đề là j nhỉ :>?

ở chỗ này 2x^2y^2, 6x^2y^2 dấu gì bn?

\(A=\dfrac{3x^2+5xy-2y^2}{3x^2-7xy+2y^2}\)

\(=\dfrac{3x^2-xy+6xy-2y^2}{3x^2-xy-6xy+2y^2}\)

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

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

\(=\dfrac{x+2y}{x-2y}\)

\(A=\frac{3x^2+5xy-2y^2}{3x^2-7xy+2y^2}=\frac{6xy-2y^2+3x^2-xy}{2y^2-6xy-xy+3x^2}\)

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

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

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

a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)

\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)

b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)

\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)

c, tương tự 

lướt

3x²+4y²=7xy

<=> 3x²-3xy+4y²-4xy=0

<=> 3x(x-y)-4y(x-y)=0

<=> (3x-4y)(x-y)=0

<=> 3x-4y=0 hoặc x-y=0

<=> 3x=4y hoặc x=y

<=> y = \(\frac{3}{4}\)x hoặc x=y

+) y = \(\frac{3}{4}\)x, ta có:

F = \(\frac{4.\frac{3}{4}x+2x}{5.\frac{3}{4}x-7x}+\)\(\frac{3x-2.\frac{3}{4}x}{10.\frac{3}{4}x-4x}\)

F = \(\frac{5x}{-\frac{13}{4}x}+\frac{\frac{3}{2}x}{\frac{7}{2}x}\)

F = \(-\frac{20}{13}+\frac{3}{7}=-\frac{101}{91}\)

+) x = y, ta có:

F = \(\frac{4x+2x}{5x-7x}+\frac{3x-2x}{10x-4x}\)

F = \(\frac{6x}{-2x}+\frac{1x}{6x}=-3+\frac{1}{6}=-\frac{17}{6}\)

Từ \(3x^2+4y^2=7xy\Rightarrow3x^2+4y^2-7xy=0\)

\(\Rightarrow3x^2-4xy-3xy+4y^2=0\)

\(\Rightarrow x\left(3x-4y\right)-y\left(3x-4y\right)=0\)

\(\Rightarrow\left(x-y\right)\left(3x-4y\right)=0\)\(\Rightarrow\left[\begin{matrix}x=y\\x=\frac{4y}{3}\end{matrix}\right.\)

*)Xét \(x=y\) ta có \(F=\frac{4y+2y}{5y-7y}+\frac{3y-2y}{10y-4y}=\frac{6y}{-2y}+\frac{y}{6y}=-3+\frac{1}{6}=-\frac{17}{6}\)

*)Xét \(x=\frac{4y}{3}\) ta có \(F=\frac{4y+2\cdot\frac{4y}{3}}{5y-7\cdot\frac{4y}{3}}+\frac{3\cdot\frac{4y}{3}-2y}{10y-4\cdot\frac{4y}{3}}=\frac{4y+\frac{8y}{3}}{5y-\frac{28y}{3}}+\frac{4y-2y}{10y-\frac{16y}{3}}=\frac{-20}{13}+\frac{3}{7}=\frac{-101}{91}\)

1A

2B

3C

4D

4A