Thực Hiện Phép Tính:\(\frac{1}{6x-4y}-\frac{1}{6x+4y}-\frac{3x}{4y^2-9x^2}\)
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18 tháng 11 2019
a)có khả năng sai đề bài
b)Liệu có sai đề bài không
c)\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)(phân số cuối có âm vì (1-x)=-(x-1)
\(=\frac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)(Hơi tắt)
\(=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{1}{x^2+x+1}\)
d)\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{x^2+2xy+x^2-2xy+4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\frac{2x^2+4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x-2y}\)
29 tháng 12 2021
f: \(=\dfrac{2x^3-10x^2-11x^2+55x+12x-60}{x-5}=2x^2-11x+12\)
29 tháng 7 2023
a: \(=\dfrac{6}{3}\cdot x\cdot\dfrac{y^2}{y}=2xy\)
b: \(=\dfrac{62}{2}\cdot\dfrac{x^4}{x^3}\cdot\dfrac{y^3}{y^2}=31xy\)
c: \(=\dfrac{-18}{6}\cdot\dfrac{x^4}{x^2}\cdot\dfrac{y^3}{y}=-3x^2y^2\)
d: \(=\dfrac{27}{9}\cdot\dfrac{x^5}{x^3}\cdot\dfrac{y^6}{y^3}=3x^2y^3\)
e: \(=\dfrac{18}{12}\cdot\dfrac{x^3}{x}\cdot\dfrac{y^4}{y^3}=\dfrac{3}{2}x^2y\)
24 tháng 5 2017
a)\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)
\(\Leftrightarrow\frac{\left(x+y\right)^2-1}{\left(x+1\right)^2-y^2}\)
\(\Leftrightarrow\frac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}\)
\(\Leftrightarrow\frac{x+y-1}{x-y+1}\)
b)\(\frac{3x^3-6x^2y+xy^2-2y^3}{9x^5-18x^4y-xy^4+2y^5}\)
\(\Leftrightarrow\frac{3x^2\left(x-2y\right)+y^2\left(x-2y\right)}{9x^4\left(x-2y\right)-y^4\left(x-2y\right)}\)
\(\Leftrightarrow\frac{\left(3x^2+y^2\right)\left(x-2y\right)}{\left(9x^4-y^4\right)\left(x-2y\right)}\)
\(\Leftrightarrow\frac{3x^2+y^2}{\left(3x^2-y^2\right)\left(3x^2+y^2\right)}\)
\(\Leftrightarrow\frac{1}{3x^2-y^2}\)
KK
23 tháng 2 2019
Ta có: 6x = 4y => x/4 = y/6
4y = 3z => y/3 = z/4 => y/6 = z/8
=> x/4 = y/6 = z/8
Đặt \(\frac{x}{4}=\frac{y}{6}=\frac{z}{8}=k\) => x = 4k; y = 6k; z = 8k
Khi đó, ta có:
M = \(\frac{2.\left(4k\right)^2+5.\left(6k\right)^2-4.\left(8k\right)^2}{7.\left(4k\right)^2-4.\left(6k\right)^2+3.\left(8k\right)^2}\)
= \(\frac{2.4^2.k^2+5.6^2.k^2-4.8^2.k^2}{7.4^2.k^2-4.6^2.k^2+3.8^2.k^2}\)
= \(\frac{k^2.\left(2.16+5.36-4.64\right)}{k^2.\left(7.16-4.36+3.64\right)}\)
= \(\frac{32+180-256}{112-144+194}\)
= \(\frac{-44}{162}=-\frac{22}{81}\)
NT
1
3 tháng 3 2017
Ta có: \(\dfrac{1+3x}{12}=\dfrac{2\left(1+3x\right)}{24}=\dfrac{2+6x}{24}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{2+6x}{24}=\dfrac{1+6x}{16}=\dfrac{2+6x-1-6x}{24-16}=\dfrac{1}{8}\)
Khi đó \(\dfrac{1+3x}{12}=\dfrac{1}{8}\)
\(\Rightarrow\left(1+3x\right)8=12\)
\(\Rightarrow8+24x=12\)
\(\Rightarrow24x=4\)
\(\Rightarrow x=\dfrac{1}{6}\)
Thay \(x=\dfrac{1}{6}\) vào đề bài:
\(\dfrac{1-5.\dfrac{1}{6}}{4y}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{\dfrac{1}{6}}{4y}=\dfrac{1}{8}\)
\(\Rightarrow4y=\dfrac{4}{3}\)
\(\Rightarrow y=\dfrac{1}{3}\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{6}\\y=\dfrac{1}{3}\end{matrix}\right.\).
\(\dfrac{1}{6x-4y}-\dfrac{1}{6x+4y}-\dfrac{3x}{4y^2-9x^2}\)
\(=\dfrac{1}{2\left(3x-2y\right)}-\dfrac{1}{2\left(3x+2y\right)}+\dfrac{3x}{\left(3x-2y\right)\left(3x+2y\right)}\)
\(=\dfrac{3x+2y-\left(3x-2y\right)+6x}{2\left(3x-2y\right)\left(3x+2y\right)}\)
\(=\dfrac{9x+2y-3x+2y}{2\left(3x-2y\right)\left(3x+2y\right)}=\dfrac{6x+4y}{2\left(3x+2y\right)\left(3x-2y\right)}=\dfrac{1}{3x-2y}\)
6x−4y1−6x+4y1−4y2−9x23x
\(= \frac{1}{2 \left(\right. 3 x - 2 y \left.\right)} - \frac{1}{2 \left(\right. 3 x + 2 y \left.\right)} + \frac{3 x}{\left(\right. 3 x - 2 y \left.\right) \left(\right. 3 x + 2 y \left.\right)}\)
\(= \frac{3 x + 2 y - \left(\right. 3 x - 2 y \left.\right) + 6 x}{2 \left(\right. 3 x - 2 y \left.\right) \left(\right. 3 x + 2 y \left.\right)}\)
\(= \frac{9 x + 2 y - 3 x + 2 y}{2 \left(\right. 3 x - 2 y \left.\right) \left(\right. 3 x + 2 y \left.\right)} = \frac{6 x + 4 y}{2 \left(\right. 3 x + 2 y \left.\right) \left(\right. 3 x - 2 y \left.\right)} = \frac{1}{3 x - 2 y}\)