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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 \(\dfrac{5x-1}{3x+2}=2\)
\(\Rightarrow5x-1=2\left(3x+2\right)\)
\(\Rightarrow5x-1-2\left(3x+2\right)=0\)
\(\Rightarrow5x-1-6x-4=0\)
\(\Rightarrow-x-5=0\)
\(\Rightarrow x=-5\)
2)
a) \(\dfrac{1}{x}.\dfrac{6x}{y}\)
\(=\dfrac{6x}{xy}\)
\(=\dfrac{6}{y}\)
b) \(\dfrac{2x^2}{y}.3xy^2\)
\(=\dfrac{2x^2.3xy^2}{y}\)
\(=\dfrac{6x^3y^2}{y}\)
\(=6x^3y\)
c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\)
\(=\dfrac{15x.2y^2}{7y^3.x^2}\)
\(=\dfrac{30xy^2}{7x^2y^3}\)
\(=\dfrac{30}{7xy}\)
d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)
\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)
\(=\dfrac{2y}{5x\left(x-y\right)}\)
a: =>-4x>16
=>x<-4
c: =>20x-25<=21-3x
=>23x<=46
=>x<=2
d: =>20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)
=>40x-100-90x+30<36-12x-30x+15
=>-50x-70<-42x+51
=>-8x<121
=>x>-121/8
Giải:
a) \(\dfrac{12x^3}{y^4}.\dfrac{2y^5}{x^2}\)
\(=\dfrac{12.2.x^3.y^5}{x^2.y^4}\)
\(=\dfrac{24.x^2.x.y^4.y}{x^2.y^4}\)
\(=24xy\)
Vậy ...
b) \(\dfrac{5x-10}{4x-8}:\dfrac{2x+4}{4-2x}\)
\(=\dfrac{5\left(x-2\right)}{4\left(x-2\right)}:\dfrac{2\left(x+2\right)}{2\left(2-x\right)}\)
\(=\dfrac{5}{4}:\dfrac{x+2}{2-x}\)
\(=\dfrac{5}{4}.\dfrac{2-x}{x+2}\)
\(=\dfrac{5\left(2-x\right)}{4\left(x+2\right)}\)
\(=\dfrac{10-5x}{4x+8}\)
Vậy ...
c) \(\dfrac{3x+10}{x+3}.\dfrac{2x+4}{x+3}\)
\(=\dfrac{3x+10}{x+3}.\dfrac{2\left(x+2\right)}{x+3}\)
\(=\dfrac{\left(3x+10\right).2.\left(x+2\right)}{\left(x+3\right)^2}\)
Vậy ...
a: \(\dfrac{5}{2x+6}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}\)
3/x^2-9=6/2(x+3)(x-3)
b: \(\dfrac{2x}{x^2-8x+16}=\dfrac{2x}{\left(x-4\right)^2}=\dfrac{6x^2}{3x\left(x-4\right)^2}\)
\(\dfrac{x}{3x^2-12x}=\dfrac{x}{3x\left(x-4\right)}=\dfrac{x\left(x-4\right)}{3x\left(x-4\right)^2}\)
c: \(\dfrac{x+y}{x}=\dfrac{\left(x+y\right)\cdot\left(x-y\right)}{x\left(x-y\right)}\)
x/x-y=x^2/x(x-y)
e: \(\dfrac{1}{x+2}=\dfrac{2x-x^2}{x\left(x+2\right)\left(2-x\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8\left(x+2\right)}{x\left(2-x\right)\left(2+x\right)}\)
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) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
\(\dfrac{\left(x+y\right)2}{x2+xy}+\dfrac{\left(x-y\right)2}{x2-xy}=-\left(\dfrac{\left(x-y\right)2}{x2-xy}\right)+\dfrac{\left(x-y\right)2}{x2-xy}=0\)
b: \(\dfrac{x^2-4x}{xy-4x-3y+12}+\dfrac{x-2}{y-4}\)
\(=\dfrac{x\left(x-4\right)}{\left(y-4\right)\left(x-3\right)}+\dfrac{x-2}{y-4}\)
\(=\dfrac{x^2-4x+x^2-5x+6}{\left(y-4\right)\left(x-3\right)}=\dfrac{2x^2-9x+6}{\left(y-4\right)\left(x-3\right)}\)
c: \(=\dfrac{y^2}{\left(y-5\right)\left(x+1\right)}+\dfrac{2}{x+1}\)
\(=\dfrac{y^2+2y-10}{\left(y-5\right)\left(x+1\right)}\)
a, \(\dfrac{x+1}{3}+\dfrac{2x-1}{3}=\dfrac{x+1+2x-1}{3}=\dfrac{3x}{3}=x\)
b, \(\dfrac{5x-2y}{x^2-y^2}+\dfrac{y-4x}{x^2-y^2}=\dfrac{5x-2y+y-4x}{\left(x+y\right)\left(x-y\right)}=\dfrac{x-y}{\left(x+y\right)\left(x-y\right)}=\dfrac{1}{x+y}\)
c, \(\dfrac{x-1}{12x}+\dfrac{2x+7}{12x}+\dfrac{6-3x}{12x}=\dfrac{x-1+2x+7+6-3x}{12x}=\dfrac{12}{12x}=\dfrac{1}{x}\)
a,\(\dfrac{x+1_{ }+2x-1}{3}\)=\(\dfrac{2x}{3}\)
b,\(\dfrac{5x-2y+y-4x}{\left(x+y\right)\left(x-y\right)}\)=\(\dfrac{x-y}{\left(x+y\left(x-y\right)\right)}\)
c,\(\dfrac{x-1+2x+7+6-3x}{12x}\)=\(\dfrac{12}{12x}\)=\(\dfrac{1}{x}\)