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Giải:
1) \(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\dfrac{55}{24}\)
\(=\dfrac{-19}{8}\)
2) \(-1,75-\left(\dfrac{-1}{9}-2\dfrac{1}{18}\right)\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+2\dfrac{1}{18}\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+\dfrac{37}{18}\)
\(=\dfrac{5}{12}\)
3) \(-\dfrac{5}{6}-\left(-\dfrac{3}{8}+\dfrac{1}{10}\right)\)
\(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\)
\(=-\dfrac{67}{120}\)
4) \(\dfrac{2}{5}+\left(-\dfrac{4}{3}\right)+\left(-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{1}{2}\)
\(=-\dfrac{43}{30}\)
5) \(\dfrac{3}{12}-\left(\dfrac{6}{15}-\dfrac{3}{10}\right)\)
\(=\dfrac{3}{12}-\dfrac{6}{15}+\dfrac{3}{10}\)
\(=\dfrac{3}{20}\)
6) \(\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}\)
\(=8\dfrac{5}{11}+3\dfrac{5}{8}-3\dfrac{5}{11}\)
\(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}\)
\(=8+\dfrac{5}{8}\)
\(=\dfrac{69}{8}\)
7) \(-\dfrac{1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}.13\dfrac{9}{11}-\dfrac{1}{4}.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}\left(13\dfrac{9}{11}+6\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+\dfrac{9}{11}+6+\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+6+1\right)\)
\(=-\dfrac{1}{4}.20=-5\)
8) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)
\(=\dfrac{4}{9}\left(-7\right)+6\dfrac{5}{9}\left(-7\right)\)
\(=-7\left(\dfrac{4}{9}+6\dfrac{5}{9}\right)\)
\(=-7\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)\)
\(=-7\left(6+1\right)\)
\(=-7.7=-49\)
Vậy ...
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
24:
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)
\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
\(\Leftrightarrow x+5=0\)
hay x=-5
a)\(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
\(\Leftrightarrow\dfrac{12x-10x-4}{12}=\dfrac{21-9x}{12}\)
\(\Leftrightarrow2x-4=21-9x\)
\(\Leftrightarrow2x-4-21+9x=0\)
\(\Leftrightarrow11x-25=0\)
\(\Leftrightarrow x=\dfrac{25}{11}\)
b)\(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{30x+9}{36}=\dfrac{36+24+32x}{36}\)
\(\Leftrightarrow30x+9=60+32x\)
\(\Leftrightarrow30x+9-60-32x=0\)
\(\Leftrightarrow-2x-51=0\)
\(\Leftrightarrow x=-\dfrac{51}{2}\)
c)\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-6\)
\(\Leftrightarrow\dfrac{2x-6x-3}{6}=\dfrac{x-36}{6}\)
\(\Leftrightarrow-4x-3=x-36\)
\(\Leftrightarrow-4x-3-x+36=0\)
\(\Leftrightarrow-5x+33=0\)
\(\Leftrightarrow x=\dfrac{33}{5}\)
d)\(\dfrac{2+x}{3}-\dfrac{1}{2}x=\dfrac{1-2x}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{8+4x-6x}{12}=\dfrac{3-6x+3}{12}\)
\(\Leftrightarrow8-2x=6-6x\)
\(\Leftrightarrow8-2x-6+6x=0\)
\(\Leftrightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Tính lại xem đúng không nha 
a) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
\(\Leftrightarrow\dfrac{24x}{24}-\dfrac{4\left(5x+2\right)}{24}=\dfrac{6\left(7-3x\right)}{24}\)
\(\Leftrightarrow24x-4\left(5x+2\right)=6\left(7-3x\right)\)
\(\Leftrightarrow24x-20x-8=42-18x\)
\(\Leftrightarrow4x-8=42-18x\)
\(\Leftrightarrow4x+18x=42+8\)
\(\Leftrightarrow22x=50\)
\(\Leftrightarrow x=\dfrac{25}{11}\)
Vậy S\(=\left\{\dfrac{25}{11}\right\}\)
a,\(x-\frac{5x+2}{6}=\frac{7-3x}{4}\)
=> \(\frac{12x}{12}-\frac{\left(5x+2\right)2}{12}=\frac{\left(7-3x\right)3}{12}\)
=>\(\frac{12x-10x-4}{12}=\frac{21-9x}{12}\)
=>(khử mẫu)
=>\(12x-10x-4=21-9x\)
=>11x=25
=>x=25/11
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>32x+60=30x+9
=>2x=-51
=>x=-51/2
c: \(\Leftrightarrow2x-3\left(2x+1\right)=x+6x\)
=>7x=2x-6x-3
=>7x=-4x-3
=>11x=-3
=>x=-3/11
d: \(\Leftrightarrow4\left(x+2\right)-6x=3\left(1-2x+1\right)\)
=>4x+8-6x=3(-2x+2)
=>-2x+8+6x-6=0
=>4x+2=0
=>x=-1/2
Có:
\(\dfrac{n}{n+2}< \dfrac{n-1}{n}\)(Vì
\(n^2< n^2+n-2\forall n>2\))
Nên ta có
\(F=\dfrac{1}{3}.\dfrac{4}{6}....\dfrac{208}{201}\)
\(\Rightarrow F< \dfrac{1}{3}.\dfrac{3}{4}.\dfrac{6}{7}...\dfrac{207}{208}\)
\(\Rightarrow F^2< \dfrac{1.4.7...208}{3.6.9.12...210}.\dfrac{1.3.6.9...207}{3.4.7.10.208}\)
\(\Rightarrow F^2=\dfrac{1}{210}.\dfrac{1}{3}\)
\(\Rightarrow F^2=\dfrac{1}{630}< \left(\dfrac{1}{25}\right)^2\)
Vậy F\(< \dfrac{1}{25}\)