Tính \(\frac{x}{y}\)biết:
\(\frac{x+2y}{4x-3y}=-2\)
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26 tháng 12 2020
a. 2x(x + y) - y(y + 2x) = 2x2 + 2xy - y2 - 2xy = 2x2 - y2
b.\(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)
Phần c nản quá.
XO
25 tháng 12 2020
a) 2x(x + y) - y(y + 2x)
= 2x2 + 2xy - y2 - 2xy
= 2x2 - y2
b) \(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)
c) \(\frac{x^3-4x^2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{x-1}\)
= \(\frac{x^3-4x^2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}+\frac{x^2+x+1}{\left(x^2+x+1\right)\left(x-1\right)}\)
= \(\frac{x^3-4x^2+2x-2+x^2+x+1}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{x^3-3x^2+3x-1}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{\left(x-1\right)^3}{\left(x^2+x+1\right)\left(x-1\right)}\)
\(=\frac{\left(x-1\right)^2}{x^2+x+1}\)
TN
1
4 tháng 2 2018
Ta có :
\(\frac{x+2y}{4x-3y}=-2\)
\(\Rightarrow\)\(x+2y=\left(-2\right).\left(4x-3y\right)\)
\(\Rightarrow\)\(x+2y=-8x+6y\)
\(\Rightarrow\)\(x+8x=6y-2y\)
\(\Rightarrow\)\(9x=4y\)
\(\Rightarrow\)\(\frac{x}{y}=\frac{4}{9}\)
Vậy tỉ số \(\frac{x}{y}=\frac{4}{9}\)
DT
1
1 tháng 2 2022
\(=\dfrac{2x+y}{2\left(x+y\right)}-\dfrac{x+2y}{x-y}+\dfrac{5}{x}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{2x^2-2xy+xy-y^2}{2\left(x+y\right)\left(x-y\right)}-\dfrac{2\left(x+2y\right)\left(x-y\right)}{2\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{2x^2-xy-y^2-2\left(x^2+xy-2y^2\right)}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{2x^2-xy-y^2-2x^2-2xy+4y^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-3xy+3y^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-9xy+9y^2-8x}{6\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-9x^2y+9xy^2-8x^2+30\left(x^2-y^2\right)}{6x\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{-9x^2y+9xy^2+22x^2-30y^2}{6x\cdot\left(x-y\right)\left(x+y\right)}\)
15 tháng 1 2016
c1:Thay số
Q=\(\frac{5+2.4-3.3}{5-2.4+3.3}\)
O=\(\frac{4}{6}\)=\(\frac{2}{3}\)
15 tháng 1 2016
\(\frac{4x}{6y}=\frac{2x+8}{3y+11}\)
\(4x\left(3y+1\right)=6y\left(2x+8\right)\)
\(12xy+4x=12xy+48y\)
\(4x-48y=0\)
\(4x=48y\)
Ta có:\(\frac{4x}{48y}\)
\(\Leftrightarrow\)\(\frac{x}{y}=\frac{1}{12}\)
31 tháng 12 2016
\(P=\frac{x+3y}{3x+y}.\frac{4x-2y}{x-y}-\frac{x+3y}{3x+y}.\frac{x-3y}{x-y}\)
\(=\frac{x+3y}{3x+y}\left(\frac{4x-2y}{x-y}-\frac{x-3y}{x-y}\right)\)
\(=\frac{x+3y}{3x+y}.\frac{3x+y}{x-y}=\frac{x+3y}{x-y}\)
TT
1
2 tháng 7 2020
Ta có: \(\frac{x+2y}{4x-3y}=-2\)
\(\Leftrightarrow x+2y=-2\left(4x-3y\right)\)
\(\Leftrightarrow x+2y=-8x+6y\)
\(\Leftrightarrow x+2y+8x-6y=0\)
\(\Leftrightarrow9x-4y=0\)
\(\Leftrightarrow9x=4y\)
hay \(\frac{x}{y}=\frac{4}{9}\)
Vậy: \(\frac{x}{y}=\frac{4}{9}\)
D
0
\(\frac{x+2y}{4x-3y}=-2\)\(\Leftrightarrow x+2y=-2\left(4x-3y\right)\Leftrightarrow x+2y=-8x+6y\Leftrightarrow x+8x=6y-2y\Leftrightarrow9x=4y\Leftrightarrow\frac{x}{y}=\frac{4}{9}\)(chuyển vế đổi dấu)