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NQ
1
13 tháng 12 2020
Ta có: \(\dfrac{-4x^2}{x^2-25}-\dfrac{2x^2+x}{x^2-25}-\dfrac{2x}{5-x}\)
\(=\dfrac{-4x^2-2x^2-x}{\left(x-5\right)\left(x+5\right)}+\dfrac{2x}{x-5}\)
\(=\dfrac{-6x^2-x}{\left(x-5\right)\left(x+5\right)}+\dfrac{2x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-6x^2-x+2x^2+10x}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4x^2+9x}{\left(x-5\right)\left(x+5\right)}\)
28 tháng 7 2021
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
28 tháng 7 2021
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
6 tháng 1 2021
a) Ta có: \(\dfrac{x}{x-1}-\dfrac{2}{x-1}\)
\(=\dfrac{x-2}{x-1}\)
b) Ta có: \(\dfrac{4+4x}{3x^2+6x}+\dfrac{x}{3x+6}\)
\(=\dfrac{4+4x}{x\left(3x+6\right)}+\dfrac{x^2}{x\left(3x+6\right)}\)
\(=\dfrac{x^2+4x+4}{3x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{3x\left(x+2\right)}\)
\(=\dfrac{x+2}{3x}\)
c) Ta có: \(\dfrac{x^2-2x}{x-1}\cdot\dfrac{1}{x}:\dfrac{x^2-4}{x^2-2x+1}\)
\(=\dfrac{x\left(x-2\right)}{x-1}\cdot\dfrac{1}{x}\cdot\dfrac{x^2-2x+1}{x^2-4}\)
\(=\dfrac{x-2}{x-1}\cdot\dfrac{\left(x-1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-1}{x+2}\)
13 tháng 6 2022
1: \(B=\left(\dfrac{4x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{16}{x+2}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\left(\dfrac{4x}{x+2}-\dfrac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\dfrac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)^2\cdot\left(x+1\right)}{16\left(x^2+x+1\right)}\)
\(=\dfrac{-\left(x+1\right)}{x^2+x+1}\)
2: Để B=0 thì -x-1=0
hay x=-1(nhận)
HL
16 tháng 6 2017
a) Có lẽ câu này sai đề rồi. Dấu chia phải là dấu nhân chứ?
Mình sửa đề luôn nhé.
\(15\dfrac{1}{4}.\left(-\dfrac{7}{5}\right)-25\dfrac{1}{4}.\left(-\dfrac{7}{5}\right)\)
= \(-\dfrac{7}{5}.\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right)\)
= \(-\dfrac{7}{5}.\left(\dfrac{61}{4}-\dfrac{101}{4}\right)\)
= \(-\dfrac{7}{5}.\left(-10\right)\)
= \(14\)
b) \(38-2y=\dfrac{4}{3}\)
\(2y=38-\dfrac{4}{3}=\dfrac{110}{3}\Rightarrow y=\dfrac{55}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 1 2022
\(\dfrac{x}{x-3}+\dfrac{-9}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{-9}{x\left(x-3\right)}=\dfrac{x^2-9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=\dfrac{x+3}{x}\)
\(\dfrac{x-5}{x^2-4x+4}:\dfrac{x^2-25}{2x-4}=\dfrac{x-5}{\left(x-2\right)^2}.\dfrac{2\left(x-2\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2}{\left(x-2\right)\left(x+5\right)}\)
AH
Akai Haruma
Giáo viên
11 tháng 1 2022
Lần sau bạn chú ý ghi đầy đủ yêu cầu của đề.
* Coi đây là bài toán rút gọn
Lời giải:
ĐKXĐ: $x\neq 2$
$\frac{x-5}{x^2-4x+4}:\frac{x^2-25}{2x-4}=\frac{x-5}{(x-2)^2}:\frac{(x-5)(x+5)}{2(x-2)}$
$=\frac{x-5}{(x-2)^2}.\frac{2(x-2)}{(x-5)(x+5)}=\frac{2}{(x-2)(x+5)}$
\(=\dfrac{4\left(x-2\right)}{x+5}\cdot\dfrac{\left(x+5\right)\left(x-5\right)}{x\left(x-2\right)}=\dfrac{4x-20}{x}\)
\(\dfrac{4x-8}{x+5}.\dfrac{25-x^2}{2x-x^2}=\dfrac{4\left(x-2\right)}{x+5}.\dfrac{\left(x+5\right)\left(x-5\right)}{x\left(x-2\right)}=\dfrac{4\left(x-5\right)}{x}\)