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Bài 3:
Gọi số học sinh là x
Theo đề, ta có: \(x\in BC\left(12;18;21\right)\)
hay x=504
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CÂU 1:
\(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
CÂU 2:
\(\dfrac{12x^3y^2}{18xy^5}=\dfrac{2x^2}{3y^3}\)
CÂU 3:
\(\dfrac{15x\left(x+5\right)^3}{20x^2\left(x+5\right)}=\dfrac{3\left(x+5\right)^2}{4x}\)
CÂU 4:
\(\dfrac{3xy+x}{9y+3}=\dfrac{x\left(3y+1\right)}{3\left(3y+1\right)}=\dfrac{x}{3}\)
CÂU 5:
\(\dfrac{3xy+3x}{9y+9}=\dfrac{3x\left(y+1\right)}{9\left(y+1\right)}=\dfrac{x}{3}\)
CÂU 6:
\(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{x\left(x-y\right)}{5y\left(y-x\right)}=\dfrac{-x\left(y-x\right)}{5y\left(y-x\right)}=\dfrac{-x}{5y}\)
CÂU 7:
\(\dfrac{2x^2+2x}{x+1}=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
CÂU 8:
\(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\\ =\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
CÂU 9:
\(\dfrac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}=\dfrac{2y}{3\left(x+y\right)^2}\)
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c) Ta có: \(\sqrt{\sqrt{x}+3}=3\)
\(\Leftrightarrow\sqrt{x}+3=9\)
\(\Leftrightarrow\sqrt{x}=6\)
hay x=36
Ta có: \(\sqrt{x-2\sqrt{x-1}}=2\)
\(\Leftrightarrow x-2\sqrt{x-1}-4=0\)
\(\Leftrightarrow x-1-2\cdot\sqrt{x-1}\cdot1+1=4\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=4\)
\(\Leftrightarrow\sqrt{x-1}-1=2\)
\(\Leftrightarrow\sqrt{x-1}=3\)
\(\Leftrightarrow x-1=9\)
hay x=10
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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