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\(ab\left(a-b\right)-ac\left(a+c\right)+bc\left(2a-b+c\right)\)
\(=ab\left(a-b\right)-ac\left(a+c\right)+bc\left[\left(a-b\right)+\left(a+c\right)\right]\)
\(=ab\left(a-b\right)-ac\left(a+c\right)+bc\left(a-b\right)+bc\left(a+c\right)\)
\(=\left(a-b\right)\left(ab+bc\right)+\left(a+c\right)\left(bc-ac\right)\)
\(=b\left(a-b\right)\left(a+c\right)-c\left(a+c\right)\left(a-b\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+c\right)\)
b: \(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
a: \(\Leftrightarrow2x^4-2x^3+2x^2+3x^3-3x^2+3x-2x^2+2x+2+a-2⋮x^2-x+1\)
=>a=2
a: =>2x^3-4x^2-3x^2+6x+4x-8+a+8 chia hết cho x-2
=>a+8=0
=>a=-8
b: =>2x^3+x^2-x^2-0,5x-0,5x+0,25+m-0,25 chia hết cho 2x+1
=>m-0,25=0
=>m=0,25
\(A=2n^2\left(2n-1\right)-3\left(2n-1\right)+2=\left(2n^2-3\right)\left(2n-1\right)+2\)
Do \(\left(2n^2-3\right)\left(2n-1\right)⋮2n-1\)
\(\Rightarrow2⋮2n-1\)
\(\Rightarrow2n-1=Ư\left(2\right)\)
Mà 2n-1 luôn lẻ \(\Rightarrow2n-1=\left\{-1;1\right\}\)
\(\Rightarrow n=\left\{0;1\right\}\)
2.
\(Q=-\left(x^2+4x+4\right)-\left(y^2-2y+1\right)+7\)
\(Q=-\left(x+2\right)^2-\left(y-1\right)^2+7\le7\)
\(Q_{max}=7\) khi \(\left(x;y\right)=\left(-2;1\right)\)