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a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow50x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\dfrac{99}{100}=1\)
\(\Leftrightarrow50x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{5000}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\dfrac{204}{205}=\dfrac{615}{205}\)
a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
Có tất cả : (99 - 1) : 1 + 1 = 99 (số x)
\(\Rightarrow99x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\dfrac{99}{100}=1\Rightarrow99x=1-\dfrac{99}{100}\)
\(\Rightarrow99x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{100.99}=\dfrac{1}{9900}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+....+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}\)
Chả hỉu gì cả.
Đăng từng bài một thôi bạn!
1)\(\left(-\dfrac{5}{13}\right)^{2017}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(-\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}.\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).1^{2016}\)
\(=-\dfrac{5}{13}\)
Bài 5:
a: \(\dfrac{x+7}{x-5}< 0\)
=>x+7>0 và x-5<0
=>-7<x<5
b: \(\dfrac{x-4}{x+9}>0\)
=>x-4>0 hoặc x+9<0
=>x>4 hoặc x<-9
c: \(\dfrac{x-1}{x+9}>1\)
\(\Leftrightarrow\dfrac{x-1-x-9}{x+9}>0\)
=>x+9>0
hay x>-9
d: \(\dfrac{x+5}{x-11}< 1\)
\(\Leftrightarrow\dfrac{x+5-x+11}{x-11}< 0\)
=>x-11<0
hay x<11
a/ \(x+\dfrac{3}{5}=\dfrac{4}{7}\)
\(x=\dfrac{4}{7}-\dfrac{3}{5}\)
\(x=-\dfrac{1}{35}\)
Vậy ....
b/ \(x-\dfrac{5}{6}=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}+\dfrac{5}{6}\)
\(x=1\)
Vậy ....
c/\(-\dfrac{5}{7}-x=\dfrac{-9}{10}\)
\(x=\dfrac{-5}{7}-\dfrac{-9}{10}\)
\(x=\dfrac{13}{70}\)
Vậy .....
d/ \(\dfrac{5}{7}-x=10\)
\(x=\dfrac{5}{7}-10\)
\(x=\dfrac{-65}{7}\)
Vậy ...
e/ \(x:\left(\dfrac{1}{9}-\dfrac{2}{5}\right)=\dfrac{-1}{2}\)
\(x:\dfrac{-13}{45}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}.\dfrac{-13}{45}\)
\(x=\dfrac{13}{90}\)
Vậy ....
f/ \(\left(\dfrac{-3}{5}+1,25\right)x=\dfrac{1}{3}\)
\(0,65.x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:0,65\)
\(x=\dfrac{20}{39}\)
Vậy ....
g/ \(\dfrac{1}{3}x+\left(\dfrac{2}{3}-\dfrac{4}{9}\right)=\dfrac{-3}{4}\)
\(\dfrac{1}{3}x+\dfrac{2}{9}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{-35}{36}\)
\(\Leftrightarrow x=\dfrac{-35}{12}\)
Vậy ...
2.Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)
\(\Rightarrow a+b+c=a+b-c\)
\(\Rightarrow a+b+c-a-b+c=0\)
\(\Rightarrow2c=0\)
\(\Rightarrow c=0\)
Vậy c=0
BT5: Ta có: f(1)=1.a+b=1 =>a+b=1 (1)
f(2)=2a+b=4 (2)
Trừ (1) cho (2) ta có: 2a+b-a-b=4-1 => a=3
Với a=3 thay vào (1) ta có: 3+b=1 => b=-2
Vậy a=3, b=-2
a) \(0,75:4,5=\dfrac{1}{15}:\left(2x\right)\)
\(\Rightarrow\) \(\dfrac{1}{6}=\dfrac{1}{30}:x\)
\(\Rightarrow\) \(x=\dfrac{1}{5}\)
a. \(0,75:4,5=\dfrac{1}{15}:\left(2x\right)\)
\(\Leftrightarrow\dfrac{1}{15}:\left(2x\right)=0,75:4,5\)
\(\Rightarrow\dfrac{1}{15}:\left(2x\right)=\dfrac{1}{6}\)
\(\Rightarrow2x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)
\(\Rightarrow x=\dfrac{2}{5}:2=\dfrac{1}{5}\)
Vậy...
b. \(\dfrac{-5}{x-2}=\dfrac{3}{-9}\)
\(\Leftrightarrow\left(x-2\right).3=\left(-5\right).\left(-9\right)\)
\(\Rightarrow\left(x-2\right).3=45\)
\(\Rightarrow\left(x-2\right)=45:3=15\)
\(\Rightarrow x=15+2=17\)
Vậy...
c. \(\dfrac{-2}{3}:x=\dfrac{1}{2}:\dfrac{3}{4}\)
\(\Rightarrow\dfrac{-2}{3}:x=\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}:\dfrac{2}{3}=-1\)
Vậy...
1. đề bạn ghi rõ lại giúp mình đc ko r mình giải lại cho
2. Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x^2}{2.3^2}=\dfrac{y^2}{5^2}=\dfrac{2x^2-y^2}{18-25}=\dfrac{-28}{-7}=4\)
\(\dfrac{x}{3}=4\Rightarrow x=12\)
\(\dfrac{y}{5}=4\Rightarrow y=20\)
Vậy x=12 và y=20
baif2 : a ) 49+x=51
x=51-49
x=2
b) x-17=15-x
x+x=17+15
2x=32
=>x=32:2
=>x=16
c) 56-6.(13-x)=116
6(13-x)=56-116
6(13-x) = -60
=>13-x=-60:6
13-x = -10
=> x=13-(-10) = 13+10
=> x=23