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14 tháng 4 2023

chỗ này nè, trừ với trừ là cộng 9 nhe

14 tháng 4 2023

\(B=\dfrac{x^2-x}{x^2-3x}-\dfrac{7x-9}{x^2-9}\)

\(B=\dfrac{x\left(x-1\right)}{x\left(x-3\right)}-\dfrac{7x-9}{x^2-3^2}\)

\(B=\dfrac{x-1}{x-3}-\dfrac{7x-9}{\left(x+3\right)\left(x-3\right)}\)

\(B=\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{7x-9}{\left(x+3\right)\left(x-3\right)}\)

\(B=\dfrac{\left(x-1\right)\left(x+3\right)-7x-9}{\left(x+3\right)\left(x-3\right)}\)

\(B=\dfrac{x^2+3x-x-3-7x+9}{\left(x+3\right)\left(x-3\right)}\)

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

\(B=\dfrac{x\left(x-5\right)+6}{\left(x+3\right)\left(x-3\right)}\)

31 tháng 12 2020

(\(3+\dfrac{x}{3-x}+\dfrac{2x}{3+x}-\dfrac{4x^2-3x-9}{x^2-9}\) ):\(\left(\dfrac{2}{3-x}-\dfrac{x-1}{3x-x^2}\right)\)\(=\left(\dfrac{3x^2-27}{\left(x-3\right)\left(x+3\right)}+\dfrac{-x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4x^2-3x-9}{\left(x-3\right)\left(x+3\right)}\right)\)\(:\left(\dfrac{2x}{x\left(3-x\right)}-\dfrac{x-1}{x\left(3-x\right)}\right)\)

\(=\dfrac{3x^2-27-x^2-3x+2x^2-6x-4x^2+3x+9}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) 

\(=\dfrac{-6x-18}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) \(=\dfrac{-6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) 

\(=\dfrac{6}{3-x}.\dfrac{x\left(x-3\right)}{x+1}\) \(=\dfrac{6x}{x+1}\)

5 tháng 11 2018

\(<=> 9x^2-6x+1+(2x+1)^2+2(3x-1)(2x-1)\)

\(<=> 9x^2-6x+1+4x^2+4x+1+(6x-2)(2x-1)\)

 \(<=> 9x^2-6x+1+4x^2+4x+1+12x^2-6x-4x+2\) 

 \(<=> 25x^2-12x+4\)

5 tháng 11 2018

có bạn nào có thể giúp mình giải câu b và d được không ạ mình cần gấp

a: =12x^4-6x^3+3x+4x^3-2x^2+1

=12x^4-2x^3-2x^2+3x+1

b: =14x^4+28x^2+6x^2+12x

=14x^4+34x^2+12x

`@` `\text {Ans}`

`\downarrow`

\((3x + 1)(4x³ - 2x² + 1)\)

`= 3x(4x^3-2x^2+1) + 4x^3 - 2x^2 + 1`

`= 12x^4 - 6x^3 + 3x + 4x^3 - 2x^2 + 1`

`= 12x^4 + (-6x^3 + 4x^3) - 2x^2 + 3x + 1`

`= 12x^4 - 2x^3 - 2x^2 + 3x + 1`

\((7x² + 3x)(2x + 4)\)

`= 7x^2(2x+4) + 3x(2x+4)`

`= 14x^3 + 28x^2 + 6x^2 + 12x`

`= 14x^3 + (28x^2 + 6x^2)+12x`

`= 14x^3 + 34x^2 + 12x`

`@` `\text {Kaizuu lv uuu}`

24 tháng 6 2021

a)      \(2x\left(5-3x^2\right)-10\left(6+x\right)\)

     \(=10x-6x^3-60-10x\)

     \(=\) \(-6x^3-60\)

24 tháng 6 2021

a) \(2x\left(5-3x^2\right)-10\left(6+x\right)\\ =2x.5-2x.3x^2-10.6-10.x\\ =10x-6x^3-60-10x\)

b) \(3\left(-x+2\right)-6\left(1-x+5x^{20}\right)\\ =-3.x+3.2-6.1+6.x-5.5x^{20}\\ =-3x+6-6+6x-25x^{20}=25x^{20}+3x\)

c) \(7x\left(2-5x^2+\dfrac{1}{2}x^3\right)-14x\left(1-2x^2\right)\\ =7x.2-7x.5x^2+7x.\dfrac{1}{2}x^3-14x.1+14x.2x^2\\ =14x-25x^3+\dfrac{7}{2}x^4-14x+28x^3=3x^2+\dfrac{7}{2}x^4\) 

18 tháng 4 2021

a, ĐK : \(x\ne1;2;3;4;5\)

b, \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)

\(=\dfrac{1}{x}-\dfrac{1}{x-5}=\dfrac{x-5-x}{x\left(x-5\right)}=\dfrac{-5}{x\left(x-5\right)}\)

a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)

b: \(P=\dfrac{1}{\left(x-1\right)\cdot x}+\dfrac{1}{\left(x-2\right)\left(x-1\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}\)

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