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Từ 3y - x = 6, ta suy ra 3y = 6 + x và x = 3y - 6
Ta có: A = \(\dfrac{x}{y-2}\)+\(\dfrac{2x-3y}{x-6}\) = \(\dfrac{x}{y-2}\)+\(\dfrac{2x-\left(6+x\right)}{x-6}\)
= \(\dfrac{x}{y-2}\)+\(\dfrac{2x-6-x}{x-6}\) = \(\dfrac{x}{y-2}\)+1 = \(\dfrac{x+y-2}{y-2}\)
= \(\dfrac{3y-6+y-2}{y-2}\) = \(\dfrac{4y-8}{y-2}\) = \(\dfrac{4\left(y-2\right)}{y-2}\) = 4
Vậy giá trị của biểu thức A là 4
Lời giải:
\(3y-x=6\Rightarrow x=3y-6\)
\(\Rightarrow \frac{x}{y-2}=\frac{3y-6}{y-2}=\frac{3(y-2)}{y-2}=3\)
\(3y-x=6\Rightarrow 3y=x+6\)
\(\Rightarrow \frac{2x-3y}{x-6}=\frac{2x-(x+6)}{x-6}=\frac{x-6}{x-6}=1\)
Do đó: \(A=\frac{x}{y-2}+\frac{2x-3y}{x-6}=3+1=4\)
Ta có : \(3y-x=6\)
\(\Rightarrow x=3y-6\)
Thay \(x=3y-6\) vào biểu thức A , ta có :
\(\Rightarrow A=\dfrac{3y-6}{y-2}+\dfrac{2\left(3y-6\right)-3y}{3y-6-6}\)
\(=\dfrac{3\left(y-2\right)}{y-2}+\dfrac{3y-12}{3y-12}=3+1=4\)
Vậy A = 4 .
a/ +) \(\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}\)\(\left(1\right)\)
+) \(\dfrac{y}{3}=\dfrac{z}{5}\Leftrightarrow\dfrac{y}{12}=\dfrac{z}{20}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
\(\Leftrightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=3\\\dfrac{y}{12}=3\\\dfrac{z}{20}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)
Vậy ..
b/ \(2x=3y=5z\)
\(\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}\)
\(\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Theo t/c dãy tỉ số bằng nhau tcos :
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=5\\\dfrac{y}{10}=5\\\dfrac{z}{6}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=75\\y=50\\z=30\end{matrix}\right.\)
Vậy..
c/ tương tự
1, \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)
\(=\dfrac{2a\left(x-1\right)^2}{5b\left(1-x\right)\left(1+x\right)}\)
\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)
2, \(\dfrac{x^2+4x+3}{2x+6}\)
\(=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}\)
\(=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
3, \(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)
\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
4, \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)
5, \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)
\(Q=2x^2+\dfrac{6}{x^2}+3y^2+\dfrac{8}{y^2}=\left(2x^2+\dfrac{2}{x^2}\right)+\left(3y^2+\dfrac{3}{y^2}\right)+\left(\dfrac{4}{x^2}+\dfrac{5}{y^2}\right)\)
\(\ge2.2+2.3+9=19\)
Dấu = xảy ra khi \(x=y=1\)
a: \(=\dfrac{1-2x+3+2y+2y-4}{6x^3y}=\dfrac{-2x+4y}{6x^3y}=\dfrac{-2\left(x-2y\right)}{6x^3y}=\dfrac{-x+2y}{3x^3y}\)
b: \(=\dfrac{x^2-2+2-x}{x\left(x-1\right)^2}=\dfrac{x\left(x-1\right)}{x\left(x-1\right)^2}=\dfrac{1}{x-1}\)
c: \(=\dfrac{3x+1+x^6-3x}{x^2-3x+1}\)
\(=\dfrac{x^6+1}{x^2-3x+1}\)
d: \(=\dfrac{x^2+38x+4+3x^2-4x-2}{2x^2+17x+1}\)
\(=\dfrac{4x^2+34x+2}{2x^2+17x+1}=2\)
Ta có : \(3y-x=6\)
\(=>x=3y-6\)
\(=>A=\dfrac{3y-6}{y-2}+\dfrac{2\left(3y-6\right)-3y}{3y-6-6}\)
\(=>A=\dfrac{3y-6}{y-2}+\dfrac{6y-12-3y}{3y-12}\)
\(=>A=\dfrac{3y-6}{y-2}+\dfrac{3y-12}{3y-12}\)
\(=>A=\dfrac{3\left(y-2\right)}{y-2}+1=3+1=4\)
Vậy A=4.
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