\(16^3_5\cdot\dfrac{-1}{3}-13^3_5\cdot\dfrac{-1}{3}+\dfrac{3}{4}\)

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\(=\dfrac{-1}{3}\left(16+\dfrac{3}{5}-13-\dfrac{3}{5}\right)+\dfrac{3}{4}=\dfrac{-1}{3}\cdot3+\dfrac{3}{4}=-1+\dfrac{3}{4}=-\dfrac{1}{4}\)

7 tháng 12 2017

Làm lại cho you đây -_- vừa nãy bấm mt nhầm,đời t nhọ vãi

1)\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+....+16\right)\)

\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+\dfrac{1+2+3+4}{4}+...+\dfrac{1+2+3+...+16}{16}\)

Xét thừa số tổng quát: \(\dfrac{1+2+3+...+t}{t}=\dfrac{\left[\left(t-1\right):1+1\right]:2.\left(t+1\right)}{t}=\dfrac{\dfrac{t}{2}\left(t+1\right)}{t}=\dfrac{\dfrac{t^2}{2}+\dfrac{t}{2}}{t}=\dfrac{t\left(\dfrac{t}{2}+\dfrac{1}{2}\right)}{t}=\dfrac{t}{2}+\dfrac{1}{2}\)

Như vậy: \(P=1+\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\left(\dfrac{3}{2}+\dfrac{1}{2}\right)+\left(\dfrac{4}{2}+\dfrac{1}{2}\right)+...+\left(\dfrac{16}{2}+\dfrac{1}{2}\right)\)

\(P=1+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+....+\dfrac{17}{2}\)

\(P=\dfrac{2+3+4+5+...+17}{2}\)

\(P=\dfrac{152}{2}=76\)

2) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}\)

\(\Rightarrow2016\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{2016}{a+b}+\dfrac{2016}{b+c}+\dfrac{2016}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{c+a}{c+a}+\dfrac{b}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{2016}{3}-1-1-1=\dfrac{2007}{3}\)

2 tháng 7 2018

1,

\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2018}-1\right)\\ A=\left(-\dfrac{1}{2}\right)\cdot\left(-\dfrac{2}{3}\right)\cdot...\cdot\left(-\dfrac{2017}{2018}\right)\\ =-\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2017}{2018}\right)\\ =-\dfrac{1}{2018}\)

AH
Akai Haruma
Giáo viên
15 tháng 11 2017

Lời giải:

Ta có:

\(\text{VT}=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=\frac{1.2.3....31}{2.4.6.8...64}\)

Xét mẫu số:

\(2.4.6.8.....62.64=(2.1)(2.2)(2.3)(2.4)....(2.31)(2.32)\)

\(=2^{32}(1.2.3....31.32)\)

Suy ra:

\(\text{VT}=\frac{1.2.3....31}{2^{32}.(1.2.3...31.32)}=\frac{1}{2^{32}.32}=\frac{1}{2^{37}}\)

Do đó \(4^x=\frac{1}{2^{37}}\Leftrightarrow 2^{2x}=\frac{1}{2^{37}}\Leftrightarrow 2^{2x+37}=1\)

\(\Leftrightarrow 2x+37=0\Leftrightarrow x=-\frac{37}{2}\)

Vậy \(x=\frac{-37}{2}\)

18 tháng 11 2017

Số 2 nó ở đâu chui ra v Violympic toán 7

Câu 1: 

c: 2x=3y

nên x/3=y/2

=>x/9=y/6

5y=3z

nên y/3=z/5

=>y/6=z/10

=>x/9=y/6=z/10

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{3x+3y-7z}{3\cdot9+3\cdot6-7\cdot10}=\dfrac{35}{-25}=-\dfrac{7}{5}\)

Do đó: x=-63/5; y=-42/5; z=-14

Bài 2:

Gọi ba số lần lượt là a,b,c

Theo đề, ta có: 4/3a=b=3/4c

\(\Leftrightarrow\dfrac{a}{\dfrac{3}{4}}=\dfrac{b}{1}=\dfrac{c}{\dfrac{4}{3}}\)

\(\Leftrightarrow\dfrac{a}{9}=\dfrac{b}{12}=\dfrac{c}{16}\)

Đặt \(\dfrac{a}{9}=\dfrac{b}{12}=\dfrac{c}{16}=k\)

=>a=9k; b=12k; c=16k

Theo đề, ta có: \(a^2+b^2+c^2=481\)

\(\Leftrightarrow81k^2+144k^2+256k^2=481\)

=>k2=1

Trường hợp 1: k=1

=>a=9; b=12; c=16

Trường hợp 2: k=-1

=>a=-9; b=-12; c=-16

 

11 tháng 8 2018

Ta có:\(C=\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)

\(\Rightarrow C< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}\)

\(\Rightarrow C^2< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}.\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)

\(\Rightarrow C^2< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.....\dfrac{199}{200}.\dfrac{200}{201}\)

\(\Rightarrow C^2< \dfrac{1}{201}\) (đpcm)

11 tháng 8 2018

good luckbanhqua

2 tháng 8 2017

Giải:

a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)

\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)

\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)

\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)

\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)

\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)

\(\Leftrightarrow x=-\dfrac{79}{210}\)

Vậy \(x=-\dfrac{79}{210}\).

b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)

\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)

\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)

\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)

\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)

\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)

\(\Leftrightarrow3x=\dfrac{67}{75}\)

\(\Leftrightarrow x=\dfrac{67}{75}:3\)

\(\Leftrightarrow x=\dfrac{67}{225}\)

Vậy \(x=\dfrac{67}{225}\).

Chúc bạn học tốt!

2 tháng 8 2017

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20 tháng 6 2018

a)x=1;2;-2(bạn nên tự giải)

b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x

=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)

=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x

=>x = 1/7680

c)=>4x - 2x = 6x - 3x

=>2x (2x-1)= 3x(2x-1)

=> 2x = 3x

=>x = 0

21 tháng 6 2018

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