Này nhỉ? 

\(R\backslash\left(-1;1\right)=\)\((-\infty;-1]\cup[1;+\infty)\)

A với D là gì ấy? Gõ muốn lủng cái tay 

Đk: \(x\ne1;x\ne0\)

a) \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right]\)

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

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

\(=\dfrac{x^2}{x-1}\)

b) \(E>1\Leftrightarrow\dfrac{x^2}{x-1}>1\) \(\Leftrightarrow\dfrac{x^2-x+1}{x-1}>0\) \(\Leftrightarrow x-1>0\) 

( do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\) )

\(\Leftrightarrow x>1\)

Vậy để E>1 thì x>1

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

\(E\in Z\Leftrightarrow x+1+\dfrac{1}{x-1}\in Z\) mà \(x\in Z\)

\(\Rightarrow x-1\inƯ\left(1\right)=\left\{-1;1\right\}\)

\(\Leftrightarrow x=0\left(ktm\right);x=2\left(tm\right)\)

Vậy \(x=2\) thì \(E\in Z\).

\(\log_2x+\log_3x+\log_4x=\log_{20}x\) (x>0)

\(\Leftrightarrow\log_220.\log_{20}x+\log_320.\log_{20}x+\log_420.\log_{20}x=\log_{20}x\)

\(\Leftrightarrow\log_{20}x\left(\log_220+\log_320+\log_420-1\right)=0\)

\(\Leftrightarrow\log_{20}x=0\)

\(\Leftrightarrow x=1\)

Vậy x=1

a) \(=x^3-8\)

b) \(=\left(2x\right)^3+1=8x^3+1\)

c) \(=1-\left(\dfrac{x}{2}\right)^3=1-\dfrac{x^3}{8}\)

d) \(=y^3-\left(\dfrac{x}{y}\right)^3\)

a) \(M=x^3+27\)

b) \(N=1-\left(3x\right)^3=1-27x^3\)

c) \(P=x^3-\left(\dfrac{1}{2}\right)^3=x^3-\dfrac{1}{8}\)

d) \(Q=\left(2x\right)^3+\left(3y\right)^3=8x^3+27y^3\)

Giả thiết: Một đường thằng cắt hai đường thẳng song song

Kết luận: Hai góc đồng vị bằng nhau, hai góc so le trong bằng nhau

\(2004^2-4^2=2004^2-4.2004+4.2004-4^2\)

\(=2004\left(2004-4\right)+4\left(2004-4\right)=\left(2004+4\right)\left(2004-4\right)=2008.2000=4016000\)

\(x^2-\left(\sqrt{3}y\right)^2=x^2-\sqrt{3}xy+\sqrt{3}xy-\left(\sqrt{3}y\right)^2\)

\(=x\left(x-\sqrt{3}y\right)+\sqrt{3}y\left(x-\sqrt{3}y\right)\)

\(=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\)

Sai mô kkk?

Ý A

Có \(2-\left(-5\right)+1=8\) số nguyên