A = (2 + √2)/(1 + √2)

= √2(√2 + 1)/(1 + √2)

= √2

C = (2√3 - √6)/(√8 - 2)

= √6(√2 - 1)/[2(√2 - 1)]

= √6/2

E = (x√x + 1)/(√x + 1)

= (√x + 1)(x - √x + 1)/(√x + 1)

= x - √x + 1

A = $\frac{2 + \sqrt{2}}{1 + \sqrt{2}}$

Để rút gọn biểu thức này, ta nhân tử và chia tử cho $1 - \sqrt{2}$:

A = $\frac{(2 + \sqrt{2})(1 - \sqrt{2})}{(1 + \sqrt{2})(1 - \sqrt{2})}$

A = $\frac{-2\sqrt{2}}{-1}$

A = $2\sqrt{2}$

C = $\frac{2\sqrt{3} - \sqrt{6}}{\sqrt{8} - 2}$

Ta nhân tử và chia tử cho $\sqrt{2}$:

C = $\frac{(2\sqrt{3} - \sqrt{6})\sqrt{2}}{(\sqrt{8} - 2)\sqrt{2}}$

C = $\frac{4\sqrt{6} - 2\sqrt{3}}{2\sqrt{2}}$

C = $\frac{2\sqrt{6} - \sqrt{3}}{\sqrt{2}}$

Ta nhân tử và chia tử cho $\sqrt{6} + \sqrt{2}$:

C = $\frac{(2\sqrt{6} - \sqrt{3})(\sqrt{6} + \sqrt{2})}{(\sqrt{2})(\sqrt{6} + \sqrt{2})}$

C = $\frac{12 - 3\sqrt{2}}{2}$

C = $6 - \frac{3\sqrt{2}}{2}$

E = $\frac{x\sqrt{x+1}}{\sqrt{x+1}}$

E = $x\sqrt{\frac{x+1}{x+1}}$

E = $x$.

\(a,\left(x-5\right)\left(2x+1\right)-2x\left(x-3\right)\\ =x.2x-5.2x+x-5-2x.x-2x.\left(-3\right)\\ =2x^2-10x+x-5-2x^2+6x\\ =2x^2-2x^2-10x+x+6x-5\\ =-3x-5\)

\(b,\left(2+3x\right)\left(2-3x\right)+\left(3x+4\right)^2\\ =\left[2^2-\left(3x\right)^2\right]+\left[\left(3x\right)^2+2.3x.4+4^2\right]\\=4-9x^2+\left(9x^2+24x+16\right)\\ =24x+20\)

`a)(2x-1)^2+(x+3)^2-5(x-7)(x+7)`

`=4x^2-4x+1+x^2+6x+9-5(x^2-49)`

`=5x^2-5x^2-4x+6x+1+9+245`

`=2x+255`

`b)(x-2)(x^2+2x+4)-(25+x^3)`

`=x^3-8-x^3-25=-33`

Lời giải:

a. 

$(2x-1)^2+(x+3)^2-5(x-7)(x+7)$

$=4x^2-4x+1+(x^2+6x+9)-5(x^2-49)$

$=5x^2+2x+10-(5x^2-245)=2x+255$

b.

$(x-2)(x^2+2x+4)-(25+x^3)=(x^3-2^3)-(25+x^3)$

$=-8-25=-33$

a: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)\)

\(=4x^2-4x+1+x^2+6x+9-5x^2+245\)

\(=2x+255\)

b: \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x^3+25\right)\)

\(=x^3-8-x^3-25\)

=-33

\(a,=\left(x+8-x+2\right)^2=10^2=100\\ b,=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ c,=x^3+1-x^3+1=2\)

a: Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+18x+x^2-9\)

\(=-18x^3-46x^2-8x+16\)

1: Ta có: \(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

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

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

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

2)

a) Thay \(x=\dfrac{9}{4}\) vào P, ta được:

\(P=\left(\dfrac{3}{2}+2\right):\left(\dfrac{3}{2}+3\right)=\dfrac{7}{2}:\dfrac{11}{2}=\dfrac{7}{11}\)

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

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

=1

Thay x=1 vào P, ta được:

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

a)      Cách 1:

\(6(y - x) - 2(x - y)\)

\( = 6y - 6x - 2x + 2y\)

\( = 8y - 8x\)

Cách 2:

\(6(y - x) - 2(x - y)\\= 6(y-x)+2(y-x)\\=(6+2).(y-x)\\=8.(y-x)\\=8y-8x\)

b)      \(3{x^2} + x - 4x - 5{x^2}\)

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

\( =  - 2{x^2} - 3x\)

Chia câu ra đi ạ

\(A=\dfrac{3}{2\left(2x-1\right)}\cdot x^2\left|2x-1\right|\cdot2\sqrt{2}\)

\(=\pm3\sqrt{2}x^2\)

\(B=\dfrac{a-b}{b^2}\cdot\dfrac{b^2\cdot\left|a\right|}{\left|a-b\right|}\)

\(=\pm\left|a\right|\)