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Bài 2:
Ta có: \(\frac{x+1}{x}=10\) hay \(\frac{x^1+1^1}{x^1}=10^1\)
Nên suy ra : \(\frac{x^5+1}{x^5}=10^5\)
= 100000 ( do 15 cũng sẽ =1 nên không viết mũ 5 cũng chả sao)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1 :
\(\text{ a) }12-2x-x^2=0\\ \Leftrightarrow2\left(6-x-x^2\right)=0\\ \Leftrightarrow6-x-x^2=0\\ \Leftrightarrow6-3x+2x-x^2=0\\ \Leftrightarrow\left(6-3x\right)+\left(2x-x^2\right)=0\\ \Leftrightarrow3\left(2-x\right)+x\left(2-x\right)=0\\ \Leftrightarrow\left(3+x\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3+x=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy \(x=-3\) hoặc \(x=2\)
\(\text{b) }\left(x^2-\dfrac{1}{2}x\right):2x-\left(3x-1\right):\left(3x-1\right)=0\\ \Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{4}-1=0\\ \Leftrightarrow\dfrac{1}{2}x-\dfrac{5}{4}=0\\ \Leftrightarrow\dfrac{1}{2}x=\dfrac{5}{4}\\ \Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}\)
Câu 2:
\(N=x^2+5y^2+2xy-2y+2005\\ N=x^2+4y^2+y^2+2xy-2y+1+2004\\ N=\left(x^2+2xy+y^2\right)+\left(4y^2-2y+1\right)+2004\\ N=\left(x+y\right)^2+\left(2y-1\right)^2+2004\\ \text{Do }\left(x+y\right)^2\ge0\forall x;y\\ \left(2y-1\right)^2\ge0\forall y\\ \Rightarrow\left(x+y\right)^2+\left(2y-1\right)^2\ge0\forall x;y\\ \Rightarrow N=\left(x+y\right)^2+\left(2y-1\right)^2+2004\ge0\forall x;y\\ \text{Dấu "=" xảy ra khi : }\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(2y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\2y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(N_{\left(Min\right)}=2004\) khi \(x=-\dfrac{1}{2};y=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có:
\(ab+cd=ab.1+cd.1\)
\(=ab\left(c^2+d^2\right)+cd\left(a^2+b^2\right)\)
\(=abc^2+abd^2+cda^2+cdb^2\)
\(=bc\left(ac+bd\right)+ad\left(bd+ac\right)\)
\(=\left(ac+bd\right)\left(bc+ad\right)=0.\left(bc+ad\right)=0\)
Tách nhỏ câu hỏi ra bạn
Bài 1 :
a, đk x khác -1 ; 0
\(x+1-2x=3\Leftrightarrow x=-2\)(tm)
b, đk : x khác 0 ; 3/2
\(x-3=10x-15\Leftrightarrow9x=12\Leftrightarrow x=\dfrac{4}{3}\)(tm)
c, đk x khác 0 ; 2
\(x^2+2x-x+2=2\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow x=0\left(ktm\right);x=-1\)
d, đk : x khác -2 ; 2
\(x^2-4x+4-3x-6=2x-22\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow x=4;x=5\)