\(\frac{2}{2x+1}-\frac{3}{1-2x}=\frac{3x+8}{4x^2-1}\)
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A
8 tháng 3 2020
\(\frac{3x}{x-1}-\frac{2x}{x-3}+\frac{4x}{\left(x-1\right)\left(x-3\right)}=0\)đkxd \(x\ne1;3\)
\(\Leftrightarrow3x^2-9x-2x^2-2x+4x=0\)
\(\Leftrightarrow x^2-7x=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-7=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=7\end{cases}\left(tm\right)}\)
8 tháng 3 2020
\(\frac{3x}{x-1}-\frac{2x}{x-3}+\frac{4x}{\left(x-1\right)\left(x-3\right)}=0\)\(ĐKXĐ:x\ne1;3\)
\(3x\left(x-3\right)\left(x+3\right)-2x\left(x-1\right)\left(x+3\right)+4x\left(x-3\right)=0\)
\(x^3-33x=0\)
\(x\left(x^2-33\right)=0\)
\(x=0;\pm\sqrt{33}\)
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) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
VM
1 tháng 2 2020
Bài 4:
a) \(\frac{2x^2-10xy}{2xy}+\frac{5y-x}{y}\)
\(=\frac{y.\left(2x^2-10xy\right)}{2xy.y}+\frac{2xy.\left(5y-x\right)}{2xy.y}\)
\(=\frac{2x^2y-10xy^2}{2xy^2}+\frac{10xy^2-2x^2y}{2xy^2}\)
\(=\frac{2x^2y-10xy^2+10xy^2-2x^2y}{2xy^2}\)
\(=\frac{0}{2xy^2}\)
\(=0.\)
b) \(\frac{2}{x+y}+\frac{1}{x-y}+\frac{3x}{x^2-y^2}\)
\(=\frac{2}{x+y}+\frac{1}{x-y}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2.\left(x-y\right)}{\left(x-y\right).\left(x+y\right)}+\frac{1.\left(x+y\right)}{\left(x-y\right).\left(x+y\right)}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2x-2y}{\left(x-y\right).\left(x+y\right)}+\frac{x+y}{\left(x-y\right).\left(x+y\right)}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2x-2y+x+y+3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{6x-y}{\left(x-y\right).\left(x+y\right)}\)
c) \(x+y+\frac{x^2+y^2}{x+y}\)
\(=\frac{x+y}{1}+\frac{x^2+y^2}{x+y}\)
\(=\frac{\left(x+y\right).\left(x+y\right)}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{\left(x+y\right)^2}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{x^2+2xy+y^2}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{x^2+2xy+y^2+x^2+y^2}{x+y}\)
\(=\frac{2x^2+2xy+2y^2}{x+y}.\)
Chúc bạn học tốt!
22 tháng 4 2020
d, (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
Đặt x2 + 4x + 8 = t ta được:
t2 + 3xt + 2x2 = 0
\(\Leftrightarrow\) t2 + xt + 2xt + 2x2 = 0
\(\Leftrightarrow\) t(t + x) + 2x(t + x) = 0
\(\Leftrightarrow\) (t + x)(t + 2x) = 0
Thay t = x2 + 4x + 8 ta được:
(x2 + 4x + 8 + x)(x2 + 4x + 8 + 2x) = 0
\(\Leftrightarrow\) (x2 + 5x + 8)[x(x + 4) + 2(x + 4)] = 0
\(\Leftrightarrow\) (x2 + 5x + \(\frac{25}{4}\) + \(\frac{7}{4}\))(x + 4)(x + 2) = 0
\(\Leftrightarrow\) [(x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\)](x + 4)(x + 2) = 0
Vì (x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\) > 0 với mọi x
\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
Vậy S = {-4; -2}
Mình giúp bn phần khó thôi!
Chúc bn học tốt!!
A
22 tháng 4 2020
c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)
⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇒x2+x+1+2x2-5=4x-4
⇔3x2-3x=0
⇔3x(x-1)=0
⇔x=0 (TMĐK) hoặc x=1 (loại)
Vậy tập nghiệm của phương trình đã cho là:S={0}
2G
2
6 tháng 3 2020
\(\frac{2}{2x+1}-\frac{3}{1-2x}=\frac{3x+8}{4x^2-1}\left(x\ne\pm\frac{1}{2}\right)\)
\(\Leftrightarrow\frac{2}{2x+1}+\frac{3}{2x-1}-\frac{3x+8}{\left(2x+1\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\frac{3x+8}{\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Leftrightarrow\frac{4x-2}{\left(2x-1\right)\left(2x+1\right)}+\frac{6x+3}{\left(2x-1\right)\left(2x+1\right)}-\frac{3x+8}{\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Leftrightarrow\frac{4x-2+6x+3-3x-8}{\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Leftrightarrow\frac{7x-7}{\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Rightarrow7x-7=0\)
\(\Leftrightarrow7\left(x-1\right)=0\)
<=> x-1=0
<=> x=1(tmđk)
Vậy x=1
\(\frac{2}{2x+1}-\frac{3}{1-2x}=\frac{3x+8}{4x^2-1}\)đkxđ \(x\ne\pm\frac{1}{2}\)
\(\Leftrightarrow4x-2+6x+3-3x-8=0\)
\(\Leftrightarrow7x-7=0\)
\(\Leftrightarrow x=1\left(tm\right)\)