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i)\(\left\{{}\begin{matrix}ab=2\\a^3+b^3=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}ab=2\\\left(a+b\right)^3-3ab\left(a+b\right)=9\end{matrix}\right.\)
\(\Rightarrow\left(a+b\right)^3-6\left(a+b\right)-9=0\)
\(\Leftrightarrow\left(a+b\right)^3-3\left(a+b\right)^2+3\left(a+b\right)^2-9\left(a+b\right)+3\left(a+b\right)-9=0\)
\(\Leftrightarrow\left(a+b-3\right)\left[\left(a+b\right)^2+3\left(a+b\right)+3\right]=0\)
\(\Leftrightarrow a+b=3\)( \(\left(a+b\right)^2+3\left(a+b\right)+3>0;\forall a,b\)
ii) \(\left\{{}\begin{matrix}a+b+ab=23\\a^2+b^2=34\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}ab=23-\left(a+b\right)\\\left(a+b\right)^2-2ab=34\end{matrix}\right.\)
\(\Rightarrow\left(a+b\right)^2-2\left[23-\left(a+b\right)\right]=34\)
\(\Leftrightarrow\left(a+b\right)^2+2\left(a+b\right)-80=0\)
\(\Leftrightarrow\left(a+b-8\right)\left(a+b+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=8\\a+b=-10\end{matrix}\right.\)
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\(\left(x^3-8\right):\left(x^2+2x+4\right)\\ =\left(x-2\right)\left(x^2+2x+4\right):\left(x^2+2x+4\right)\\ =x-2\)
bài 2
a)
\(2xy^2-4y\\ =2y\left(xy-2\right)\)
b)
\(x^2-6xy+9y^2\\ =\left(x-3y\right)^2\)
c)
\(x^2+x-y^2+y\\ =\left(x^2-y^2\right)+\left(x+y\right)\\ =\left(x-y\right)\left(x+y\right)+\left(x+y\right)\\ =\left(x+y\right)\left(x-y+1\right)\)
d)
\(x^2+4x+3\\ =x^2+3x+x+3\\ =x\left(x+3\right)+\left(x+3\right)\\ =\left(x+3\right)\left(x+1\right)\)
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Bài 10:
e: \(\Leftrightarrow x\left(x+2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
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Ta có: ΔABC∼ΔDEF
nên DE/AB=EF/BC=DF/AC
=>9/6=EF/10=DF/14
=>EF/10=DF/14=3/2
=>EF=15cm; DF=21cm
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a, \(2x=5\Leftrightarrow x=\dfrac{5}{2}\)
b, \(2x-1=4x-8\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\)
c, \(3x+9-6=2x+4\Leftrightarrow x=1\)
d, \(\left[{}\begin{matrix}2x+1=0\\-3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
e, đk : x khác 0 ; 3
\(2x+8x-24=16\Leftrightarrow10x=40\Leftrightarrow x=4\left(tm\right)\)
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1: Ta có: \(a^2+2ab+b^2-12a-12b+50\)
\(=\left(a+b\right)^2-12\left(a+b\right)+50\)
\(=2^2-12\cdot2+50\)
=54-24
=30