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5 tháng 12 2023

a) A = (x - 5)(x² + 5x + 25) - (x - 2)(x + 2) + x(x² + x + 4)

= x³ - 125 - x² + 4 + x³ + x² + 4x

= (x³ + x³) + (-x² + x²) + 4x + (-125 + 4)

= 2x³ + 4x - 121

b) Tại x = -2 ta có:

A = 2.(-2)³ + 4.(-2) - 121

= 2.(-8) - 8 - 121

= -16 - 129

= -145

c) x² - 1 = 0

x² = 1

x = -1; x = 1

*) Tại x = -1 ta có:

A = 2.(-1)³ + 4.(-1) - 121

= 2.(-1) - 4 - 121

= -2 - 125

= -127

*) Tại x = 1 ta có:

A = 2.1³ + 4.1 - 121

= 2.1 + 4 - 121

= 2 - 117

= -115

18 tháng 12 2023

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) Ta có: \(P=\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\cdot\dfrac{1-x^2}{2}\)

\(=\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\cdot\dfrac{-\left(x-1\right)\left(x+1\right)}{2}\)

\(=\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x+2\right)\left(x-1\right)}{2\left(x+1\right)}\)

\(=\dfrac{2\left(x-2\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2\cdot\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{2x-4-\left(x^2-2x+1\right)\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{2x-4-\left(x^3+2x^2-2x^2-4x+x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{2x-4-\left(x^3-3x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{2x-4-x^3+3x-2}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-x^3+5x-6}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-\left(x^3-5x+6\right)}{2\left(x-1\right)\left(x+1\right)}\)

 

3 tháng 10 2021

1) \(4-25x^2=0\)

\(\Rightarrow\left(2-5x\right)\left(2+5x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x=2\\5x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

2) Tính thì phải cho giá trị của x.

\(A=x^3-3x^2+3x-1\)

\(=\left(x-1\right)^3\)

\(=\left[{}\begin{matrix}\left(\dfrac{5}{2}-1\right)^3=\dfrac{27}{8}\\\left(-\dfrac{5}{2}-1\right)^3=-\dfrac{343}{8}\end{matrix}\right.\)

9 tháng 3 2022

chịu

30 tháng 5 2017

ko biết

30 tháng 5 2017
  1. a/  [x/x^2-4 -2(x+2)/x^2-4 +x-2/x^2-4]:[x^2-4/x+2 +10-x^2/x+2] =(x-2x-4+x-2/x^2-4):(x^2-4+10-x^2/x+2) = - 6/x^2-4 nhân với x+2/x^2-4+10-x^2= - 6/(x+2)(x-2) nhân với x+2/6= - 1/x-2.

c/đễ A<0  <=>  -1/X-2 <0  <=> x-2<0  <=>x<2 

20 tháng 6 2021

a)

A=\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5x-5}\)

\(\Leftrightarrow\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5\left(x-1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+1\\x=0-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

MTC: 5(x-1)(x+1)

\([\dfrac{5\left(x+1\right)\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}-\dfrac{5\left(x-1\right)\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}]\div\dfrac{2x\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow[5\left(x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)]\div2x\left(x+1\right)\)

\(\Leftrightarrow[5\left(x+1\right)^2-5\left(x-1\right)^2]\div2x^2+2x\)

\(\Leftrightarrow[5\left(x^2+2x+1\right)-5\left(x^2-2x+1\right)]\div2x^2+2x\)

\(\Leftrightarrow(5x^2+10x+5-5x^2+10x-5)\div2x^2+2x\)

\(\Leftrightarrow20x\div\left(2x^2+2x\right)\)

\(\Leftrightarrow10x+10\)

18 tháng 10 2018

a) -4x2+2x

b) -4x2+2x=0

x(-4x+2)=0

=> x=0 hoặc -4x+2=0

                     -4x = -2

                        x=1/2(đpcm)

c) Thay x=-1/4 vào -4x2+2x ta có : -4 (-1/4)2 +2(-1/4) = ... (tự tính )

9 tháng 7 2020

a) A = (x - 3)(x + 1) - (2x - 1)^2 - (x + 2)(x - 2)

A = x^2 - 2x - 3 - 4x^2 + 4x - 1 - x^2 + 4

A = -4x^2 + 2x

b) 4x^2 - 2x = 0

<=> 2x(2x - 1) = 0

<=> 2x = 0 hoặc 2x - 1 = 0

<=> x = 0 hoặc x = 1/2

c) với x = -1/4, ta có:

4(-1/4)^2 - 2(-1/4) = 3/4

7 tháng 5 2017

A> A="X-2<X+2>+X-2 / X2-4" / "X2-4+10-X2 / X+2"

A="X-2X-4+X-2 / X2-4" / " -6/X+2"

A=-6/X2-4 / -6/X+2

CÒN CÂU B THÌ CHIA THÀNH 2 TH MÀ TÍNH NHÉ