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26 tháng 12 2020
a. 2x(x + y) - y(y + 2x) = 2x2 + 2xy - y2 - 2xy = 2x2 - y2
b.\(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)
Phần c nản quá.
XO
25 tháng 12 2020
a) 2x(x + y) - y(y + 2x)
= 2x2 + 2xy - y2 - 2xy
= 2x2 - y2
b) \(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)
c) \(\frac{x^3-4x^2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{x-1}\)
= \(\frac{x^3-4x^2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}+\frac{x^2+x+1}{\left(x^2+x+1\right)\left(x-1\right)}\)
= \(\frac{x^3-4x^2+2x-2+x^2+x+1}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{x^3-3x^2+3x-1}{\left(x^2+x+1\right)\left(x-1\right)}=\frac{\left(x-1\right)^3}{\left(x^2+x+1\right)\left(x-1\right)}\)
\(=\frac{\left(x-1\right)^2}{x^2+x+1}\)
TT
19 tháng 3 2020
1.(4x+2)(x2 +1)=0
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}4x+2=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=-2\\x^2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{2}\\x=-1\end{matrix}\right.\)
2.(x-1)(2x+7)(x2+2)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\2x+7=0\\x^2-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=\frac{-7}{2}\\x=-1,4142\end{matrix}\right.\)
S
14 tháng 12 2018
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
14 tháng 12 2018
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
5 tháng 12 2016
\(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)
\(=\frac{1-4x^2}{x^2+4x}.\frac{3x}{x-4x}\)
=\(\frac{\left(1-4x^2\right)3x}{\left(x^2+4x\right)\left(2-4x\right)}=\frac{3x-12x^3}{2x^2-4x^3+8x-16x^2}\)
4 tháng 2 2020
Bài 1:
d)ĐKXĐ: \(x\ne8\)
Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)
\(\Leftrightarrow\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3x-24}=0\)
\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)
MTC=24(x-8)
\(\Leftrightarrow\frac{36}{24\left(x-8\right)}+\frac{72x-480}{24\left(x-8\right)}+\frac{3x-24}{24\left(x-8\right)}-\frac{104x-816}{24\left(x-8\right)}=0\)
\(\Leftrightarrow36+72x-480+3x-24-104x+816=0\)
\(\Leftrightarrow348-29x=0\)
\(\Leftrightarrow-29x+348=0\)
\(\Leftrightarrow x=\frac{-348}{-29}=12\)
Vậy: x=12
e) ĐKXĐ: \(x\ne\pm1\)
Ta có: \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)
\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4x+4}+\frac{12x-1}{4-4x}=0\)
\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}+\frac{12x-1}{4\left(1-x\right)}=0\)
MTC=4(x+1)(x-1)
\(\Leftrightarrow\frac{24}{4\left(x-1\right)\left(x+1\right)}+\frac{20x^2-20}{4\left(x-1\right)\left(x+1\right)}-\frac{8x^2-9x+1}{4\left(x-1\right)\left(x+1\right)}-\frac{12x^2-11x-1}{4\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2+11x+1=0\)
\(\Leftrightarrow20x+4=0\)
\(\Leftrightarrow20x=-4\)
\(\Leftrightarrow x=-\frac{4}{20}=-0,2\)(loại)
Vậy: x không có giá trị
g) Ta có: \(\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\frac{1}{2}\)
\(\Leftrightarrow\frac{\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}}{\frac{x-1}{x-1}+\frac{x+1}{x-1}}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}}{\frac{2x}{x-1}}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-1}{2x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{4x\cdot\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\cdot2x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{2}=0\)
MTC=2(x+1)
\(\Leftrightarrow\frac{2}{2\left(x+1\right)}-\frac{x+1}{2\left(x+1\right)}=0\)
\(\Leftrightarrow2-x+1=0\)
\(\Leftrightarrow1-x=0\)
\(\Leftrightarrow x=1\)(loại vì không thỏa mãn ĐKXĐ)
Vậy: x không có giá trị
15 tháng 12 2017
a) \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)
b) \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}=\frac{3+6x}{2x+8}\)
Bài làm
\(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}:\frac{2\left(1-2x\right)}{3x}\)
\(=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3x\left(1+2x\right)}{2x\left(x+4\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}\)