TG
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
19 tháng 2 2017
Chú ý: \(a^2-1=\left(a-1\right)\left(a+1\right)\)
Áp dụng:
\(A=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}...\frac{49.51}{50^2}=\frac{2.3.4^2.5^2...49^2.50.51}{3^2.4^2.5^2...50^2}=\frac{2.51}{3.50}=\frac{51}{75}\)
Q
29 tháng 3 2017
\(6,25.x.\left(\dfrac{8}{5}:1\dfrac{1}{2}+\dfrac{1}{50}\right)=\dfrac{7}{50}\left(2,91+0,09\right).4\)
\(6,25.x.\left(\dfrac{8}{5}:1\dfrac{1}{2}+\dfrac{1}{50}\right)=\dfrac{7}{50}.3.4\)
\(6,25.x.\left(\dfrac{8}{5}:1\dfrac{1}{2}+\dfrac{1}{50}\right)=\dfrac{7}{50}.12\)
\(6,25.x.\left(\dfrac{8}{5}:1\dfrac{1}{2}+\dfrac{1}{50}\right)=\dfrac{42}{25}\)
\(6,25.x.\dfrac{163}{150}=\dfrac{42}{25}\)
\(\dfrac{25}{4}.x=\dfrac{42}{25}:\dfrac{163}{150}\)
\(\dfrac{25}{4}.x=\dfrac{252}{163}\)
\(x=\dfrac{252}{163}:\dfrac{25}{4}\)
\(x=\dfrac{1008}{4075}\)
1 tháng 8 2020
a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\)
hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\)
Vậy: \(x=\frac{9}{10}\)
b) Ta có: \(5\frac{4}{7}:x=13\)
\(\Leftrightarrow\frac{39}{7}:x=13\)
\(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\)
Vậy: \(x=\frac{3}{7}\)
c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\)
\(\Leftrightarrow x\cdot\frac{14}{5}=84\)
\(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\)
Vậy: x=30
d) Ta có: \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{-1}{15}\)
hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\)
Vậy: x=-5
e) Ta có: \(8\frac{2}{3}:x-10=-8\)
\(\Leftrightarrow\frac{26}{3}:x=2\)
hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\)
Vậy: \(x=\frac{13}{3}\)
g) Ta có: \(x+30\%=-1.3\)
\(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\)
hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\)
Vậy: \(x=\frac{-8}{5}\)
i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\)
\(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\)
\(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\)
\(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\)
Vậy: x=-9
k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\)
\(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\)
hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\)
Vậy: x=30
m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\)
\(\Leftrightarrow\left|2x-1\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)
A
1
LG
17 tháng 5 2017
\(\Rightarrow\)\(\left[3x+\dfrac{3}{2}+3-5x\right].\dfrac{1}{2}=-x:\dfrac{31}{5}\)
\(\Rightarrow\)\(\left[-2x+\dfrac{9}{2}\right].\dfrac{1}{2}=-x.\dfrac{5}{31}\)
\(\Rightarrow\)\(\dfrac{-2x}{2}+\dfrac{9}{4}=-x.\dfrac{5}{31}\)
\(\Rightarrow\)\(-x+\dfrac{9}{4}=-x.\dfrac{5}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{-x5}{31}+x\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{-x5+31x}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{x\left(-5+31\right)}{31}\)
\(\Rightarrow\dfrac{9}{4}=\dfrac{x.26}{31}\)
\(\Rightarrow\dfrac{9}{4}.31=x.26\)\(\Rightarrow\dfrac{279}{4}=x.26\)
\(\Rightarrow x=\dfrac{279}{4}:26=\dfrac{279}{4}.\dfrac{1}{26}\)
\(\Rightarrow x=\dfrac{279}{104}\)
Vậy x=\(\dfrac{279}{104}\)
25 tháng 9 2017
Ta có : \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\Leftrightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\\left(x-5\right)=-1;1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=4;6\end{cases}}\)
Vậy x = {4;5;6}.
b) Ta có : ax = a50
=> x = 50
d) Ta có : 1 + 2 + 3 + ..... + x = 222111
=> \(\frac{\left[\left(x-1\right):1+1\right]\left(x+1\right)}{2}=222111\)
=> \(\frac{x\left(x+1\right)}{2}=222111\)
=> x(x + 1) = 444222
=> x(x + 1) = 666.667
=> x = 666
Vậy x = 666.
HL
5 tháng 7 2017
a) \(x-25\%x=0,5\)
\(\dfrac{3}{4}x=0,5\)
x = \(\dfrac{2}{3}\)
b) \(\left(50\%x+5\dfrac{1}{4}\right).\dfrac{-2}{3}=2\dfrac{5}{6}\)
\(\left(0,5x+\dfrac{21}{4}\right)=\dfrac{-17}{4}\)
\(0,5x=\dfrac{-19}{2}\)
x = -19
c) \(\left(1\dfrac{1}{3}-25\%-\dfrac{5}{12}\right)+2x=1,5:\dfrac{3}{5}\)
\(\dfrac{2}{3}+2x=\dfrac{5}{2}\)
\(2x=\dfrac{11}{6}\)
x= \(\dfrac{11}{12}\)
9 tháng 3 2019
Bài 1:
\(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
= \(\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
= \(\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
=\(\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
=\(\frac{1}{26}+\frac{1}{27}+....+\frac{1}{26}\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)
......????
2 tháng 5 2019
\(1\frac{3}{8}:x=-5\frac{1}{2}\)
\(\frac{11}{8}:x=\frac{-11}{2}\)
\(x=\frac{11}{8}:\frac{-11}{2}\)
\(x=\frac{-1}{4}\)
9 bạn nhé
chúc bạn học tốt!!
9 NHA BẠN