Câu 1 :
\(\left(x-2\right)^2-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right)\)
\(=\left(x^2-2\cdot x\cdot2+2^2\right)-x\left(x^2-1^2\right)+\left(6x^2-18x\right)\)
\(=\left(x^2-4x+4\right)-\left(x^2-x\right)+6x^2-18x\)
\(=x^2-4x+4-x^2+x+6x^2-18x\)
\(=\left(x^2-x^2+6x^2\right)-\left(4x-x+18x\right)+4\)
\(=6x^2-21x+4\)
Bài 1 :
\(\left(x-2\right)^2-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right)\)
\(=x^2-4x+2^2-x^3+x+6x^2-18x\)
\(=x^3+7x^2-21x+4\)
Bài 2 :
Câu a : \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-3^3+2x^2-x^3+4x-2x^2=1\)
\(\Leftrightarrow-3^3+4x=1\)
\(\Leftrightarrow4x=28\)
\(\Rightarrow x=7\)
Câu 2 :
\(\text{a) }\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\\ \Leftrightarrow x\left(x^2+3x+9\right)-3\left(x^2+3x+9\right)+x\left(2^2-x^2\right)=1\\ \Leftrightarrow x^3+3x^2+9x-3x^2-9x-27+4x-x^3=1\\ \Leftrightarrow\left(x^3-x^3\right)+\left(3x^2-3x^2\right)+\left(9x-9x+4x\right)-27=1\\ \Leftrightarrow4x-27=1\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\\ \text{Vậy }x=7\)
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