1/x+1/y=1/6
10/x+1/y=1
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MV
10 tháng 7 2017
a)
\(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=255\\ x+x-1+x-2+...+x-50=255\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+50\right)\\ 51x-1275=255\\ 51x=1530\\ x=30\)
e)
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ x+x+1+x+2+...+x+30=1240\\ \left(x+x+x+...+x\right)+\left(1+2+3+...+30\right)=1240\\ 31x+465=1240\\ 31x=775\\ x=25\)
f)
\(\left(x-1\right)+\left(x-2\right)+...+\left(x-19\right)+\left(x-20\right)=-610\\ x-1+x-2+...+x-19+x-20=-610\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+20\right)=-610\\ 20x-210=-610\\ 20x=-400\\ x=-20\)
10 tháng 7 2017
b) Ta có : \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x-\frac{1}{3}\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=-\frac{1}{6}\end{cases}}\)
10 tháng 7 2017
b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{4}\\x-\frac{1}{3}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{1}{12}\end{cases}}\)
d) \(\frac{x+5}{2}=\frac{8}{x+5}\)
\(\Rightarrow\left(x+5\right)^2=16\)
\(\Rightarrow\orbr{\begin{cases}x+5=16\\x+5=-16\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-21\end{cases}}}\)
CC
1
LA
1
21 tháng 3 2015
(x-1)+(x-2)+...+(x-20)=-610
=> (x+x+x+x+...+x)- (1+2+3+4+...+20)=-610
=> 20x-210=-610
20x=-610+210
20x=-400
x=-400:20
x=-20
HT
8
9 tháng 1 2018
( x-1)+(x-2)+....+ (x-19)+(x-20)=610.
=>(x+x+x+...x)+(1+2+3+...+20)=610
=> 20x + 210 = 610
=> 20x =610-210
=> 20x =400
=> x =400:20
=> x = 20
C
6
4 tháng 7 2023
\(\left(x-1\right)+\left(x-2\right)+....+\left(x-20\right)=610\)
\(x-1+x-2+...+x-20=610\)
\(\left(x+x+...+x\right)-\left(1+2+...+20\right)=610\)
Từ 1 đến 20 có 20 số hạng
=> Tổng từ 1 đến 20 là: \(\left(20+1\right)\times20\div2=210\)
=> \(x\times20+210=610\)
\(x\times20=610-210\)
\(x\times20=400\)
\(x=400\div20\)
\(x=20\)
4 tháng 7 2023
\(\left(x-1\right)+\left(x-2\right)+....+\left(x-20\right)=610\)
\(\left(x+x+...+x\right)-\left(1+2+3+...+20\right)=610\)
Ta thấy từ 1 đến 20 có 20 số hạng
=> Tổng từ 1 đến 20 là: \(\left(20+1\right)\times20\div2=210\)
=> \(x\times20-210=610\)
\(x\times20=610+210\)
\(x\times20=820\)
\(x=820\div20\)
\(x=410\)
N
11 tháng 1 2023
\(a.\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}-2=-1\\\dfrac{4}{x}+\dfrac{3}{y}-2=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a-b-2=-1\\4a+3b-2=5\end{matrix}\right.\) (với \(\dfrac{1}{x}=a-\dfrac{1}{y}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{10}{7}\\b=\dfrac{3}{7}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{10}{7}\Rightarrow x=\dfrac{7}{10}\\\dfrac{1}{y}=\dfrac{3}{7}\Rightarrow y=\dfrac{7}{3}\end{matrix}\right.\)
\(b.\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{5}{\left(x+y\right)}=2\\\dfrac{3}{x}+\dfrac{1}{\left(x+y\right)}=\dfrac{17}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2a+5b=2\\3a+b=\dfrac{17}{10}\end{matrix}\right.\) (với \(\dfrac{1}{x}=a-\dfrac{1}{x+y}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{1}{5}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{2}\Rightarrow x=2\\\dfrac{1}{x+y}=\dfrac{1}{5}\Rightarrow y=3\end{matrix}\right.\)
\(c.\left\{{}\begin{matrix}\dfrac{2}{x-1}+\dfrac{1}{y+1}=7\\\dfrac{5}{x-1}-\dfrac{2}{y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+b=7\\5a-2b=4\end{matrix}\right.\) (với \(\dfrac{1}{x-1}=a-\dfrac{1}{y+1}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}=2\Rightarrow x=\dfrac{3}{2}\\\dfrac{1}{y+1}=3\Rightarrow y=-\dfrac{2}{3}\end{matrix}\right.\)
\(d.\left\{{}\begin{matrix}\dfrac{2}{\sqrt{x-1}}-\dfrac{1}{\sqrt{y-1}}=1\\\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{y-1}}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\a+b=2\end{matrix}\right.\) (với \(\dfrac{1}{\sqrt{x-1}}=a-\dfrac{1}{\sqrt{y-1}}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-1}}=1\Rightarrow x=2\\\dfrac{1}{\sqrt{y-1}}=1\Rightarrow y=2\end{matrix}\right.\)
Theo bài ra ta có: \(\frac{1}{\frac{x+1}{y}}=\frac{1}{y.\left(x+1\right)}=\frac{1}{6}\)
\(\Rightarrow\left(x+1\right)y=6\)
Bạn tự kẻ bảng nha
b,\(\frac{10}{\frac{x+1}{y}}=\frac{10}{\left(x+1\right)y}=\frac{10}{10}\)
\(\Rightarrow y.\left(x+1\right)=10\)
Bn kẻ bảng giùm nha