Tìm y:
6-8+10-12+...-y=-200
Thanks m.n :m.n giải thích hộ mik nhé cảm ưn
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Ta có: \(\frac{36}{x+6}+\frac{36}{x-6}=\frac{9}{2}\)
ĐKXĐ: \(\left\{\begin{matrix}x\ne-6\\x\ne6\end{matrix}\right.\)
\(\Leftrightarrow36\left(\frac{1}{x+6}+\frac{1}{x-6}\right)=\frac{9}{2}\)
\(\Leftrightarrow36\left(\frac{x-6+x+6}{\left(x+6\right)\left(x-6\right)}\right)=\frac{9}{2}\)
\(\Leftrightarrow36.\frac{2x}{x^2-36}=\frac{9}{2}\)
\(\Leftrightarrow\frac{72x}{x^2-36}=\frac{9}{2}\)
\(\Leftrightarrow72x.2=9.\left(x^2-36\right)\)
\(\Leftrightarrow8x.2=x^2-36\) ( chia cả hai vế cho 9 )
\(\Leftrightarrow16x=x^2-36\)
\(\Leftrightarrow16x-x^2+36=0\)
\(\Leftrightarrow-x^2+16x+36=0\)
\(\Leftrightarrow x^2-16x-36=0\)
\(\Leftrightarrow x^2+2x-18x-36=0\)
\(\Leftrightarrow x\left(x+2\right)-18\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-18\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+2=0\\x-18=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-2\\x=18\end{matrix}\right.\)
Vậy: \(x=-2;18\)
\(\frac{36}{x+6}+\frac{36}{x-6}=4,5\left(đkxđ:x\ne\pm6\right)\)
\(\Leftrightarrow\frac{36\left(x-6\right)+36\left(x+6\right)}{\left(x-6\right)\left(x+6\right)}=\frac{4,5\left(x-6\right)\left(x+6\right)}{\left(x-6\right)\left(x+6\right)}\)
\(\Leftrightarrow\frac{36\left(x-6+x+6\right)}{\left(x-6\right)\left(x+6\right)}=\frac{4,5\left(x^2-36\right)}{\left(x-6\right)\left(x+6\right)}\)
\(\Rightarrow36\times2x=4,5\left(x^2-36\right)\)
\(\Leftrightarrow72x=4,5x^2-162\)
\(\Leftrightarrow4,5x^2-72x-162=0\)
\(\Leftrightarrow4,5x^2+9x-81x-162=0\)
\(\Leftrightarrow4,5x\left(x+2\right)-81\left(x+2\right)=0\)
\(\Leftrightarrow\left(4,5x-81\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}4,5x-81=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=18\\x=-2\end{matrix}\right.\left(TMĐKXĐ\right)\)
\(-23.63+23.21-58.23\\ =23.\left(-63\right)+23.21-58.23\\ =23\left(-63+21-58\right)\\ =23.\left(-100\right)=-2300\)
-23 . 63 + 23 . 21 - 58 . 23
= 23 . (-63 + 21 - 58) = 23 . (-100) = -2300
Ta thấy rằng từ 6 đến y có:\(\frac{y-6}{2}+1=\frac{y-4}{2}\)số
Nên sẽ có: \(\frac{y-4}{4}\)cặp
Theo đề bài ta có:
\(6-8+10-12+...-y=-200\)
\(\Rightarrow\left(6-8\right)+\left(10-12\right)+...+\left(y-2-y\right)=-200\)
\(\Rightarrow-2-2-...-2=-200\) (có \(\frac{y-4}{4}\)số -2)
\(\Rightarrow-2.\frac{y-4}{4}=-200\)
\(\Rightarrow y-4=400\)
\(\Rightarrow y=404\)
bạn giải thích vì sao làm dc như vậy đi.cảm ơn bạn