1/

\(\frac{x}{y}=\frac{3}{4}\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{4x}{12}=\frac{5y}{20}=\frac{4x-5y}{-8}\) (1)

\(\frac{x}{3}=\frac{y}{4}=\frac{3x}{9}=\frac{4y}{16}=\frac{3x+4y}{25}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{4x-5y}{-8}=\frac{3x+4y}{25}\Rightarrow\frac{4x-5y}{3x+4y}=\frac{-8}{25}\)

2/

\(M-N=3x\left(x-y\right)-\left(y-x\right)\left(y+x\right)=\)

\(=3x\left(x-y\right)+\left(x-y\right)\left(y+x\right)=\left(x-y\right)\left(4x+y\right)\)

Mà \(x-y\) chia hết cho 11 nên \(M-N\) chia hết cho 11

a) ta có: \(\frac{x}{y}=\frac{3}{4}\Rightarrow4x=3y\)

\(D=\frac{4x-5y}{3x+4y}=\frac{3y-5y}{3y+4y-x}=\frac{-2y}{7y-x}=\frac{-2y}{7y-y3:4}\)

\(=\frac{-2y}{\frac{25}{4}y}=-2y:\left(\frac{25}{4}y\right)=-\frac{8}{25}\)

b) ta có: M=3x.(x-y) chia hết cho 11

N = y² - x² = y² - xy - x² + xy = y.(y-x) - x.(x-y) = (y-x).(y+x) = - (x-y).(y+x) chia hết cho 11

=> M-N chia hết cho 11 (đpcm)

\(B=\left(y^2-x^2\right)=\left(y-x\right)\left(y+x\right)\)

\(A-B=\left(y-x\right)\left(2x-y\right)\).Do \(\left(x-y\right)⋮11\Rightarrow-1\left(x-y\right)⋮11\Rightarrow y-x⋮11\)

Đặt y - x = 11k.Ta có: \(A-B=11k\left(2x-y\right)⋮11^{\left(đpcm\right)}\)

Có :\(\left(x-y\right)⋮11\)=> M\(⋮11\)

       N= \(y^2-x^2\) = \(-\text{(}x^2-y^2\text{)}=-\text{[}\left(x-y\right).\left(x+y\right)\text{]}\)=> N\(⋮11\)

=> M-N \(⋮11\)

Vậy \(M-N⋮11\)(đpcm)

Câu 2: 

a: \(n^2-2n+5⋮n-1\)

\(\Leftrightarrow n^2-n-n+1+4⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{2;0;3;-1;5;-3\right\}\)

b: \(4x^2-6x-16⋮x-3\)

\(\Leftrightarrow4x^2-12x+6x-18+2⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{4;2;5;1\right\}\)

Câu 3: 

a: \(\left(3x-8\right)\left(7x+10\right)-\left(2x-15\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left(3x-8\right)\left(7x+10-2x+15\right)=0\)

\(\Leftrightarrow\left(3x-8\right)\left(5x+25\right)=0\)

=>x=8/3 hoặc x=-5

b: \(\dfrac{\left(x^4-2x^2-8\right)}{x-2}=0\)(ĐKXĐ: x<>2)

\(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2+2\right)=0\)

=>x+2=0

hay x=-2

a)\(M=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)=6x^2+23x-55-6x^2-23x-21=-76\)

b) \(N=\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x^2+2x+1=9\)

Ta có \(xy\le\dfrac{\left(x+y\right)^2}{4}\).

Do đó ta có: \(x+y+xy=x+y-2xy+3xy\le x+y-2xy+\dfrac{3}{4}\left(x+y\right)^2\)

\(\Rightarrow x^2+y^2\le x+y-2xy+\dfrac{3}{4}\left(x+y\right)^2\)

\(\Leftrightarrow\dfrac{1}{4}\left(x+y\right)^2-\left(x+y\right)\le0\)

\(\Leftrightarrow\left(x+y\right)\left[\dfrac{1}{4}\left(x+y\right)-1\right]\le0\)

\(\Leftrightarrow0\le x+y\le4\).

Do đó m = 0, n = 4.

Vậy m² + n² = 16. Chọn A.