1) Tìm x biết: a) a2x+ 4x = 3a4-48 ( a là hằng số) b) a2x + 5ax + 25 = a2
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\(\text{Ta có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0.\)
\(\Leftrightarrow bc+ac+ab=0\Rightarrow\hept{\begin{cases}bc=-ac-ab\\ac=-bc-ab\\ab=-bc-ac\end{cases}}\)
\(\Rightarrow BT\text{hức}=\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}\)
\(=\frac{bc}{a^2-ac-ab+bc}+\frac{ac}{b^2-bc-ab+ac}+\frac{ab}{c^2-bc-ac+ab}\)
\(=\frac{bc}{a\left(a-b\right)-c\left(a-b\right)}+\frac{ac}{b\left(b-a\right)-c\left(b-a\right)}+\frac{ab}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\frac{bc}{\left(a-c\right)\left(a-b\right)}-\frac{ac}{\left(b-c\right)\left(a-b\right)}+\frac{ab}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc\left(b-c\right)-ac\left(a-c\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{b^2c-bc^2-a^2c+ac^2+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{c\left(b^2-a^2\right)-c^2\left(b-a\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+ab\left(a+b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left(c^2-ac-bc+ab\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{c\left(c-b\right)-a\left(c-b\right)}{\left(b-c\right)\left(a-c\right)}=\frac{\left(a-c\right)\left(b-c\right)}{....}=1\)
Lâu ko lm đổi dấu hơi thừa ra!! ko hiểu chỗ nào thì ib mk giải thích cho
\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right).\)
\(\Rightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+4\left(x^2-5x-2012\right)^2=0\)
\(\Leftrightarrow\left[\left(2x^2+x-2013\right)-2\left(x^2-5x-2012\right)\right]^2=0\)(Hằng đẳng thức)
\(\Leftrightarrow2x^2+x-2013-2x^2+10x+4024=0\)
\(\Leftrightarrow11x=-2011\)
\(\Leftrightarrow x=\frac{-2011}{11}\)
\(\left(x^5+x^3+x^2+1\right):\left(x^3+1\right)\)
\(=\left[x^2\left(x^3+1\right)+1\left(x^3+1\right)\right]:\left(x^3+1\right)\)
\(=\left(x^3+1\right)\left(x^2+1\right):\left(x^3+1\right)\)
\(=x^2+1\)