdot\left(1+\sqrt{5}\right)^2}\)

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

\(=\dfrac{-3}{2}\sqrt{5}+3+4\sqrt{5}+4\)

\(=\dfrac{5}{2}\sqrt{5}+7\)

b) Ta có: \(\left(1+\dfrac{1}{\tan^225^0}\right)\cdot\sin^225^0-\tan55^0\cdot\tan35^0\)

\(=\dfrac{\tan^225^0+1}{\tan^225^0}\cdot\sin25^0-1\)

\(=\left(\dfrac{\sin^225^0}{\cos^225^0}+1\right)\cdot\dfrac{\cos^225^0}{\sin^225^0}\cdot\sin25^0-1\)

\(=\dfrac{\sin^225^0+\cos^225^0}{\cos^225^0}\cdot\dfrac{\cos^225^0}{\sin25^0}-1\)

\(=\dfrac{1}{\sin25^0}-1\)

\(=\dfrac{1-\sin25^0}{\sin25^0}\)

`a)ĐK:` \(\begin{cases}x \ge 0\\x-\sqrt{x} \ne 0\\x-1 \ne 0\\\end{cases}\)

`<=>` \(\begin{cases}x \ge 0\\x \ne 0\\x \ne 1\\\end{cases}\)

`<=>` \(\begin{cases}x>0\\x \ne 1\\\end{cases}\)

`b)A=(sqrtx/(sqrtx-1)-1/(x-sqrtx)):(1/(1+sqrtx)+2/(x-1))`

`=((x-1)/(x-sqrtx)):((sqrtx-1+2)/(x-1))`

`=(x-1)/(x-sqrtx):(sqrtx+1)/(x-1)`

`=(sqrtx+1)/sqrtx:1/(sqrtx-1)`

`=(x-1)/sqrtx`

`c)A>0`

Mà `sqrtx>0AAx>0`

`<=>x-1>0<=>x>1`

a, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

b, Ta có : \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

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

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1}{\sqrt{x}-1}=\dfrac{x-1}{\sqrt{x}}\)

c, Ta có : \(A>0\)

\(\Leftrightarrow x-1>0\)

\(\Leftrightarrow x>1\)

Vậy ...

a: =2+6*(-1)^2019+2026

=2028-6

=2022

b: \(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}...\cdot\dfrac{625}{624}\)

\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{4^2}{\left(4-1\right)\left(4+1\right)}...\cdot\dfrac{625}{\left(25-1\right)\left(25+1\right)}\)

\(=\dfrac{2\cdot3\cdot4\cdot...\cdot49}{1\cdot2\cdot3\cdot...\cdot48}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot49}{3\cdot4\cdot5\cdot...\cdot50}\)

\(=\dfrac{49}{1}\cdot\dfrac{2}{50}=\dfrac{98}{50}=\dfrac{49}{25}\)

a) Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

b) Để \(A< -\dfrac{1}{3}\) thì \(A+\dfrac{1}{3}< 0\)

\(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{3}< 0\)

\(\Leftrightarrow\dfrac{-9+\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}< 0\)

\(\Leftrightarrow\sqrt{x}-6< 0\)

\(\Leftrightarrow x< 36\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 36\\x\ne9\end{matrix}\right.\)

a: Ta có: \(A=\left(\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1}{x-1}\)

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

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+1}{1}\)

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

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1; x\neq 25$

a) 

\(A=\frac{4\sqrt{x}}{\sqrt{x}-5}:\left[\frac{(\sqrt{x}-2)(\sqrt{x}+2)+\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+2}+\frac{5-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}\right]\)

\(=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{x-4+\sqrt{x}-1+5-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}\)

\(=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{\sqrt{x}}{\sqrt{x}+2}=\frac{4\sqrt{x}}{\sqrt{x}-5}.\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{4(\sqrt{x}+2)}{\sqrt{x}-5}\)

b) Tại $x=81$ thì $\sqrt{x}=9$.

Khi đó: $A=\frac{4(9+2)}{9-5}=11$

c) $A< 4\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}-5}< 1$

$\Leftrightarrow \frac{7}{\sqrt{x}-5}< 0\Leftrightarrow \sqrt{x}-5< 0$

$\Leftrightarrow 0\leq x< 25$. Kết hợp với ĐKXĐ suy ra: $0\leq x< 25; x\neq 1$