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16 tháng 7 2018

\(\left(x+5\right)\left(x^2-5x+25\right)=x^3+125\)

2 tháng 12 2015

phân tích lần ra , rồi rút gọn

2 tháng 12 2015

\(\left(\frac{3x-5}{x^2-5x}-\frac{x+5}{5x-25}\right):\frac{x^2-25}{x}\)

\(=\left[\frac{3x-5}{x\left(x-5\right)}-\frac{x+5}{5\left(x-5\right)}\right].\frac{x}{x^2-25}\)

\(=\left[\frac{\left(3x-5\right).5}{x\left(x-5\right).5}-\frac{\left(x+5\right).x}{5\left(x-5\right).x}\right].\frac{x}{x^2-25}\)

\(=\left[\frac{15x-25}{5x\left(x-5\right)}-\frac{x^2+5x}{5x\left(x-5\right)}\right].\frac{x}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{15x-25-x^2-5x}{5x\left(x-5\right)}.\frac{x}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{-x^2+10x-25}{5x\left(x-5\right)}.\frac{x}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{-\left(x-5\right)^2.x}{5x\left(x-5\right)\left(x-5\right)\left(x+5\right)}\)

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

12 tháng 12 2020

a) Ta có: \(B=\dfrac{x^2}{5x+25}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)

\(=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)

\(=\dfrac{x^3}{5x\left(x+5\right)}+\dfrac{10\left(x+5\right)^2}{5x\left(x+5\right)}+\dfrac{250+25x}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10x^2+100x+250+250+25x}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10x^2+125x+500}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+5x^2+5x^2+25x+100x+500}{5x\left(x+5\right)}\)

\(=\dfrac{x^2\left(x+5\right)+5x\left(x+5\right)+100\left(x+5\right)}{5x\left(x+5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x^2+5x+100\right)}{5x\left(x+5\right)}\)

\(=\dfrac{x^2+5x+100}{5x}\)

b) Thay x=-2 vào biểu thức \(B=\dfrac{x^2+5x+100}{5x}\), ta được:

\(B=\dfrac{\left(-2\right)^2+5\cdot\left(-2\right)+100}{-5\cdot2}=\dfrac{4+100-10}{-10}=\dfrac{94}{-10}=-\dfrac{94}{10}=\dfrac{-47}{5}\)

Vậy: Khi x=-2 thì \(B=-\dfrac{47}{5}\)

19 tháng 12 2021

b: \(=\dfrac{x^3+6x^2-25}{x\left(x+5\right)\left(x-2\right)}+\dfrac{x+5}{x\left(x-2\right)}\)

\(=\dfrac{x^3+6x^2-25+x^2+10x+25}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x^3+7x^2+10x}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x+2}{x-2}\)

7 tháng 7 2019

Đề bài là: \(\frac{3\text{x}+5}{x^2-5\text{x}+25}-\frac{x}{25-5\text{x}}\)

hay: \(\frac{3\text{x}+5}{\frac{x^2-5\text{x}+25-x}{25-5\text{x}}}\)

thế bạn? lolang

8 tháng 7 2019

\(\frac{3x+5}{x^2-5x}+\frac{25-x}{25-5x}\)

8 tháng 6 2018

\(P=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)

\(=\left[\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right]:\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)

\(=\left[\frac{x^2}{x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\right]:\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)

\(=\frac{x^2-\left(x^2-10x+25\right)}{x\left(x-5\right)\left(x+5\right)}:\frac{10x-25}{x\left(x+5\right)}+\frac{x}{5-x}\)

\(=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{10x-25}+\frac{x}{5-x}\)

\(=\frac{1}{x-5}-\frac{x}{x-5}\)

\(=\frac{1-x}{x-5}=-\frac{x-1}{x-5}=-\frac{x-5+4}{x-5}=-1-\frac{4}{x-5}\)

Để P nguyên <=> x - 5 thuộc Ư(4) = {1;-1;2;-2;4;-4}

Ta có bảng:

x - 51-12-24-4
x647391

Vậy....

8 tháng 6 2018

\(ĐKXĐ:x\ne0;x\ne\pm5;x\ne\frac{5}{2}\)

8 tháng 12 2021

\(a,=\dfrac{15x+25-25x+x^2}{5x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\\ b,=\dfrac{x^2-x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2+x-6}{x^2+x-6}=1\)

8 tháng 12 2021

\(a,\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)

\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{25-x}{5\left(5-x\right)}\)

\(=\dfrac{-3x-5}{x\left(5-x\right)}+\dfrac{25-x}{5\left(5-x\right)}\)

\(=\dfrac{5\left(-3x-5\right)}{5x\left(5-x\right)}+\dfrac{x\left(25-x\right)}{5x\left(5-x\right)}\)

\(=\dfrac{-15x-25+25x-x^2}{5x\left(5-x\right)}\)

\(=\dfrac{10x-25-x^2}{5x\left(5-x\right)}\)

\(=\dfrac{-\left(5-x\right)^2}{5x\left(5-x\right)}\)

\(=\dfrac{-5+x}{5x}\)

\(b,\dfrac{x+1}{x+3}+\dfrac{x-7}{x^2+x-6}+\dfrac{1}{x-2}\)

\(=\dfrac{x+1}{x+3}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{1}{x-2}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)

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

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

\(=\dfrac{x^2+x-6}{x^2-2x+3x-6}\)

\(=\dfrac{x^2+x-6}{x^2+x-6}\)

\(=1\)

Đề có sai không bạn?

14 tháng 2 2020

=\(\frac{3x+5}{-x.\left(-x+5\right)}\)+\(\frac{25-x}{-5x+25}\)

=\(\frac{1x-25-x^2}{5x.\left(-\left(x-5\right)\right)}\)

=\(\frac{-\left(x^2-10x+25\right)}{5x.\left(-\left(x-5\right)\right)}\)

=\(\frac{x-5}{5x}\)