\(A=\dfrac{6x}{5x-20}-\dfrac{x}{x^2-8x+16}\)

\(ĐKXĐ:x\ne\pm4\)

\(\Leftrightarrow A=\dfrac{6x}{5\left(x-4\right)}-\dfrac{x}{\left(x-4\right)^2}\)

\(\Leftrightarrow A=\dfrac{6x^2-24x-5x}{5\left(x-4\right)^2}\)

\(\Leftrightarrow\dfrac{6x^2-29x}{5\left(x-4\right)^2}\)

\(\Leftrightarrow\dfrac{x\left(6x-29\right)}{5\left(x-4\right)^2}\)

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

\(ĐKXĐ:x\ne0;x\ne\pm1\)

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

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

\(\Leftrightarrow A=\dfrac{x+1}{x-1}\)

\(A=\left[\dfrac{6x+1}{x^2-6x}+\dfrac{6x-1}{x^2+6x}\right].\dfrac{x^2-36}{x^2+1}\)

\(ĐKXĐ:x\ne0;x\ne\pm6\)

\(\Leftrightarrow A=\left[\dfrac{6x+1}{x\left(x-6\right)}+\dfrac{6x-1}{x\left(x+6\right)}\right].\dfrac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(\Leftrightarrow A=\left[\dfrac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}\right].\dfrac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(\Leftrightarrow A=\left[\dfrac{6x^2+37x+6+6x^2-37x+6}{x\left(x-6\right)\left(x+6\right)}\right].\dfrac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(\Leftrightarrow A=\dfrac{12\left(x^2+1\right)}{x\left(x-6\right)\left(x+6\right)}.\dfrac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(\Leftrightarrow A=\dfrac{12}{x}\)

đề là mô thế bợn ơi!!!!!!!!!!

a: \(=\dfrac{6x}{5\left(x-4\right)}-\dfrac{x}{\left(x-4\right)^2}\)

\(=\dfrac{6x^2-24x-5x}{5\left(x-4\right)^2}=\dfrac{6x^2-29x}{5\left(x-4\right)^2}\)

b: \(=\dfrac{4}{x+2}+\dfrac{3}{x-2}-\dfrac{5x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2-2x+4}{x^3+8}\)

\(=\dfrac{4x-8+3x+6-5x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{1}{x+2}\)

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

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

c: \(\left(\dfrac{x}{x-1}-\dfrac{x+1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\)

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

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

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

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

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

\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)

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

Lời giải:

a) ĐKXĐ: \(\left\{\begin{matrix} x^2-6x\neq 0\\ x^2+6x\neq 0\\ x^2+1\neq 0\end{matrix}\right.\Leftrightarrow x\neq 0; x\neq \pm 6\)

b)

\(A=\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right).\frac{x^2-36}{x^2+1}=\frac{(6x+1)(x+6)+(6x-1)(x-6)}{x(x-6)(x+6)}.\frac{x^2-36}{x^2+1}\)

\(=\frac{6x^2+37x+6+6x^2-37x+6}{x(x-6)(x+6)}.\frac{(x-6)(x+6)}{x^2+1}=\frac{12(x^2+1)}{x(x-6)(x+6)}.\frac{(x-6)(x+6)}{x^2+1}=\frac{12}{x}\)

`a)5y(2y-1)-(3y+2)(3-3y)`

`=10y^2-5y+(3y+2)(3y-3)`

`=10y^2-5y+9y^2-9y+6y-6`

`=19y^2-8y-6`

`b)(6x+1)^2-2(6x+1)(6x-1)+(6x-1)^2`

`=(6x+1-6x+1)^2`

`=2^2=4`

`c)(2x+3)^2-2(2x+3)(x-20+(x-2)^2`

`=(2x+3-x+2)^2`

`=(x+5)^2`

`=x^2+10x+25`

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

Rút gọn nha các cậu

\(A=\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\times\frac{x^2-36}{12x^2+12}\)

\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\times\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)

\(A=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}\times\frac{1}{12\left(x^2+1\right)}\)

\(A=\frac{12\left(x^2+1\right)}{x}\times\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)

a ) ( 6x + 1 )² + ( 6x - 1 )²  - 2 . ( 6x + 1 )( 6x - 1 )

= ( 6x + 1 )² - 2 . ( 6x + 1 )( 6x - 1 ) + ( 6x - 1 )²

= ( 6x + 1 - 6x + 1 )²

= 2² = 4

b ) x . ( 2x² - 3 ) - x² . ( 5x + 1 ) + x²

= 2x³ - 3x - 5x³ - x² + x²

= ( 2x³ - 5x³ ) - 3x - ( x² - x² )

= - 3x³ - 3x

= - 3x . ( x² + 1)

a) Ta có: \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)

\(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)

\(=\left(6x+1-6x+1\right)^2=2^2=4\)

b) Ta có: \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)

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

\(=-3x-3x^3\)

c) Ta có: \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=24-11x\)

d) Ta có: \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)