chỉ mik vs
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
\(=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
=> \(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)
Khi đó \(\hept{\begin{cases}\frac{y+z+1}{x}=2\\\frac{x+z+2}{y}=2\\x+y+z=\frac{1}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+2=3y\\x+y+z=\frac{1}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{1}{2}+1=3x\\\frac{1}{2}+2=3y\\x+y+z=\frac{1}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=-\frac{5}{6}\end{cases}}\)
Theo tính chất dãy tỉ số bằng nhau
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)
\(=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}-y-z;y=\frac{1}{2}-x-z;z=\frac{1}{2}-y-x\)
*, \(\frac{y+z+1}{x}=2\Rightarrow2x=y+z+1\)
\(\Leftrightarrow2x=\frac{1}{2}-x-z+z+1\Leftrightarrow2x=\frac{1}{2}-x+1\Leftrightarrow3x=\frac{3}{2}\Leftrightarrow x=\frac{1}{2}\)
*, \(\frac{x+z+2}{y}=2\Rightarrow x+z+2=2y\)
\(\Leftrightarrow2y=\frac{1}{2}-y-z+z+2\Leftrightarrow2y=\frac{1}{2}-y+2\Leftrightarrow3y=\frac{5}{2}\Leftrightarrow y=\frac{5}{6}\)
*, \(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\Leftrightarrow\frac{1}{2}+\frac{5}{6}+z=\frac{1}{2}\Leftrightarrow z=-\frac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a). x^4 = x^7
=> x = \(\orbr{\begin{cases}0\\1\end{cases}}\)
b) (2y+1)^6 = (2y+1)^8
=> \(\orbr{\begin{cases}2y+1=0\\2y+1=1\end{cases}}\)
=> \(\orbr{\begin{cases}y=-\frac{1}{2}\\y=0\end{cases}}\)
`a,`
`x^4=x^7`
`->x^4-x^7=0`
`->x^4 (1 - x^3)=0`
TH1 : `x^4=0 ->x=0`
TH2 : `1-x^3=0 ->x^3=1^3 ->x=1`
Vậy `x=0,x=1`
`b,`
`(2y+1)^6 = (2y+1)^8`
`-> (2y+1)^6 - (2y+1)^8=0`
`-> (2y+1)^6 [1-(2y+1)^2]=0`
TH1 : `(2y+1)^6 =0 ->2y+1=0 ->y=(-1)/2`
TH2 : `1 - (2y+1)^2=0`
`-> (2y+1)^2=1`
`-> 2y+1=1` hoặc `2y+1=-1`
`->y=0` hoặc `y=-1`
Vậy `y=0,y=-1,y=(-1)/2`
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời câu hỏi chính dùm
Vào câu hỏi của ng khác xong hỏi là zô zuyên nhé
Ht
____ Vanilla ___
\(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{1}{2008}\)
\(=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)\)
\(=\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\)
Khi đó \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}{2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)}=\frac{1}{2009}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải :
\(\left|-3,45\right|+\frac{1,5}{3}\)
\(=3,45+0,5\)
\(=3,95\)
=-1x1x-1x1x-1x1x-1x1x-1x1=-1
\(=\left(-1\right).1.\left(-1\right).1.\left(-1\right).1.\left(-1\right).1.\left(-1\right).1=-1\)