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Có phải đề bài như thế này không bạn:
\(m=a^3+b^3+3AB.\left(a^2+b^2\right)+6.a^2b^2\left(a+b\right).\)
a: \(\Leftrightarrow\dfrac{y+5}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)
\(\Leftrightarrow2\left(y+5\right)^2-\left(y-5\right)^2=y^2+25y\)
=>\(2y^2+20y+50-y^2+10y-25=y^2+25y\)
=>30y+25=25y
=>5y=-25
=>y=-5(loại)
b: \(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
=>x^2+x+x^2-3x-4x=0
=>2x^2-6x=0
=>2x(x-3)=0
=>x=0(nhận) hoặc x=3(loại)
c: =>x^2-9-6(2x+7)=-13(x+3)
=>x^2-9-12x-42+13x+39=0
=>x^2+x-6=0
=>(x+3)(x-2)=0
=>x=2(nhận) hoặc x=-3(loại)
a: =>3x-2x=-11-9
=>x=-20
c: \(\Leftrightarrow\left(2x+3\right)\left(x^2+3\right)=2\left(2x+3\right)\)
=>2x+3=0
hay x=-3/2
\(a.\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\\\Leftrightarrow \frac{3\left(x-5\right)}{12}-\frac{24}{12}x+\frac{12}{12}=\frac{4x}{12}-\frac{2\left(2-x\right)}{12}\\\Leftrightarrow 3\left(x-5\right)-24x+12=4x-2\left(2-x\right)\\\Leftrightarrow 3x-15-24x+12=4x-4+2x\\ \Leftrightarrow3x-15-24x+12-4x+4-2x=0\\ \Leftrightarrow-27x+1=0\\ \Leftrightarrow-27x=-1\\ \Leftrightarrow x=\frac{1}{27}\)
\(b.\left(2x-1\right)^2=\left(x-2\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)
\(c.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{-3}{25-x^2}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{x^2-25}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=3\\\Leftrightarrow x^2+5x+5x+25-\left(x^2-5x-5x+25\right)=3\\\Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25=3\\ \Leftrightarrow20x=3\\ \Leftrightarrow x=\frac{3}{20}\)
\(d.x^2-x-12=0\\\Leftrightarrow x^2-4x+3x-12=0\\\Leftrightarrow \left(x^2-4x\right)+\left(3x-12\right)=0\\ \Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
1: \(\dfrac{x}{3}-\dfrac{2x+1}{2}=x-6-x\)
=>2x-3(2x+1)=-36
=>2x-6x-3=-36
=>-4x=-33
=>x=33/4
2: \(3x-15=2x\left(x-5\right)\)
=>(x-5)(2x-3)=0
=>x=3/2 hoặc x=5
3: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
=>x(x+3)+(x-2)(x+1)=2x(x+1)
=>x^2+3x+x^2+x-2-2x^2-2x=0
=>2x-2=0
=>x=1
\(\dfrac{3x-6}{2x\left(x-2\right)}=\dfrac{3\left(x-2\right)}{2x\left(x-2\right)}=\dfrac{3}{2x}\)
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