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2: =>(3x-7)^2=25/144=(5/12)^2
=>3x-7=5/12 hoặc 3x-7=-5/12
=>3x=5/12+7=89/12 hoặc 3x=7-5/12=79/12
=>x=89/36 hoặc x=79/36
3:Sửa đề: |2x-3|=|x+1|
=>2x-3=x+1 hoặc 2x-3=-x-1
=>x=4 hoặc 3x=2
=>x=2/3 hoặc x=4
4: =>3x+1=5 hoặc 3x+1=-5
=>3x=4 hoặc 3x=-6
=>x=-2 hoặc x=4/3
1: =>\(2x-7=\sqrt[3]{\dfrac{26}{63}}\)
=>\(2x=\sqrt[3]{\dfrac{26}{63}}+7\)
=>\(x=\dfrac{1}{2}\cdot\left(\sqrt[3]{\dfrac{26}{63}}+7\right)\)
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1) \(6x+3=0\Leftrightarrow6x=-3\Leftrightarrow x=-\dfrac{1}{2}\)
2) \(-5x-7=0\Leftrightarrow-5x=7\Leftrightarrow x=-\dfrac{7}{5}\)
3) \(\left(3x-2\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\5-x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=5\end{matrix}\right.\)
4) \(x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
6) \(x^4+8=0\)( vô lý do \(x^4+8\ge8>0\))
Vậy \(S=\varnothing\)
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\(5x=2y\Rightarrow\frac{x}{2}=\frac{y}{5}\)
\(3x=5z\Rightarrow\frac{y}{5}=\frac{z}{3}\)
\(\Rightarrow\frac{x}{2}=\frac{y}{5}=\frac{z}{3}\Rightarrow\frac{3x}{6}=\frac{2y}{10}=\frac{z}{3}=\frac{3x+2y-z}{6+10-3}=\frac{26}{13}=2\)
\(\Rightarrow x=4;y=10;z=6\)
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\(A-B=\left(1+\dfrac{3}{4}-\dfrac{3}{4}\right)x^4+\left(-\dfrac{1}{8}x^3+\dfrac{1}{8}x^3\right)+\left(-\dfrac{5}{4}+\dfrac{9}{4}\right)x^2+\left(\dfrac{2}{5}x-\dfrac{2}{5}x\right)+\dfrac{4}{7}+\dfrac{3}{7}\)
\(=x^4+x^2+1>0\forall x\)
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4) xy-5x+y=10
=> x(y-5)+(y-5)=15
=> (y-5)(x-1)=15
từ đây lập bảng ra nhé , chắc bạn biết