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c: Ta có: \(\left(2x-3\right)^2-\left(2x-3\right)\left(x-10\right)=7\)
\(\Leftrightarrow4x^2-12x+9-2x^2+20x+3x-30=7\)
\(\Leftrightarrow11x=28\)
hay \(x=\dfrac{28}{11}\)
d: Ta có: \(\left(3x-4\right)^2-9\left(x-3\right)\left(x+3\right)=8\)
\(\Leftrightarrow9x^2-24x+16-9x^2+81=8\)
\(\Leftrightarrow-24x=-89\)
hay \(x=\dfrac{89}{24}\)
f: Ta có: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x=-1\)
hay \(x=-\dfrac{1}{8}\)
\(a,\Leftrightarrow x^3-4x^2+4x=0\\ \Leftrightarrow x\left(x^2-4x+4\right)=0\\ \Leftrightarrow x\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ b,\Leftrightarrow4\left(x-1\right)=3x+6\left(2x-3\right)\\ \Leftrightarrow4x-4=3x+12x-18\\ \Leftrightarrow11x=14\Leftrightarrow x=\dfrac{14}{11}\)
a/ \(x^3-4x^2=-4x\)
\(\Leftrightarrow x^3-4x^2+4x=0\)
\(\Leftrightarrow x\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b/ \(\dfrac{x-1}{3}=\dfrac{x}{4}+\dfrac{2x-3}{2}\)
\(\Leftrightarrow8\left(x-1\right)=6x+12\left(2x-3\right)\)
\(\Leftrightarrow8x-8=6x+24x-36\)
\(\Leftrightarrow8x-8=30x-36\)
\(\Leftrightarrow8x-30x=8-36\)
\(\Leftrightarrow-22x=-28\)
\(\Leftrightarrow x=\dfrac{14}{11}\)
\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)
1:
a: =>(|x|+4)(|x|-1)=0
=>|x|-1=0
=>x=1; x=-1
b: =>x^2-4>=0
=>x>=2 hoặc x<=-2
d: =>|2x+5|=2x-5
=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0
=>x=0(loại)
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
