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13 tháng 11 2017
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
TD
1
HN
17 tháng 12 2017
= ( \(\dfrac{8}{\left(x+4\right)\left(x-4\right)}\)+ \(\dfrac{x-4}{\left(x+4\right)\left(x-4\right)}\)) : \(\dfrac{x+2}{x^{2^{ }}-4x}\)
= \(\dfrac{1}{x-4}\): \(\dfrac{x+2}{x\left(x-4\right)}\)
= \(\dfrac{1}{x-4}\). \(\dfrac{x\left(x-4\right)}{x+2}\)= \(\dfrac{x}{x+2}\)
14 tháng 6 2022
a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}\)
b: \(=\dfrac{x+2}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{2-x}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(2-x\right)}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\dfrac{2x+4-4}{\left(2-x\right)\left(x+2\right)}\)
\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)
29 tháng 11 2022
a: \(B=\left(\dfrac{4x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\left(\dfrac{4x}{x+2}-\dfrac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\dfrac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)^2\cdot\left(x+1\right)}{16\left(x^2+x+1\right)}\)
\(=\dfrac{-16}{16\left(x^2+x+1\right)}\cdot\left(x+1\right)=-\dfrac{x+1}{x^2+x+1}\)
b: \(B=\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x+2}{x^2+x+1}\)
\(P=A+B=\dfrac{-x-1+x+2}{x^2+x+1}=\dfrac{1}{x^2+x+1}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< =1:\dfrac{3}{4}=\dfrac{4}{3}\)
Dấu = xảy ra khi x=-1/2
Y
26 tháng 8 2018
a) \(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
\(=\dfrac{5x-1-5x+7}{3x+2-3x+1}\)
\(=\dfrac{-1+7}{2+1}\)
\(=\dfrac{6}{3}\)
\(=2\)
Với \(\dfrac{5x-1}{3x+2}=2\)
\(\Rightarrow5x-1=2\left(3x+2\right)\)
\(\Rightarrow5x-1-2\left(3x+2\right)=0\)
\(\Rightarrow5x-1-6x-4=0\)
\(\Rightarrow-x-5=0\)
\(\Rightarrow x=-5\)
16 tháng 12 2018
\(\dfrac{2015.\left(x-y\right)^2}{x^2-2xy+y^2}\) =\(\dfrac{2015.\left(x-y\right)^2}{\left(x-y\right)^2}=2015\)
\(\dfrac{x^3}{x+3}+\dfrac{3x^2}{x+3}=\dfrac{x^3+3x^2}{x+3}=\dfrac{x^2\left(x+3\right)}{x+3}=x^2\)
\(\dfrac{4}{x^2-4x}+\dfrac{x-8}{4x-16}=\dfrac{4}{x\left(x-4\right)}+\dfrac{x-8}{4\left(x-4\right)}=\dfrac{16+x^2-8}{4x\left(x-4\right)}=\dfrac{8-x^2}{4x\left(x-4\right)}\dfrac{\left(4-x\right)\left(4+x\right)}{-4x\left(4-x\right)}=\dfrac{4+x}{-4x}\)
2 tháng 8 2022
a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)
=>-12x-4=2x-10
=>-14x=-6
hay x=3/7
b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)
=>15x-9-10x+2=-4
=>5x-7=-4
=>5x=3
hay x=3/5(loại)
c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\)
=>4x=2
hay x=1/2(nhận)
NN
1 tháng 12 2017
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\) MTC: \(xy\left(x-2y\right)\left(x+2y\right)\)
\(=\dfrac{2x.y\left(x-2y\right)}{xy\left(x+2y\right)\left(x-2y\right)}+\dfrac{y.x\left(x+2y\right)}{xy\left(x-2y\right)\left(x+2y\right)}+\dfrac{4.xy}{xy\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-4xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y-2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) \(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{1}{x-y}-\dfrac{3xy}{x^3-y^3}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\) MTC: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{\left(x-y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{\left(x^2+xy+y^2\right)-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
GH
20 tháng 2 2018
a)\(\dfrac{3x+2}{3x-2}-\dfrac{6}{2+3x}=\dfrac{9x^2}{9x^2-4}\left(ĐKXĐ:x\ne\pm\dfrac{2}{3}\right)\)
\(\Leftrightarrow\dfrac{3x+2}{3x-2}-\dfrac{6}{3x+2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(3x+2\right)^2-6\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Rightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow9x^2-6x+16-9x^2=0\)
\(\Leftrightarrow-6x=-16\)
\(\Leftrightarrow x=\dfrac{8}{3}\) (thỏa mãn ĐKXĐ)
Vậy .................
GH
20 tháng 2 2018
b) \(\dfrac{5-x}{4x^2-8x}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8x-16}\left(ĐKXĐ:x\ne0;x\ne2\right)\)
\(\Leftrightarrow\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{2\left(5-x\right)+7\left(x-2\right)}{8x\left(x-2\right)}=\dfrac{4\left(x-1\right)+x}{8x\left(x-2\right)}\)
\(\Rightarrow10-2x+7x-14=4x-4+x\)
\(\Leftrightarrow5x-4=5x-4\)
\(\Leftrightarrow0x=0\) (vô số nghiệm)
Vậy \(S=R\backslash\left\{0;2\right\}\)
=))
Em cần làm gì với biểu thức này?