Bài 1 : 

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

b, \(x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)=x\left[\left(x+y\right)^2-9\right]=x\left(x+y-3\right)\left(x+y+3\right)\)

Bài 2 : 

a, \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)=8x^3-4x^2y+2xy^2-4x^2y+2xy^2-y^3\)

\(=8x^3-8x^2y+4xy^2-y^3\)

b, \(6x^5y^2-9x^4y^3:3x^3y^2=\left(6x^5y^2:3x^3y^2\right)+\left(-9x^4y^3:3x^3y^2\right)=2x^2-3xy\)

Bài làm

ĐKXĐ : x ≠ 2

Ta có : \(\frac{x^3-2x^2+4}{x-2}=\frac{x^2\left(x-2\right)+4}{\left(x-2\right)}=\frac{x^2\left(x-2\right)}{x-2}+\frac{4}{x-2}=x^2+\frac{4}{x-2}\)

Vì x nguyên => x² nguyên

=> Để phân thức có giá trị nguyên thì \(\frac{4}{x-2}\)có giá trị nguyên

=> \(4⋮\left(x-2\right)\)

=> \(\left(x-2\right)\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Các giá trị trên đều tmđk x ≠ 2

Vậy x ∈ { -2 ; 0 ; 1 ; 3 ; 4 ; 6 }

để bth trên có gtr nguyên thì x^3-2x^2+4 chia hết x-2

ta có: x^3-2x^2+4 = x^2(x-2)+4

vì x^2(x-2) chia hết cho x-2 với mọi x

nên để x^3-2x^2+4 chia hết cho x-2

thì 4 chia hết cho x-2

=> x-2 thuộc vào ước của 4
=>x-2 thuộc {+-1;+-2;+-4;0}

=> x = {2;3;1;4;0;6;-2}

Dễ ợt đâu :))

 \(2^{51}-1=\left(2+2^2+2^3+.....+2^{51}\right)-\left(1+2+2^2+....+2^{50}\right)\)

Ta có :

\(2+2^2+2^3+....+2^{51}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+....+\left(2^{49}+2^{50}+2^{51}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+....+2^{49}\left(1+2+2^2\right)\)

\(=2.7+2^4.7+....+2^{49}.7\)

\(=7\left(2+2^4+....+2^{49}\right)⋮7\)(1)

Chứng minh tương tự ta cũng có : \(\left(1+2+2^2+....+2^{50}\right)⋮7\)(2)

Từ (1) ; (2) \(\Rightarrow\left(2+2^2+2^3+.....+2^{51}\right)-\left(1+2+2^2+....+2^{50}\right)⋮7\)

Hay \(2^{51}-1⋮7\)(đpcm)

A B C H

Xét tam giác ABH ta có : \(AB^2=BH^2+AH^2\Rightarrow15^2=12^2+AH^2\)

\(AH^2=AB^2-BH^2=15^2-12^2=81\Rightarrow AH=9\)cm 

Xét tam giác ACH ta có : \(AC^2=AH^2+HC^2\Rightarrow AC^2=AH^2+HC^2\)

\(HC^2=AC^2-AH^2=41^2-9^2=1600\Rightarrow HC=40\)cm 

Ta có : \(BC=CH+HB=40+12=52\)cm

\(\Rightarrow S_{ABC}=\frac{1}{2}AH.BC=\frac{1}{2}.9.52=234\)cm²

a, \(x^2-\frac{x^4}{x^2}-1=x^2-x^2-1=-1\)

b, \(\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}=\frac{x+9}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x\left(x+3\right)}\)

\(=\frac{x\left(x+9\right)}{x\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\frac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}=\frac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x\left(x-3\right)}\)

c, \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}=\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x\left(1-2x\right)}\)

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

\(=\frac{1-2x-3x+6x^2-6x^2+4x+3x-2}{2x\left(1-2x\right)}=\frac{-1+2x}{2x\left(1-2x\right)}=\frac{-\left(1-2x\right)}{2x\left(1-2x\right)}=\frac{-1}{2x}\)

d, viết lại đề đy nhé 

e, \(\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x\left(1-x\right)}{9-x^2}=\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x-2x^2}{\left(3-x\right)\left(x+3\right)}\)

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

\(=\frac{x^2+3x+x+3-x+3+x^2+3x+2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\frac{8x+6}{\left(x-3\right)\left(x+3\right)}\)