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a) x - 3/97 + x - 2/98 = x - 1/99 + x/100
<=> x + 1/99 + 1 + x + 2/98 + 1 + x + 3/97 + 1 + (x + 4/96 + 1 + x + 5/95 + 1 + x + 10/90 + 1) = 0
<=> x + 100/99 + x + 100/98 + x + 100/97 + (x + 100/96 + x + 100/95 + x + 100/90) = 0
<=> (x + 100)(1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90) = 0
Mà 1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90 khác 0
=> x + 100 = 0
=> x = -100
c) (1/1.2 + 1/2.3 + ... + 1/99.100) - 2x = 1/2
<=> (1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100) - 2x = 1/2
<=> (1 - 1/100) - 2x = 1/2
<=> 99/100 - 2x = 1/2
<=> -2x = 1/2 - 99/100
<=> -2x = -49/100
<=> x = 49/200
=> x = 49/200
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Rightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
x=-5
vì (-5)+5=0
mà 0 chia cho số mấy cũng bằng 0 nên x= -5.
Ta có : 3x = 5y = 8z => \(\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{5}}=\frac{z}{\frac{1}{8}}\)
Đặt \(\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{5}}=\frac{z}{\frac{1}{8}}=k\)
=> \(\hept{\begin{cases}\frac{x}{\frac{1}{3}}=k\\\frac{y}{\frac{1}{5}}=k\\\frac{z}{\frac{1}{8}}=k\end{cases}}\)
=> \(x=\frac{1}{3}k,y=\frac{1}{5}k,z=\frac{1}{8}k\)
=> \(x+y+z=\frac{1}{3}k+\frac{1}{5}k+\frac{1}{8}k\)
=> \(\frac{79}{120}k=158\)
=> \(k=240\)
Do đó : \(x=\frac{1}{3}k=\frac{1}{3}\cdot240=80\)
\(y=\frac{1}{5}k=\frac{1}{5}\cdot240=48\)
\(z=\frac{1}{8}k=\frac{1}{8}\cdot240=30\)
Vậy x = 80,y = 48,z = 30
\(\left|x-1\right|+\left|y+2\right|+\left|z-3\right|=0\)
Ta có: \(\hept{\begin{cases}\left|x-1\right|\ge0\forall x\\\left|y+2\right|\ge0\forall x\\\left|z-3\right|\ge0\forall x\end{cases}\Rightarrow\left|x-1\right|+\left|y+2\right|+\left|z-3\right|\ge0\forall x;y;z}\)
Mà \(\left|x-1\right|+\left|y+2\right|+\left|z-3\right|=0\)
\(\hept{\begin{cases}\left|x-1\right|=0\\\left|y+2\right|=0\\\left|z-3\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\\z=3\end{cases}}\)
Vậy \(x=1;y=-2;z=3\)
Suy ra xA=x+x^2+x^3+...+x^101
xA-A=x^101-1
A(x-1)=x^101-1
A=(x^101-1)/(x-1)
\(A=1+x+x^2+...+x^{99}+x^{100}\Rightarrow x.A=x+x^2+x^3+...+x^{100}+x^{101}\)
\(\Rightarrow x.A-A=\left(x+x^2+x^3+...+x^{100}+x^{101}\right)-\left(1+x+x^2+...+x^{99}+x^{100}\right)\)
\(\Rightarrow\left(x-1\right).A=x^{101}-1\Rightarrow A=\frac{x^{101}-1}{x-1}\) (đpcm)
\(\frac{x}{98}+\frac{x-1}{99}+\frac{x-2}{100}+\frac{1-3}{101}=-4\)
<=> \(\frac{x}{98}+1+\frac{x-1}{99}+1+\frac{x-2}{100}+1+\frac{x-3}{101}+1=0\)
<=> \(\frac{x+98}{98}+\frac{x+98}{99}+\frac{x+98}{100}+\frac{x+98}{101}=0\)
<=> \(\left(x+98\right)\left(\frac{1}{98}+\frac{1}{99}+\frac{1}{100}+\frac{1}{101}\right)=0\)
<=> \(x+98=0\) (do 1/98 + 1/99 + 1/100 + 1/101 khác 0)
<=> \(x=-98\)
Vậy...