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\(a,\left(x-2\right)^2-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Vậy:....
\(b,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25^2+9=30\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy :....
\(c,\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)\(\Leftrightarrow x^3+27-x\left(x^2-4\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=15-27=-12\)
\(\Leftrightarrow x=-3\)
vậy : .....
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3x^2-5x+2=0
<=>3x2-3x-2x+2=0
<=>3x.(x-1)-2.(x-1)=0
<=>(x-1)(3x-2)=0
<=>x-1=0 hoặc 3x-2=0
<=>x=1 hoặc 3x=2
<=>x=1 hoặc x=2/3
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\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=x+1-\left(x-9\right)\)
\(\Rightarrow6x^2+27x+4x+18-6x^2-x-12x-2=10\)
\(\Rightarrow18x+16=10\)
\(\Rightarrow18x=-6\)
\(\Rightarrow x=-\frac{6}{18}=-\frac{1}{3}\)
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\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)
\(2x\left(x-2\right)-\left(x-2\right)^2=\left(x-2\right)\left[2x-\left(x-2\right)\right]=\left(x-2\right)\left(2x-x+2\right)=\left(x-2\right)\left(x+2\right)\)
\(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2=\left(2x-5y\right)^2\)
\(x^2+3x-x-3=x\left(x+3\right)-\left(x+3\right)=\left(x-1\right)\left(x+3\right)\)
\(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
\(2y\left(x+2\right)-3x-6=2y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(2y-3\right)\)
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\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Rightarrow3x^2-3^2-3x^2+15x=1\)
\(\Rightarrow3x^2-9-3x^2+15x=1\)
\(\Rightarrow-9+15x=1\)
\(\Rightarrow15x=-8\)
\(\Rightarrow x=\frac{-8}{15}\)
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