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là A = ...........
Theo bài ra , ta có :
\(9x^2+4y^2=20xy\)
\(\Leftrightarrow9x^2-18xy-2xy+4y^2=0\)
\(\Leftrightarrow9x\left(x-2y\right)-2y\left(x-2y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(9x-2y\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-2y=0\\9x-2y=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2y\\9x=2y\end{matrix}\right.\)
Thay x = 2y vào A ta đk :
\(A=\frac{3x-2y}{3x+2y}=\frac{3.2y-2y}{3.2y+2y}=\frac{4y}{8y}=\frac{4}{8}=\frac{1}{2}\)
Vậy \(A=\frac{1}{2}\)
Chúc bạn học tốt =))
Ta có: A=\(\frac{3x-2y}{3x+2y}\)
=>A2=\(\frac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}\)=\(\frac{9x^2-12xy+4y^2}{9x^2+12xy+4y^2}\)=\(\frac{\left(9x^2+4y^2\right)-12xy}{\left(9x^2+4y^2\right)+12xy}\)=\(\frac{20xy-12xy}{20xy+12xy}\)=\(\frac{8xy}{32xy}\)=\(\frac{1}{4}\)
=>\(\left\{\begin{matrix}A=\frac{1}{2}\\A=\frac{-1}{2}\end{matrix}\right.\)
Do 2y<3x<0
=>\(\frac{3x-2y}{3x+2y}\)<0
=>A=\(\frac{-1}{2}\)
is ............
\(\left\{{}\begin{matrix}x-y-z=0\\x+2y-10z=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}y=3z\\x=y+z=4z\\x+2y=10z\end{matrix}\right.\)
\(B=\dfrac{2x^2+4xy}{y^2+z^2}=\dfrac{2x\left(x+2y\right)}{9z^2+z^2}=\dfrac{2.4z.10z}{10.z^2}=8\)
là: M = ...........
Ta có: \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2a+2b+2c}{a+b+c}=2\)
\(\Rightarrow\) a + b = 2c; b + c = 2a; c + a = 2b
\(\Rightarrow\) M = \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
= \(\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{a+c}{a}\right)\)
= \(\frac{2c}{b}\times\frac{2a}{c}\times\frac{2b}{a}\)
= 8
Vậy: M = 8.
mình lớp 6