a) \(\dfrac{2}{3}.x-\dfrac{1}{2}.x=\dfrac{5}{12}\)

=> \(\left(\dfrac{2}{3}-\dfrac{1}{2}\right).x=\dfrac{5}{12}\)

=> \(\left(\dfrac{4}{6}-\dfrac{3}{6}\right).x=\dfrac{5}{12}\)

=> \(\dfrac{1}{6}\) . x = \(\dfrac{5}{12}\)

=> \(x=\dfrac{5}{12}:\dfrac{1}{6}\)

=> x =\(\dfrac{5}{12}.\dfrac{6}{1}\)

=> x = \(\dfrac{5}{2}\)

Vậy x = \(\dfrac{5}{2}\)

Lời giải:

a, \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\Rightarrow\left(\dfrac{2}{3}-\dfrac{1}{2}\right)x=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{1}{6}x=\dfrac{5}{12}\)

\(\Rightarrow x=\dfrac{5}{12}.6\)

\(\Rightarrow x=\dfrac{5}{2}\)

\(a,\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\Leftrightarrow\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}\Leftrightarrow\dfrac{2}{3}x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{3}:\dfrac{2}{3}=1\)\(b,5\dfrac{4}{7}:x=13\Leftrightarrow\dfrac{39}{7}:x=13\Leftrightarrow x=\dfrac{39}{7}:13=\dfrac{3}{7}\)\(c,\left(2\dfrac{4}{5}x-50\right):\dfrac{2}{3}=51\Leftrightarrow\left(\dfrac{14}{5}x-50\right).\dfrac{3}{2}=51\Leftrightarrow\dfrac{21}{5}x-75=51\Leftrightarrow\dfrac{21}{5}x=51+75=126\Leftrightarrow x=126:\dfrac{21}{5}=30\)

d,\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{-1}{2}\\2x=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

a, \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

=> \(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

=> \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b, Dễ ==> tự lm

c,( \(2\dfrac{4}{5}x-50\)) \(:\dfrac{2}{3}=51\)

=> \(2\dfrac{4}{5}x-50=51\cdot\dfrac{2}{3}=34\)

=> \(\dfrac{14}{5}x=34+50=84\)

=> \(x=84:\dfrac{14}{5}=30\)

d, \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\-2x=-\dfrac{2}{3}\Rightarrow x=\dfrac{1}{3}\end{matrix}\right.\)

e, \(\dfrac{2}{3}-\dfrac{1}{2}x=\dfrac{5}{12}\)

=> \(-\dfrac{1}{2}x=\dfrac{5}{12}-\dfrac{2}{3}=-\dfrac{1}{4}\)

=> \(x=-\dfrac{1}{4}:\left(-\dfrac{1}{2}\right)=\dfrac{1}{2}\)

Bài 1: Tìm x biết:

a) \(\dfrac{6}{5}-2\left|1-3x\right|=1\dfrac{2}{3}\)

\(2\left|1-3x\right|=\dfrac{6}{5}-1\dfrac{2}{3}\)

\(2\left|1-3x\right|=\dfrac{-7}{15}\)

\(\left|1-3x\right|=\dfrac{-7}{15}:2\)

\(\left|1-3x\right|=\dfrac{-7}{30}\)

\(\left|1-3x\right|\in N\) nhưng \(\dfrac{-7}{30}\notin N\)

\(\Rightarrow x=\varnothing\)

b) \(\left(2,8x+50\right):\dfrac{-3}{2}=51\)

\(\left(2,8x+50\right)=51.\dfrac{-3}{2}\)

\(2,8x+50=\dfrac{-153}{2}\)

\(2,8x=\dfrac{-153}{2}-50\)

\(2,8x=\dfrac{-253}{2}\)

\(x=\dfrac{-253}{2}:2,8\)

\(x=\dfrac{-1265}{28}\)

c) \(\dfrac{x-2}{-2}=\dfrac{x+4}{3}\)

\(\Rightarrow\left(x-2\right).3=-2.\left(x+4\right)\)

\(x.3-2.3=\left(-2\right).x+\left(-2\right).4\)

\(3x-6=\left(-2\right)x+\left(-8\right)\)

\(3x-\left(-2\right)x=6+\left(-8\right)\)

\(5x=-2\)

\(x=\left(-2\right):5\)

\(x=\dfrac{-2}{5}\)

d) \(4\left(3-2x\right)-5\left(x-1\right)=12\)

\(4.3-4.2x-5x+5.1=12\)

\(12-8x-5x+5=12\)

\(12+\left(-8\right)x+\left(-5\right)x+5=12\)

\(12+\left(-13\right)x+5=12\)

\(\left(-13\right)x=12-12-5\)

\(\left(-13\right)x=-5\)

\(x=\left(-5\right):\left(-13\right)\)

\(x=\dfrac{5}{13}\)

Bài 2: Chứng minh:

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\) (đpcm)

bài 1 tự làm na mk giải giúp bài 2 đặt vế trái là A A<1/2.3+1/3.4+1/4.5+......+1/99.100 A<1/2-1/3+1/3-1/4+1/4-1/5+....+1/99-1/100 A<1/2-1/100<1/2 vậy A<1/2

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8

a: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5

=>x=3/5:2/3=3/5x3/2=9/10

b: \(\Leftrightarrow x\cdot2.8-50=34\)

=>2,8x=84

=>x=30

c: \(\Leftrightarrow\dfrac{1}{6}x=\dfrac{5}{12}\)

hay x=5/2

d: \(\Leftrightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}+\dfrac{7}{4}=\dfrac{41}{4}\)

=>2x-3/4=41/4 hoặc 2x-3/4=-41/4

=>2x=44/4=11 hoặc 2x=-19/2

=>x=11/2 hoặc x=-19/4

Lời giải:

a)

\(\frac{\frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\frac{2}{11}}{\frac{8}{3}-\frac{8}{5}+\frac{8}{7}-\frac{8}{9}+\frac{8}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}{8\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}\) \(=\frac{2}{8}=\frac{1}{4}\)

b)

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{50}-1\right)\left(\frac{1}{51}-1\right)\)

\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}....\frac{1-50}{50}.\frac{1-51}{2}=\frac{(-1)(-2)(-3)...(-49)(-50)}{2.3.4....50.51}\)

\(=\frac{(-1)^{50}.1.2.3....49.50}{2.3.4...50.51}=\frac{1}{51}\)

Pt\(\Leftrightarrow\dfrac{x+98}{2}+1+\dfrac{x+96}{4}+1+\dfrac{x+65}{35}+1=\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+49}{51}+1\)

\(\Leftrightarrow\dfrac{x+100}{2}+\dfrac{x+100}{4}+\dfrac{x+100}{35}-\dfrac{x+100}{97}-\dfrac{x+100}{95}-\dfrac{x+100}{51}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{35}-\dfrac{1}{97}-\dfrac{1}{35}-\dfrac{1}{51}\right)=0\)

\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)

Vậy...

\(1,\left(dk:x\ne0,-1,4\right)\)

\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)

\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)

\(\Leftrightarrow-x=-44\)

\(\Leftrightarrow x=44\left(tm\right)\)

\(2,\left(đk:x\ne4\right)\)

\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)

\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)

\(\Leftrightarrow28-12-6x-9+5x-20=0\)

\(\Leftrightarrow-x=13\)

\(\Leftrightarrow x=-13\left(tm\right)\)