a) \(\left(x+4\right)^2-x\left(x+10\right)=0\)
\(\Rightarrow x^2+8x+16-x^2-10x=0\\ \Rightarrow-2x=-16\\ \Rightarrow x=8\)
b) \(\left(x-5\right)^2-\left(x+4\right)\left(x-6\right)=0\)
\(\Rightarrow x^2-10x+25-x^2+6x-4x+24=0\\ \Rightarrow-8x=-49\\ \Rightarrow x=\dfrac{49}{8}\)
c) \(4x\left(x-2\right)=4\left(x-3\right)\left(x+3\right)\)
\(\Rightarrow4x^2-8x=4x^2-36\\ \Rightarrow-8x=-36\\ \Rightarrow x=\dfrac{9}{2}\)
d) \(x\left(x^2+3x-2\right)=x^3+100+3x^2\)
\(\Rightarrow x^3+3x^2-2x=x^3+100+3x^2\\ \Rightarrow-2x=100\\ \Rightarrow x=-50\)
e) \(6x\left(x-1\right)=\left(2x+1\right)\left(3x-2\right)\)
\(\Rightarrow6x^2-6x=6x^2-x-2\\ \Rightarrow5x=2\\ \Rightarrow x=\dfrac{2}{5}\)
f) \(\left(x+1\right)^3-\left(x^3+3x^2+4\right)=0\)
\(\Rightarrow x^3+3x^2+3x+1-x^3-3x^2-4=0\\ \Rightarrow3x=3\\ \Rightarrow x=1\)
a) \(\left(x+4\right)^2-x\left(x-10\right)=0\)
\(x^2+8x+16-x^2+10x=0\)
\(18x+16=0\)
\(18x=-16\)
Vậy \(x=\dfrac{-16}{18}=\dfrac{-8}{9}\)
_______________________
b) \(\left(x-5\right)^2-\left(x+4\right)\left(x+6\right)=0\)
\(x^2-10x+25-\left(x^2+6x+4x+24\right)=0\)
\(x^2-10x+25-x^2-6x-4x-24=0\)
\(-20x+1=0\)
\(-20x=-1\)
Vậy \(x=\dfrac{1}{20}\)
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c) \(4x\left(x-2\right)=4\left(x-3\right)\left(x+3\right)\)
\(4x^2-8x=4\left(x^2-9\right)\)
\(4x^2-8x=4x^2-36\)
\(4x^2-4x^2-8x=-36\)
\(-8x=-36\)
Vậy \(x=\dfrac{36}{8}=\dfrac{9}{2}\)
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d) \(x\left(x^2+3x-2\right)=x^3+100+3x^2\)
\(x^3+3x^2-2x=x^3+100+3x^2\)
\(x^3-x^3+3x^2-3x^2-2x=100\)
\(-2x=100\)
Vậy \(x=-50\)
_______________________________
e) \(6x\left(x-1\right)=\left(2x+1\right)\left(3x-2\right)\)
\(6x^2-6x=6x^2-4x+3x-2\)
\(6x^2-6x^2-6x+4x+3x=-2\)
Vậy \(x=-2\)
______________________________
f) \(\left(x+1\right)^3-\left(x^3+3x^2+4\right)=0\)
\(x^3+3x^2+3x+1-x^3-3x^2-4=0\)
\(3x-3=0\)
\(3x=3\)
Vậy \(x=1\)
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