\(a,\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0;x\ne1\right)\)

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

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

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

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

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

\(b,\) Thay \(x=4-2\sqrt{3}\) vào biểu thức trên, ta được:

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

\(=\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}+1\)

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

\(=\left|\sqrt{3}-1\right|+1\)

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

\(=\sqrt{3}\)

Vậy: ...

\(\text{#}Toru\)

\(a\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{x+\sqrt{x}+2}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\\ =\left(\dfrac{\sqrt{x}-1}{x-1}+\dfrac{x+\sqrt{x}+2}{x-1}\right).\sqrt{x}-1\\ =\dfrac{x+\sqrt{2}+1}{x-1}.\sqrt{x}-1\\ =\sqrt{x}+1\\ b,tacóx=4-2\sqrt{3}=\left(\sqrt{3}-\sqrt{1}\right)^2thãy=\sqrt{3}-\sqrt{1}vàobiểuthức,tađược\\ \sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-1=\sqrt{3}-1-1=\sqrt{3}-2\)

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

Thay \(x=6-2\sqrt{5}\) vào B ta có:

\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ =\dfrac{\sqrt{6-2\sqrt{5}}-1}{\sqrt{6-2\sqrt{5}}+1}\\ =\dfrac{\sqrt{5-2\sqrt{5}+1}-1}{\sqrt{5-2\sqrt{5}+1}+1}\\ =\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}\\ =\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1+1}\\ =\dfrac{\sqrt{5}-2}{\sqrt{5}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{5}\\ =\dfrac{5-2\sqrt{5}}{5}\)

a/ ĐKXĐ: x khác -1

\(P=\left(\dfrac{4}{x+1}-1\right):\dfrac{9-x^2}{x^2+2x+1}=\left(\dfrac{4}{x+1}-\dfrac{x+1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(3-x\right)\left(3+x\right)}\)

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

b/ |x + 1| = 2

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

Với x = 1 P = \(\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)

c/ \(\dfrac{x+1}{x+3}=\dfrac{x+3-2}{x+3}=\dfrac{x+3}{x+3}-\dfrac{2}{x+3}=1-\dfrac{2}{x+3}\)

ĐỂ P nguyên thì \(\dfrac{2}{x+3}\in Z\Leftrightarrow x+3\inƯ\left(2\right)\)

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

=> \(x=\left\{-5;-4;-2;-1\right\}\) (tm)

Vậy............

Bài 1: 

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

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

\(=\dfrac{2x}{x-1}\)

A(1/2^2022)=1/2^2022+1/2^4044+...+1/2^(2022^2021)

=>2^2022*A=1+1/2^2022+...+1/2^(2022^2020)

=>A*(2^2022-1)=1-1/2^(2022^2021)

=>\(A=\dfrac{2^{2022^{2021}}-1}{2^{2022}-1}\)

+Giá trị của biểu thức đại số 3x²-5x+1=9

+Giá trị của biểu thức đại số x²y=48

- ta có x = -1

thay vào ta có: 3 . \(\left(-1\right)^2\) - 5.(-1) + 1

=> 3 . 1 + 5 +1

=> 3 + 5 + 1

=> 9

- ta có x = -4 , y = 3

thay vào ta có : \(\left(-4\right)^2\). 3

=> 16 . 3

=> 48

Bạn nào có kết quả khác thì góp ý cho mk nhé

A=\(\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)  +\(\frac{x-2}{\left(x-1\right).\left(x^2+x+1\right)}\) 

A=\(\frac{2x-3}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{11.2-3}{\left(11-1\right)\left(11^2+11+1\right)}\)=\(\frac{19}{1330}\)

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

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

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

\(=\frac{2.11-3}{11^3-1}=\frac{19}{1330}=\frac{1}{70}\)

A = (\(x-y\)).(\(x^2\) + \(xy\) + y²) + 2y³

A = \(x^3\) - y³ + 2y³

A = \(x^3\) + y³

Thay \(x=\dfrac{2}{3}\); y = \(\dfrac{1}{3}\) vào biểu thức

A = \(x\)³ +  y³ ta có:

A = (\(\dfrac{2}{3}\))³ + (\(\dfrac{1}{3}\))³

A = \(\dfrac{8}{27}\) + \(\dfrac{1}{27}\)

A = \(\dfrac{9}{27}\)

A = \(\dfrac{1}{3}\) 

Thay x=31 vào biểu thức ta có:

31³-30 x 31²-32 x 31+1 = 29731