[ x nhân 1/2 +2] :3/5 + 4 =9
x nhân 2/3 - 1/5=2/5+1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
10 tháng 1 2023
a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)
b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)
c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)
d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)
BD
15 tháng 6 2023
\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)
\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)
\(=\dfrac{2}{5}\times\dfrac{3}{7}\)
\(=\dfrac{6}{35}\)
\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(x+\dfrac{1}{6}=\dfrac{3}{4}\)
\(x=\dfrac{9}{12}-\dfrac{2}{12}\)
\(x=\dfrac{7}{12}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)
\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)
\(\dfrac{1}{2020}+x=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{1}{2020}\)
\(x=\dfrac{1010}{2020}-\dfrac{1}{2020}\)
\(x=\dfrac{1009}{2020}\)
15 tháng 6 2023
\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)
\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)
\(=\dfrac{2}{5}\times\dfrac{3}{7}\)
\(=\dfrac{6}{35}\)
\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}-x\)
\(\Rightarrow\dfrac{3}{4}-x=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1\times2\times3\times4\times...\times2019}{2\times3\times4\times5\times...\times2020}+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2020}+x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-\dfrac{1}{2020}=\dfrac{1009}{2020}\)
4 tháng 10 2018
22 + (x - 3) = 52
4 + (x - 3) = 25
x - 3 = 25 - 4
x - 3 = 21
x = 21 + 3
x = 24
6 tháng 10 2018
22 + ( x - 3 ) = 52
<=> 4 + ( x - 3 ) = 25
<=> x - 3 = 21
<=> x = 24
9x - 2 . 32 = 34
<=>9x - 2 = 34 : 32
<=>9x - 2 = 9
<=>9x = 11
<=> x = 11/9
10x + 22 . 5 = 102
<=> 10x + 4 . 5 = 100
<=> 10x + 4 = 20
<=> 10x = 16
<=> x = 1,6
14 tháng 5 2022
X x \(\dfrac{3}{4}\)+ X x\(\dfrac{1}{5}\)+ X x \(\dfrac{1}{20}\)+ X= 1000
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
Câu 1:
a. $3\frac{2}{3}+2\frac{1}{2}=(3+2)+(\frac{2}{3}+\frac{1}{2})=5+\frac{7}{6}=6+\frac{1}{6}=6\frac{1}{6}$
b. \(2\frac{1}{2}\times 3\frac{2}{5}=\frac{5}{2}\times \frac{17}{5}=\frac{17}{2}\)
c.
\(3\frac{1}{3}: 4\frac{1}{4}=\frac{10}{3}: \frac{17}{4}=\frac{40}{51}\)
d.
\(3\frac{1}{2}+4\frac{5}{7}-5\frac{5}{14}=(3+4-5)+(\frac{1}{2}+\frac{5}{7}-\frac{5}{14})=2+\frac{6}{7}=2\frac{6}{7}\)
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
Câu 2:
a. $x\times \frac{2}{7}=\frac{6}{11}$
$x=\frac{6}{11}: \frac{2}{7}=\frac{21}{11}$
b. $x: \frac{3}{2}=\frac{1}{4}$
$x=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$
28 tháng 8 2019
\(\frac{1}{9}.3^4.3^x=3^7\)
\(\Leftrightarrow3^x=3^7:\frac{1}{9}:3^4=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
1 tháng 10 2020
1) \(x^3+2x-3\)
\(=\left(x^3-x^2\right)+\left(x^2-x\right)+\left(3x-3\right)\)
\(=x^2\left(x-1\right)+x\left(x-1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+3\right)\)
2) \(x^3-6x+4\)
\(=\left(x^3-2x^2\right)+\left(2x^2-4x\right)-\left(2x-4\right)\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x-2\right)\)
1 tháng 10 2020
3) \(x^3-2x^2+1\)
\(=\left(x^3-x^2\right)-\left(x^2-x\right)-\left(x-1\right)\)
\(=x^2\left(x-1\right)-x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-x-1\right)\)
4) \(x^3+5x^2-12\)
\(=\left(x^3+2x^2\right)+\left(3x^2+6x\right)-\left(6x+12\right)\)
\(=x^2\left(x+2\right)+3x\left(x+2\right)-6\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+3x-6\right)\)
Ta có : \(\left[x\cdot\frac{1}{2}+2\right]:\frac{3}{5}+4=9.\)
=> \(\left[x\cdot\frac{1}{2}+2\right]:\frac{3}{5}=9-4.\)
=> \(\left[x\cdot\frac{1}{2}+2\right]:\frac{3}{5}=5.\)
=> \(x\cdot\frac{1}{2}+2=5\cdot\frac{3}{5}.\)
=> \(x\cdot\frac{1}{2}+2=3.\)
=> \(x\cdot\frac{1}{2}=3+2.\)
=> \(x\cdot\frac{1}{2}=5.\)
=> \(x=5:\frac{1}{2}.\)
=> \(x=5\cdot2.\)
=> \(x=10.\)
Vậy x = 10 .
a)\(\left[x\times\frac{1}{2}+2\right]:\frac{3}{5}+4=9\) b)\(x\times\frac{2}{3}-\frac{1}{5}=\frac{2}{5}+1\)
\(\left[x\times\frac{1}{2}+2\right]:\frac{3}{5}=9-4\) \(x\times\frac{2}{3}-\frac{1}{5}=\frac{7}{5}\)
\(\left[x\times\frac{1}{2}+2\right]:\frac{3}{5}=5\) \(x\times\frac{2}{3}=\frac{7}{5}+\frac{1}{5}\)
\(x\times\frac{1}{2}+2=5\times\frac{3}{5}\) \(x\times\frac{2}{3}=\frac{8}{5}\)
\(x\times\frac{1}{2}+2=3\) \(x=\frac{8}{5}:\frac{2}{3}\)
\(x\times\frac{1}{2}=3-2\) \(x=\frac{8}{5}\times\frac{3}{2}\)
\(x\times\frac{1}{2}=1\) \(x=\frac{12}{5}\)
\(x=1:\frac{1}{2}\)
\(x=2\)