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a: \(A=4x-3x^2+20-15x-9x^2-12x-4+\left(2x+1\right)^3-\left(8x^3-1\right)\)
\(=-12x^2-23x+16+8x^3+12x^2+6x+1-8x^3+1\)
\(=-17x+18\)
a: |x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=-2(nhận) hoặc x=4(loại)
Khi x=-2 thì \(A=\dfrac{4+4}{-2-4}=\dfrac{8}{-6}=\dfrac{-4}{3}\)
b: ĐKXĐ: x<>4; x<>-4
\(B=\dfrac{-\left(x+4\right)}{x-4}+\dfrac{x-4}{x+4}-\dfrac{4x^2}{\left(x-4\right)\left(x+4\right)}\)
\(=\dfrac{-x^2-8x-16+x^2-8x+16-4x^2}{\left(x-4\right)\left(x+4\right)}=\dfrac{-4x^2-16x}{\left(x-4\right)\left(x+4\right)}\)
=-4x/x-4
c: A+B
=-4x/x-4+x^2+4/x-4
=(x-2)^2/(x-4)
A+B>0
=>x-4>0
=>x>4
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: \(A=\left[\left(\dfrac{4x}{x+2}+\dfrac{8x^2}{4-x^2}\right)\right]:\left[\dfrac{x-1}{x^2-2x}-\dfrac{2}{x}\right]\)
\(=\left(\dfrac{4x}{x+2}-\dfrac{8x^2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x-1}{x\left(x-2\right)}-\dfrac{2}{x}\right)\)
\(=\dfrac{4x\left(x-2\right)-8x^2}{\left(x+2\right)\left(x-2\right)}:\dfrac{x-1-2\left(x-2\right)}{x\left(x-2\right)}\)
\(=\dfrac{-8x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-2\right)}{x-1-2x+4}\)
\(=\dfrac{-8x^2}{\left(x+2\right)\cdot\left(-x+3\right)}\)
\(=\dfrac{8x^2}{\left(x-3\right)\left(x+2\right)}\)
b: \(x^2+2x=15\)
=>\(x^2+2x-15=0\)
=>(x+5)(x-3)=0
=>\(\left[{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
Thay x=-5 vào A, ta được:
\(A=\dfrac{8\cdot\left(-5\right)^2}{\left(-5-3\right)\left(-5+2\right)}=\dfrac{8\cdot25}{\left(-8\right)\cdot\left(-3\right)}=\dfrac{25}{3}\)
c: |A|>A
=>A<0
=>\(\dfrac{8x^2}{\left(x-3\right)\left(x+2\right)}< 0\)
=>(x-3)(x+2)<0
TH1: \(\left\{{}\begin{matrix}x-3>0\\x+2< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-3< 0\\x+2>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 3\\x>-2\end{matrix}\right.\)
=>-2<x<3
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}-2< x< 3\\x\notin\left\{0;2\right\}\end{matrix}\right.\)
a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài
