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Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
a) 3x=9y-1 => 3x= 32(y-1) => x=2(y-1)=2y-2
8y=2x+8 => 23y=2x+8 => 3y=x+8
3y-(2y-2)=x+8-x
y+2=8 => y=6
x+8=3y=3.6=18 => x=10
a) Ta có:
+) \(3^x=9^{y-1}\)
\(\Rightarrow3^x=3^{2.\left(y-1\right)}\)
\(\Rightarrow x=2.\left(y-1\right)\left(1\right)\)
+) \(8^y=2^{x+8}\)
\(\Rightarrow2^{3y}=2^{x+8}\)
\(\Rightarrow3y=x+8\left(2\right)\)
Thay (1) vào (2) ta được:
\(3y=\left[2.\left(y-1\right)\right]+8\)
\(\Rightarrow3y=2y-2+8\)
\(\Rightarrow3y=2y+6\)
\(\Rightarrow y=6\)
\(\Rightarrow x=2.\left(6-1\right)=10\)
Vậy \(x=10;y=6\)
Có: \(x+y+9=xy-7\)
\(\Leftrightarrow x+16=y\left(x-1\right)\)
\(\Leftrightarrow\frac{x+16}{x-1}=y\)
\(\Leftrightarrow y=1+\frac{17}{x-1}\in Z\Leftrightarrow x-1\inƯ\left(17\right)\)
Bn giải x ra rồi tính y
b) \(x^3y=xy^3+1997\)
\(\Leftrightarrow xy\left(x-y\right)\left(x+y\right)=1997\)
Phân tích 1997=1*1997 và ngược lại chia TH giải
a) => (2x -1)6 - (2x - 1)8 = 0 => (2x - 1)6 - (2x - 1)6.(2x - 1)2 = 0
=> (2x - 1)6.[1 - (2x - 1)2] = 0 => (2x - 1)6 = 0 hoặc 1 - (2x - 1)2 = 0
+) (2x - 1)6 = 0 => 2x - 1 = 0 => x = 1/2
+) 1 - (2x - 1)2 = 0 => (2x - 1)2 = 1 => 2x - 1 = 1 hoặc 2x - 1 = - 1
2x - 1 = 1 => x = 1
2x - 1 = - 1 => x = 0
Vậy x = 0 ; x = 1/2; x = 1
b) => 5x + 5.5x = 650
=> 6.5x = 650 => 5x = 650 : 6 = 325/3 => không có số x nào thỏa mãn
Vậy............
Vi 8x = 5y , 7y = 12z
=>\(\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{8}\\\dfrac{y}{12}=\dfrac{z}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{60}=\dfrac{y}{96}\\\dfrac{y}{96}=\dfrac{z}{56}\end{matrix}\right.\)
=> \(\dfrac{x}{60}=\dfrac{y}{96}=\dfrac{z}{56}\)
Ap dung tinh chat day ti so bang nhau co
\(\dfrac{x}{60}=\dfrac{y}{96}=\dfrac{z}{56}=\dfrac{x+y+z}{60+96+56}=\dfrac{-318}{212}=\dfrac{-3}{2}\)
\(\dfrac{x}{60}=\dfrac{-3}{2}\Rightarrow x=60.\dfrac{-3}{2}=-90\)
\(\dfrac{y}{96}=\dfrac{-3}{2}\Rightarrow y=96.\dfrac{-3}{2}=-144\)
\(\dfrac{z}{56}=\dfrac{-3}{2}\Rightarrow z=56.\dfrac{-3}{2}=-84\)
Vay x= -90, y= -144 va z=-84
c: =>|x-2009|=2009-x
=>x-2009<=0
=>x<=2009
d: =>2x-1=0 và y-2/5=0 và x+y-z=0
=>x=1/2 và y=2/5 và z=x+y=1/2+2/5=9/10
a: 8x=5y; 7y=12z
=>x/5=y/8; y/12=z/7
=>x/15=y/24=z/14
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{15}=\dfrac{y}{24}=\dfrac{z}{14}=\dfrac{x+y+z}{15+24+14}=-\dfrac{318}{53}=-6\)
=>x=-90; y=-144; z=-84
\(xy+5x+y=8\)
\(\Rightarrow xy+5x+y+5=13\)
\(\Rightarrow x\left(y+5\right)+1\left(y+5\right)=13\)
\(\Rightarrow\left(x+1\right)\left(y+5\right)=13\)
\(\Rightarrow x+1;y+5\inƯ\left(13\right)\)
\(Ư\left(13\right)=\left\{\pm1;\pm13\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=1\Rightarrow x=0\\y+5=13\Rightarrow y=8\\x+1=-1\Rightarrow x=-2\\y+5=-13\Rightarrow y=-18\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=13\Rightarrow x=12\\y+5=1\Rightarrow y=-4\\x+1=-13\Rightarrow x=-14\\y+5=-1\Rightarrow y=6\end{matrix}\right.\)