Tính giá trị của biểu thức: \(\frac{16x^2-40xy}{8x^2-24xy}\). Biết \(\frac{x}{y}=\frac{10}{3}\)
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\(\frac{x}{y}=10\Rightarrow x=10y\)
\(M=\frac{16x^2-40xy}{8x^2-24xy}=\frac{8x\left(2x-5y\right)}{8x\left(x-3y\right)}=\frac{2x-5y}{x-3y}\)
\(=\frac{2.10y-5y}{10y-3y}=\frac{15}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có : \(y=\frac{3}{10}x\left(1\right)\)
Thay (1) và PT đc \(N=\frac{16x^2-12x^2}{8x^2-\frac{36}{5}x^2}=\frac{4x^2}{\frac{4}{5}x^2}=5\)
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a)\(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-b+c\right)}=\dfrac{a+b-c}{a-b+c}\)Giá trị của biểu thức trên tại \(a=4;b=-5;c=6\) là:
\(\dfrac{4-5-6}{4-\left(-5\right)+6}=-\dfrac{7}{15}\)
b: \(=\dfrac{8x\left(2x-5y\right)}{8x\left(x-3y\right)}=\dfrac{2x-5y}{x-3y}\)
Đặt x/10=y/3=k
=>x=10k; y=3k
\(A=\dfrac{2\cdot10k-5\cdot3k}{10k-3\cdot3k}=\dfrac{5k}{k}=5\)
c: \(C=\left(\dfrac{x^3-y^3-x^3-y^3}{\left(x+y\right)\left(x-y\right)}\right):\dfrac{x^2-y^2-x^2}{x+y}\)
\(=\dfrac{-2y^3}{\left(x+y\right)\left(x-y\right)}\cdot\dfrac{x+y}{-y^2}=\dfrac{2y}{x-y}\)
\(=\dfrac{20}{9-10}=-20\)
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\(M=\frac{16x^2-40xy}{8x^2-24xy}\)ĐK:\(x^2-3xy\ne0;y\ne0\)
\(\frac{x}{y}=10\Rightarrow x=10y\)
\(M=\frac{2x^2-5xy}{x^2-3xy}\)=\(\frac{15}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=\(\frac{16x^2-40xy}{8x^2-24xy}=\frac{8x(2x-5y)}{ 8x(x-3y)} =\frac{2x-5y}{x-3y} \)
\(\frac{x}{y}=\frac{10}{3}<=>10y=3x <=>y=\frac{3}{10}x \)
=>A=(\(2x-\frac{3}{2}x):(x-\frac{9}{10}x) \)
=\(\frac{1}{2}x:\frac{1}{10}x=\frac{1}{2}x.\frac{10}{x}=5 \)
![](https://rs.olm.vn/images/avt/0.png?1311)