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\(C=\left(\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\right)\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
ĐKXĐ: \(x\ne1\)
\(C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right)]\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
\(\Leftrightarrow C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\right)]\div[\dfrac{(x-1)\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}-\dfrac{(x^2-2)(x-1)}{(x^2+x+1)\left(x-1\right)}]\)
\(\Rightarrow C=\left[2x^2+1-1\left(x^2+x+1\right)\right]\div\left[\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2\right)\right]\)
\(\Rightarrow C=(2x^2+1-x^2-x-1)\div\left[\left(x-1\right)\left(x^2+x+1-x^2+2\right)\right]\)
\(\Rightarrow C=\left(x^2-x\right)\div\left[\left(x-1\right)\left(x+3\right)\right]\)
a: \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x-3|=2
=>x-3=2 hoặc x-3=-2
=>x=5(nhận) hoặc x=1(loại)
Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)
c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow2x^2-x+1=0\)
hay \(x\in\varnothing\)
f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)
-Vậy \(A_{min}=4\)
a: \(A=\dfrac{x^2+1}{x}+\dfrac{x^3-1}{x^2-x}+\dfrac{x^4-x^3+x-1}{x-x^3}\)
\(=\dfrac{x^2+1}{x}+\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x-1\right)}-\dfrac{x^3\left(x-1\right)+\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+1}{x}+\dfrac{x^2+x+1}{x}-\dfrac{\left(x-1\right)\left(x^3+1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+1+x^2+x+1}{x}-\dfrac{x^2-x+1}{x}\)
\(=\dfrac{2x^2+x+2-x^2+x-1}{x}=\dfrac{x^2+2x+1}{x}=\dfrac{\left(x+1\right)^2}{x}\)
b: \(x^2+x=12\)
=>\(x^2+x-12=0\)
=>(x+4)(x-3)=0
=>\(\left[{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)
Thay x=3 vào A, ta được:
\(A=\dfrac{\left(3+1\right)^2}{3}=\dfrac{16}{3}\)
Khi x=-4 thì \(A=\dfrac{\left(-4+1\right)^2}{-4}=\dfrac{9}{-4}=-\dfrac{9}{4}\)
c: \(A-4=\dfrac{\left(x+1\right)^2}{x}-4\)
\(=\dfrac{\left(x+1\right)^2-4x}{x}\)
\(=\dfrac{x^2+2x+1-4x}{x}=\dfrac{x^2-2x+1}{x}=\dfrac{\left(x-1\right)^2}{x}\)>0 với mọi x>0
=>A>4
Bài 1: ĐKXĐ:`x + 3 ne 0` và `x^2+ x-6 ne 0 ; 2-x ne 0`
`<=> x ne -3 ; (x-2)(x+3) ne 0 ; x ne2`
`<=>x ne -3 ; x ne 2`
b) Với `x ne - 3 ; x ne 2` ta có:
`P= (x+2)/(x+3) - 5/(x^2 +x -6) + 1/(2-x)`
`P = (x+2)/(x+3) - 5/[(x-2)(x+3)] + 1/(2-x)`
`= [(x+2)(x-2)]/[(x-2)(x+3)] - 5/[(x-2)(x+3)] - (x+3)/[(x-2)(x+3)]`
`= (x^2 -4)/[(x-2)(x+3)] - 5/[(x-2)(x+3)] - (x+3)/[(x-2)(x+3)]`
`=(x^2 - 4 - 5 - x-3)/[(x-2)(x+3)]`
`= (x^2 - x-12)/[(x-2)(x+3)]`
`= [(x-4)(x+3)]/[(x-2)(x+3)]`
`= (x-4)/(x-2)`
Vậy `P= (x-4)/(x-2)` với `x ne -3 ; x ne 2`
c) Để `P = -3/4`
`=> (x-4)/(x-2) = -3/4`
`=> 4(x-4) = -3(x-2)`
`<=>4x -16 = -3x + 6`
`<=> 4x + 3x = 6 + 16`
`<=> 7x = 22`
`<=> x= 22/7` (thỏa mãn ĐKXĐ)
Vậy `x = 22/7` thì `P = -3/4`
d) Ta có: `P= (x-4)/(x-2)`
`P= (x-2-2)/(x-2)`
`P= 1 - 2/(x-2)`
Để P nguyên thì `2/(x-2)` nguyên
`=> 2 vdots x-2`
`=> x -2 in Ư(2) ={ 1 ;2 ;-1;-2}`
+) Với `x -2 =1 => x= 3` (thỏa mãn ĐKXĐ)
+) Với `x -2 =2 => x= 4` (thỏa mãn ĐKXĐ)
+) Với `x -2 = -1=> x= 1` (thỏa mãn ĐKXĐ)
+) Với `x -2 = -2 => x= 0`(thỏa mãn ĐKXĐ)
Vậy `x in{ 3 ;4; 1; 0}` thì `P` nguyên
e) Từ `x^2 -9 =0`
`<=> (x-3)(x+3)=0`
`<=> x= 3` hoặc `x= -3`
+) Với `x=3` (thỏa mãn ĐKXĐ) thì:
`P = (3-4)/(3-2)`
`P= -1/1`
`P=-1`
+) Với `x= -3` thì không thỏa mãn ĐKXĐ
Vậy với x= 3 thì `P= -1`
a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))
\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)
\(C=\left[\dfrac{x\left(3x+5\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right]\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)
\(C=\dfrac{3x^2+5x-x^2-2x+x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)
\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)
\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)
\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)
b) Thay x = 2003 ta có:
\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)
c) \(C>0\) khi:
\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)
\(\Leftrightarrow x-3>0\)
\(\Leftrightarrow x>3\) (đpcm)
a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)
b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)
\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}=\dfrac{x+1}{2x}\)
c: Thay x=2017 vào C, ta được:
\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)
1: \(B=\dfrac{2x+1-x^2+2x^2-3x-1}{x\left(2x+1\right)}=\dfrac{x^2-x}{x\left(2x+1\right)}=\dfrac{x-1}{2x+1}\)
2: \(C=A:B\)
\(=\dfrac{x-1}{x^2}:\dfrac{x-1}{2x+1}=\dfrac{2x+1}{x^2}\)
\(C+1=\dfrac{2x+1+x^2}{x^2}=\dfrac{\left(x+1\right)^2}{x^2}>=0\)
=>C>=-1
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