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a) (-xy)10 : (-xy)5 = (-xy)2
b) (5x2y4) : 10x2y = y3/2
c) (15x4y3z2) : (5x2y2z2) = 3x2y
d) \(\frac{3}{4}x^3y^3:\left(-\frac{1}{2}x^2y^2\right)=\frac{\frac{3}{4}x^3y^2}{-\frac{1}{2}x^2y^2}=-\frac{3}{2}x\)
e) 6x3y5 : 12x3y2 = y3/2
f) (25x5 - 5x4 + 10x2) : (5x2)
= 5x3 - x2 + 2
g) \(\frac{2}{3}xy\cdot\left(2x^2y-3xy+y^2\right)=\frac{4}{3}x^3y^2-2x^2y^2+\frac{2}{3}xy^3\)
h) \(\frac{3}{4}x^3y^5z:\left(5x^2y^2z\right)=\frac{\frac{3}{4}x^3y^5z}{5x^2y^2z}=\frac{3}{20}xy^3\)
Bài 1:
\(A=\frac{a+b}{b+c}.\)
Ta có:
\(\frac{b}{a}=2\Rightarrow\frac{b}{2}=\frac{a}{1}\) (1)
\(\frac{c}{b}=3\Rightarrow\frac{c}{3}=\frac{b}{1}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{b}{2}=\frac{c}{6}.\)
\(\Rightarrow\frac{a}{1}=\frac{b}{2}=\frac{c}{6}=\frac{a+b}{3}=\frac{b+c}{8}.\)
\(\Rightarrow A=\frac{a+b}{b+c}=\frac{3}{8}\)
Vậy \(A=\frac{a+b}{b+c}=\frac{3}{8}.\)
Bài 2:
a) \(\frac{72-x}{7}=\frac{x-40}{9}\)
\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow648+280=7x+9x\)
\(\Rightarrow928=16x\)
\(\Rightarrow x=928:16\)
\(\Rightarrow x=58\)
Vậy \(x=58.\)
b) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right).\left(x+4\right)=5.20\)
\(\Rightarrow\left(x+4\right).\left(x+4\right)=100\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow x+4=\pm10.\)
\(\Rightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-4\\x=\left(-10\right)-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)
Vậy \(x\in\left\{6;-14\right\}.\)
Chúc bạn học tốt!
Bài 2:
a, \(\frac{72-x}{7}=\frac{x-40}{9}\)
\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)
\(\Rightarrow9.72-9.x=7.x-7.40\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow-9x-7x=-280-648\)
\(\Rightarrow-16x=-648\)
\(\Rightarrow x=58\)
Vậy \(x=58\)