\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)
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LH
2
HN
10 tháng 8 2021
\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)
\(\Leftrightarrow\dfrac{1680.\left(x-3\right)+1560.\left(x-3\right)-1456.\left(x-3\right)-1365.\left(x-3\right)}{21840}=0\)
\(\Leftrightarrow\left(x-3\right).\left(1680+1560-1456-1365\right)=0\)
\(\Leftrightarrow\left(x-3\right).419=0\)
\(\Leftrightarrow419x=1257\)
\(\Leftrightarrow x=3\)
AH
Akai Haruma
Giáo viên
10 tháng 8 2021
Lời giải:
\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\((x-3)\left(\frac{1}{13}+\frac{1}{14}\right)=(x-3)\left(\frac{1}{15}+\frac{1}{16}\right)\)
\((x-3)\left[\left(\frac{1}{13}+\frac{1}{14}\right)-\left(\frac{1}{15}+\frac{1}{16}\right)\right]=0\)
Ta thấy:
\(\frac{1}{13}>\frac{1}{15}; \frac{1}{14}>\frac{1}{16}\Rightarrow \frac{1}{13}+\frac{1}{14}> \frac{1}{15}+\frac{1}{16}\)
Do đó biểu thức trong ngoặc vuông lớn hơn $0$ hay khác $0$
$\Rightarrow x-3=0$
$\Leftrightarrow x=3$
17 tháng 7 2021
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
17 tháng 7 2021
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
14 tháng 6 2021
Giải:
\(9-3\times\left(x-9\right)=6\)
\(3\times\left(x-9\right)=9-6\)
\(3\times\left(x-9\right)=3\)
\(x-9=3:3\)
\(x-9=1\)
\(x=1+9\)
\(x=10\)
\(4+6\times\left(x+1\right)=70\)
\(6\times\left(x+1\right)=70-4\)
\(6\times\left(x+1\right)=66\)
\(x+1=66:6\)
\(x+1=11\)
\(x=11-1\)
\(x=10\)
\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\)
\(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\)
\(\dfrac{x}{13}=\dfrac{4}{13}\)
\(\Rightarrow x=4\)
\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{3}{7}\)
\(\Rightarrow x=7\)
\(5\times\left(3+7\times x\right)=40\)
\(3+7\times x=40:5\)
\(3+7\times x=8\)
\(7\times x=8-3\)
\(7\times x=5\)
\(x=5:7\)
\(x=\dfrac{5}{7}\)
\(x\times6+12:3=120\)
\(x\times6+4=120\)
\(x\times6=120-4\)
\(x\times6=116\)
\(x=116:6\)
\(x=\dfrac{58}{3}\)
\(x\times3,7+x\times6,3=120\)
\(x\times\left(3,7+6,3\right)=120\)
\(x\times10=120\)
\(x=120:10\)
\(x=12\)
\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\)
\(\left(360-x\right):0,25=400\)
\(360-x=400.0,25\)
\(360-x=100\)
\(x=360-100\)
\(x=260\)
\(71+65\times4=\dfrac{x+140}{x}+260\)
\(\left(x+140\right):x+260=71+260\)
\(x:x+140:x+260=331\)
\(1+140:x+260=331\)
\(140:x=331-1-260\)
\(140:x=70\)
\(x=140:70\)
\(x=2\)
\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\)
\(10\times x+\left(1+4+7+...+28\right)=155\)
Số số hạng \(\left(1+4+7+...+28\right)\) :
\(\left(28-1\right):3+1=10\)
Tổng dãy \(\left(1+4+7+...+28\right)\) :
\(\left(1+28\right).10:2=145\)
\(\Rightarrow10\times x+145=155\)
\(10\times x=155-145\)
\(10\times x=10\)
\(x=10:10\)
\(x=1\)
Đều theo cách lớp 5 nha em!
24 tháng 9 2023
`#3107`
`7/13 \div x = 14/39`
`=> x = 7/13 \div 14/39`
`=> x = 3/2`
Vậy, `x = 3/2`
________
`x \times 3/5 = 14/15`
`=> x = 14/15 \div 3/5`
`=> x = 14/9`
Vậy, `x=14/9.`
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
\(\dfrac{7}{13}:x=\dfrac{14}{39}\)
\(x=\dfrac{7}{13}:\dfrac{14}{39}\)
\(x=\dfrac{7}{13}\times\dfrac{39}{14}\)
\(x=\dfrac{3}{2}\)
___
\(x\times\dfrac{3}{5}=\dfrac{14}{15}\)
\(x=\dfrac{14}{15}:\dfrac{3}{5}\)
\(x=\dfrac{14}{15}\times\dfrac{5}{3}\)
\(x=\dfrac{14}{9}\)
VT
1
9 tháng 1 2022
\(\dfrac{x+1}{15}+\dfrac{x+2}{14}=\dfrac{x+3}{13}+\dfrac{x+4}{12}\\ \Rightarrow\left(\dfrac{x+1}{15}+1\right)+\left(\dfrac{x+2}{14}+1\right)-\left(\dfrac{x+3}{13}+1\right)-\left(\dfrac{x+4}{12}+1\right)=0\\ \dfrac{x+16}{15}+\dfrac{x+16}{14}-\dfrac{x+16}{13}-\dfrac{x+16}{12}=0\\ \left(x+16\right)\left(\dfrac{1}{15}+\dfrac{1}{14}-\dfrac{1}{13}-\dfrac{1}{12}\right)=0\\ x+16=0\\ x=-16\)
H
20 tháng 9 2023
\(\dfrac{15-x}{14}=\dfrac{5}{7}\\ \left(15-x\right)\times7=5\times14\\ \left(15-x\right)\times7=70\\ 15-x=70:7\\ 15-x=10\\ x=15-10\\ x=5\)
__
\(\dfrac{x}{9}=\dfrac{17}{3}\\ x\times3=17\times9\\ x\times3=153\\ x=153:3\\ x=51\)
__
\(15-\dfrac{x}{2}=\dfrac{3}{7}\\ \dfrac{x}{2}=15-\dfrac{3}{7}\\ \dfrac{x}{2}=\dfrac{102}{7}\\ x\times7=102\times2\\ x\times7=204\\ x=204:7\\ x=\dfrac{204}{7}\)
17 tháng 4 2021
a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
17 tháng 4 2021
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
26 tháng 6 2022
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
13 tháng 7 2023
a: =>x-5/14=6/14-1/14=5/14
=>x=10/14=5/7
b; =>2/9:x=3/6+4/6=7/6
=>x=2/9:7/6=2/9*6/7=4/21
c: =>32:x=8
=>x=4
NN
13 tháng 7 2023
`@` `\text {Answer}`
`\downarrow`
`a,`
`x - 5/14 = 3/7 - 1/14`
`x - 5/14 = 5/14`
`=> x = 5/14 + 5/14`
`=> x = 5/7`
Vậy, `x = 5/7`
`b,`
`2/9 \div x = 1/2 + 2/3`
`2/9 \div x = 7/6`
`x = 2/9 \div 7/6`
`x = 4/21`
Vậy, `x = 4/21`
`c,`
\(\dfrac{6}{32\div x}=\dfrac{12}{16}\)
`6/(32 \div x) = 3/4`
`32 \div x = 6 \div 3/4`
`32 \div x = 8`
` x = 32 \div 8`
`x = 4`
Vậy, `x = 4`
`(x-3)/13+(x-3)/14=(x-3)/15+(x-3)/16`
`=>(x-3).(1/13+1/14-1/15-1/16)=0`
`=>x-3=0`
`=>x=3`
\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)
\(\Leftrightarrow\dfrac{x-3}{13}+\dfrac{x-3}{14}-\dfrac{x-3}{15}-\dfrac{x-3}{16}=0\)
\(\Leftrightarrow\left(x-3\right).\left(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
\(\Leftrightarrow x-3=0\)( Vì \(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}>0\))
\(\Leftrightarrow x-3=0\Rightarrow x=3\)