Giúp mình vớiiii

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

b: \(=\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+5}{\sqrt{x}-2}\)

a, \(\dfrac{x-3\sqrt{x}+2}{x-\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

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

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

\(=x^2-4x+4-4x^2+4x-1+3x^2-15x-x+5\)

\(=-16x+8\)

b: Ta có: \(\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

=27x-55

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

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

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

\(=4x^2\)

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

\(=3x^2+12x+12-3x^2-16x-5+x^2+10x+25\)

\(=x^2+6x+32\)

\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)

\(a.x^3+8-x^3+2=10\)

\(b.x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)=x^2+10x+25-16x^3-28x^2-36x-2x^3+18x+x^2-9=-18x^3-26x^2-8x+16=\)

`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`

`b)` Với `x ne -1;x ne -5` có:

`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`

`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`

`A=[x^2-3x-4]/[(x+1)(x+5)]`

`A=[(x+1)(x-4)]/[(x+1)(x+5)]`

`A=[x-4]/[x+5]`

`c)` Với `x ne -5; x ne -1; x ne 4` có:

`P=A.B=[x-4]/[x+5].[-10]/[x-4]`

           `=[-10]/[x+5]`

Để `P` nguyên `<=>[-10]/[x+5] in ZZ`

    `=>x+5 in Ư_{-10}`

Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`

`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)

\(A=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{x-25}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)

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

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

Bài 1:

a. 

$=(x^3+2^3)-(x^3-2)=2^3+2=10$

b.

$=(x^2+10x+25)-4x(4x^2+12x+9)-(2x-1)(x^2-9)$
$=x^2+10x+25-16x^3-48x^2-36x-(2x^3-18x-x^2+9)$

$=-18x^3-46x^2-8x+16$

2.

a. 

$301^2=(300+1)^2=300^2+2.300+1=90000+600+1$

$=90601$

b.

$198^2=(200-2)^2=4(100-1)^2=4(100^2-2.100+1)$

$=4(10000-200+1)=4.9801=39204$

c.

$93.107=(100-7)(100+7)=100^2-7^2$

$=10000-49=9951$

d.

$127^2+146.127+73^2$

$=127^2+2.73.127+73^2$
$=(127+73)^2=200^2=40000$