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: \(x=-2;18\)
\(\frac{36}{x+6}+\frac{36}{x-6}=4,5\left(đkxđ:x\ne\pm6\right)\)
\(\Leftrightarrow\frac{36\left(x-6\right)+36\left(x+6\right)}{\left(x-6\right)\left(x+6\right)}=\frac{4,5\left(x-6\right)\left(x+6\right)}{\left(x-6\right)\left(x+6\right)}\)
\(\Leftrightarrow\frac{36\left(x-6+x+6\right)}{\left(x-6\right)\left(x+6\right)}=\frac{4,5\left(x^2-36\right)}{\left(x-6\right)\left(x+6\right)}\)
\(\Rightarrow36\times2x=4,5\left(x^2-36\right)\)
\(\Leftrightarrow72x=4,5x^2-162\)
\(\Leftrightarrow4,5x^2-72x-162=0\)
\(\Leftrightarrow4,5x^2+9x-81x-162=0\)
\(\Leftrightarrow4,5x\left(x+2\right)-81\left(x+2\right)=0\)
\(\Leftrightarrow\left(4,5x-81\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}4,5x-81=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=18\\x=-2\end{matrix}\right.\left(TMĐKXĐ\right)\)
a: \(\frac{52}{17}>\frac{51}{17}=3\)
\(3=\frac{121}{41}>\frac{120}{41}\)
Do đó: \(\frac{52}{17}>\frac{120}{41}\)
b: \(\frac34+\frac14:\left(\frac{7}{12}-\frac16\right)\)
\(=\frac34+\frac14:\left(\frac{7}{12}-\frac{2}{12}\right)\)
\(=\frac34+\frac14:\frac{5}{12}\)
\(=\frac34+\frac14\times\frac{12}{5}=\frac34+\frac35=\frac{15}{20}+\frac{12}{20}=\frac{27}{20}\)
c: \(372,463\cdot998+744,926\)
\(=372,463\cdot998+372,463\cdot2\)
\(=372,463\times\left(998+2\right)=372,463\times1000=372463\)
d: Số số hạng trong dãy số 2;4;6;...;100 là:
\(\left(100-2\right):2+1=98:2+1=49+1=50\) (số)
\(2-4+6-8+10-12+\cdots+98-100+102\)
\(=\left(2-4\right)+\left(6-8\right)+\cdots+\left(98-100\right)+102\)
=(-2)+(-2)+...+(-2)+102
\(=-2\cdot\frac{50}{2}+102=-50+102=52\)
e: (y+112)-113=79
=>y+112-113=79
=>y-1=79
=>y=79+1=80
f: \(\frac34-y=\frac12\)
=>\(y=\frac34-\frac12=\frac14\)
g: \(\left(\frac45-2\times y\right)+\frac16=\frac56\)
=>\(\frac45-2\times y=\frac56-\frac16=\frac46=\frac23\)
=>\(2\times y=\frac45-\frac23=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}\)
=>\(y=\frac{2}{15}:2=\frac{1}{15}\)
h: (y+1)+(y+2)+...+(y+50)=1750
=>50y+(1+2+...+50)=1750
=>\(50y+50\times\frac{51}{2}=1750\)
=>50y+1275=1750
=>50y=1750-1275=475
=>\(y=\frac{475}{50}=9,5\)
\(-23.63+23.21-58.23\\ =23.\left(-63\right)+23.21-58.23\\ =23\left(-63+21-58\right)\\ =23.\left(-100\right)=-2300\)
-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