Ta có: \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right).\left(\frac{x+2003}{x}\right)\)    \(\left(ĐK:x\ne\pm1;x\ne0\right)\)

    \(\Leftrightarrow A=\left(\frac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x^2-4x-1\right)}{\left(x-1\right).\left(x+1\right)}\right).\left(\frac{x+2003}{x}\right)\)

    \(\Leftrightarrow A=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right).\left(x+1\right)}\right).\left(\frac{x+2003}{x}\right)\)

    \(\Leftrightarrow A=\left(\frac{x^2-1}{x^2-1}\right).\left(\frac{x+2003}{x}\right)\)

    \(\Leftrightarrow A=\frac{x+2003}{x}\)

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right)\left(\frac{x+2003}{x}\right)\)

\(=\left(\frac{\left(x+1\right)^2-\left(x-1\right)^2+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\left(\frac{x-2003}{x}\right)\)

\(=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\left(\frac{x-2003}{x}\right)\)

\(=\left(\frac{x^2-1}{\left(x-1\right)\left(x+1\right)}\right)\left(\frac{x-2003}{x}\right)=\frac{x-2003}{x}\)

Các bạn giúp mình câu này nhanh nha 

A B C H E F M N

Theo tính chất đường thẳng song song : 

\(AK=KI=IH\)( gt )

=> AE = EM = MB 

=> AF = FN = NC 

Theo bài ra ta có : \(\frac{MN}{BC}=\frac{AM}{MB}=\frac{2MB}{MB}=2\)cm 

\(\frac{EF}{BC}=\frac{AE}{EB}=\frac{AE}{2AE}=\frac{1}{2}\)cm 

hay \(2EF=BC\)(*) 

Ta có : \(S_{ABC}=\frac{1}{2}AH.BC=90\)( gt ) 

\(\Delta AMN\)có EF là đường trung bình ( AE = EM ; AF = FN ) 

Suy ra : EF // MN và EF = 1/2 MN 

Ta có :  \(S_{MNEF}=\frac{\left(EF+MN\right).IK}{2}\)mà \(IK=\frac{1}{3}AH\)

\(=\frac{\left(EF+MN\right).\frac{AH}{3}}{2}=\frac{\left(EF+2EF\right).\frac{AH}{3}}{2}\)

\(=\frac{EF.AH}{2}\)mà \(2EF=BC\)cmt (*)

\(=\frac{\frac{BC}{2}.AH}{2}=\frac{BC.AH}{4}\)

Vậy \(S_{MNEF}=\frac{180}{4}=45\)cm²

\(x+\left(x-2\right)\left(2x-1\right)=2\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=2-x\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)+\left(x-2\right)=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow x=0;2\)

Vậy tập nghiệm phương trình là S = { 0 ; 2 } 

x + ( x - 2 )( 2x + 1 ) = 2

<=> x + 2x² - 3x - 2 - 2 = 0

<=> 2x² - 2x - 4 = 0

<=> x² - x - 2 = 0

<=> x² - 2x + x - 2 = 0

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

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

<=> x = 2 hoặc x = -1

Vậy phương trình có tập nghiệm S = { 2 ; -1 }

\(3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)

Ta có: \(3x^2-6xy+3y^2-12z^2\)

    \(=3.\left(x^2-2xy+y^2-4z^2\right)\)

    \(=3.\left[\left(x-y\right)^2-4z^2\right]\)

    \(=3.\left(x-y-2z\right).\left(x-y+2z\right)\)

Theo bài ra ta có : 

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

Gớm Tú ơi, làm gì mà Dis nhiều thế :)) Nghiếp khiếp vậy mày:))))

a) \(A=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

\(=2\sqrt{3}:\sqrt{3}-5\sqrt{27}:\sqrt{3}+4\sqrt{12}:\sqrt{3}\)

\(=2\sqrt{3:3}-5\sqrt{27:3}+4\sqrt{12:3}\)

\(=2\sqrt{1}-5\sqrt{9}+4\sqrt{4}=2.1-5.3+4.2=2-15+8=-5\)

\(B=\frac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\frac{\left(2+\sqrt{3}\right).\left(\sqrt{2-\sqrt{3}}\right)^2}{\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}\)

\(=\frac{\left(2+\sqrt{3}\right).\left(2-\sqrt{3}\right)}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}=\frac{4-3}{\sqrt{4-3}}=\frac{1}{\sqrt{1}}=1\)

b) \(ĐKXĐ:x\ge\frac{7}{2}\)

Thay \(A=-5\), \(B=1\)vào biểu thức ta được:

\(1-3\sqrt{2x-7}=-5\)\(\Leftrightarrow3\sqrt{2x-7}=6\)

\(\Leftrightarrow\sqrt{2x-7}=2\)\(\Leftrightarrow2x-7=4\)

\(\Leftrightarrow2x=11\)\(\Leftrightarrow x=\frac{11}{2}\)( thỏa mãn ĐKXĐ )

Vậy \(x=\frac{11}{2}\)

Ta có: \(2x^2+5x+3=0\)

\(\Leftrightarrow\left(2x^2+2x\right)+\left(3x+3\right)=0\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-1\end{cases}}\)

\(2x^2+5x+3=0\)

\(\Leftrightarrow2x^2+2x+3x+3=0\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)=0\Leftrightarrow x=-\frac{3}{2}orx=-1\)

Vậy nghiệm của phương trình là x = -3/2 ; -1