(2x-1)2-(x-3)=0
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\(12x+15x^2y=3x.4+3x.5xy=3x\left(4+5xy\right)\)
\(\text{Vậy } 12x + 15x^2y=3x(4+5xy)\)
a/
\(a+b+c=2\)
\(\Rightarrow\left(a+b+c\right)^2=4\)
\(a^2+b^2+c^2+2ab+2bc+2ca=4\)
\(2+2\left(ab+bc+ca\right)=4\)
\(\Rightarrow ab+bc+ca=1\left(đpcm\right)\)
b/
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ca+ab}{abc}=\frac{1}{abc}\left(đpcm\right)\)
+) \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(=\frac{c+a+b}{abc}=\frac{0}{abc}=0\)
+) \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(=\frac{\left(bc+ac+ab\right)^2}{\left(abc\right)^2}=\frac{b^2c^2+2bc\left(ac+ab\right)+\left(ac+ab\right)^2}{a^2b^2c^2}\)
\(=\frac{b^2c^2+2abc^2+2ab^2c+a^2c^2+2a^2bc+a^2b^2}{a^2b^2c^2}\)
\(=\frac{b^2c^2+a^2c^2+a^2b^2+2abc\left(a+b+c\right)}{a^2b^2c^2}\)
\(=\frac{b^2c^2+a^2c^2+a^2b^2}{a^2b^2c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(N=\frac{\left(4x^2y-8xy^2\right)}{\left(2xy-6y^35xy^2\left(xy^2\right)\right)}\)
\(N=7x-10y\)
HT