Bài 6:
a/ \(A=\left(k-4\right)\left(k^2+4k+16\right)-\left(128+k^3\right)\)
\(=k^3-64-128-k^3\)
\(=-192\)
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b/ \(B=\left(2m+3n\right)\left(4m^2-6mn+9n^2\right)-\left(3m-2n\right)\left(9m^2+6mn+4n^2\right)\)
\(=8m^3+27n^3-27m^3+8n^3\)
\(=35n^3-19m^3\)
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Bài 7:
a/ \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)
\(\Leftrightarrow9x+7=16\)
\(\Leftrightarrow x=1\)
Vậy: x=1
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b/ \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Leftrightarrow x^3+8-x^3+2x=15\)
\(\Leftrightarrow2x+8=15\)
\(\Leftrightarrow x=\dfrac{7}{2}\)
Vậy: \(x=\dfrac{7}{2}\)
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