a/317+(600-x)=517
b/325-(x+16)=425:17
c/25*(x-1)=0
d/713:(2x+17)=23
e/(x+a/3)*7/4=5-7/6
K
Khách
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CT
4
3 tháng 7 2016
a) \(435-\left(x+16\right)=425:17\)
\(435-\left(x+16\right)=25\)
\(\Rightarrow x+16=435-25=410\)
\(\Rightarrow x=410-16=394\)
b) \(\left(x+\frac{4}{3}\right)\cdot\frac{7}{4}=5-\frac{7}{6}\)
\(\left(x+\frac{4}{3}\right)\cdot\frac{7}{4}=\frac{23}{6}\)
\(\Rightarrow x+\frac{4}{3}=\frac{23}{6}:\frac{7}{4}=\frac{46}{21}\)
\(\Rightarrow x=\frac{46}{21}-\frac{4}{3}=\frac{6}{7}\)
Ủng hộ mk nka!!!^_^^_^^_^
3 tháng 7 2016
a) 435 - ( x + 16 ) = 25
x+ 16 = 435 - 25
x + 16 = 410
x = 410 - 16
x = 394
b) (x+4/3 ) * 7/4 = 23/6
x+4/3= 23/6 : 7/4
x+4/3 = 46/21
x = 46/21 - 4/3
x = 6/7
MH
5
RM
24 tháng 7 2017
a) \(435-\left(x+16\right)=425:17\)
\(435-\left(x+16\right)=25\)
\(x+16=435-25\)
\(x+16=410\)
\(x=410-16\)
\(x=394\)
b) \(\left(x+\frac{3}{4}\right)\times\frac{7}{4}+\frac{7}{6}=5\)
\(\left(x+\frac{3}{4}\right)\times\frac{7}{4}=5-\frac{7}{6}\)
\(\left(x+\frac{3}{4}\right)\times\frac{7}{4}=\frac{23}{6}\)
\(x+\frac{3}{4}=\frac{23}{6}:\frac{7}{4}\)
\(x+\frac{3}{4}=\frac{46}{21}\)
\(x=\frac{46}{21}-\frac{3}{4}\)
\(x=\frac{121}{84}\)
24 tháng 7 2017
435 - ( x + 16 ) = 425 : 17
435 - ( x + 16 ) = 25
x + 16 = 435 - 25
x + 16 = 410
x = 410 - 16
x = 394
TN
28 tháng 7 2018
\(2\frac{3}{4}.\frac{1}{2}-\frac{1}{2}.\frac{3}{7}+\frac{1}{3}\)
\(=\frac{11}{4}.\frac{1}{2}-\frac{1}{2}.\frac{3}{7}+\frac{1}{3}\)
\(=\frac{11}{8}-\frac{3}{14}+\frac{1}{3}\)
\(=\frac{251}{168}\)
T
28 tháng 7 2018
Bài 1 : a, thực hiện phép tính :
\(2\frac{3}{4}×\frac{1}{2}-\frac{1}{2}×\frac{3}{7}+\frac{1}{3}\)
\(=\frac{11}{4}×\frac{1}{2}-\frac{1}{2}×\frac{3}{7}+\frac{1}{3}\)
\(=\frac{1}{2}×\left(\frac{11}{4}-\frac{3}{7}\right)+\frac{1}{3}\)
\(=\frac{1}{2}×\frac{65}{28}+\frac{1}{3}\)
\(=\frac{65}{56}+\frac{1}{3}\)
\(=\frac{251}{168}\)
b , Tìm x biết :
a, 435- ( x + 16 ) = 425 : 17
435 - ( x + 16 ) = 25
x + 16 = 435 - 25
x + 16 = 410
x = 410 - 16
x = 394
Vậy x = 394
b, ( x + 3/4 ) × 7/4 = 5 - 7/6
( x + 3/4 ) × 7/4 = 23/6
x + 3/4 = 23/6 : 7/4
x + 3/4 = 23/6 × 4/7
x + 3/4 = 46/21
x = 46/21 - 3/4
x = 121/84
Vậy x = 121/84
17 tháng 12 2022
a: \(\Leftrightarrow4^{x-5}\cdot17=68\)
=>4^x-5=4
=>x-5=1
=>x=6
b: \(\Leftrightarrow\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}+\dfrac{2}{3}=1\)
=>|2x-1|=1/3
=>2x-1=1/3 hoặc 2x-1=-1/3
=>x=2/3 hoặc x=1/3
c: =>|2x-2|=|3x+15|
=>3x+15=2x-2 hoặc 3x+15=-2x+2
=>x=-17 hoặc x=-13/5
BT
2 tháng 6 2017
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
9 tháng 7 2023
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6-0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
f) \(25+\left(15-x\right)=30\)
\(\Rightarrow25+15-x=30\)
\(\Rightarrow40-x=30\)
\(\Rightarrow x=40-30\)
\(\Rightarrow x=10\)
g) \(43-\left(24-x\right)=20\)
\(\Rightarrow43-24+x=20\)
\(\Rightarrow19+x=20\)
\(\Rightarrow x=20-19\)
\(\Rightarrow x=1\)
h) \(2\left(x-5\right)-17=25\)
\(\Rightarrow2\left(x-5\right)=17+25\)
\(\Rightarrow x-5=21\)
\(\Rightarrow x=21+5\)
\(\Rightarrow x=26\)
i) \(3\left(x+7\right)-15=27\)
\(\Rightarrow3\left(x+7\right)=27+15\)
\(\Rightarrow x+7=14\)
\(\Rightarrow x=14-7\)
\(\Rightarrow x=7\)
j) \(15+4\left(x-2\right)=95\)
\(\Rightarrow4\left(x-2\right)=95-15\)
\(\Rightarrow4\left(x-2\right)=80\)
\(\Rightarrow x-2=20\)
\(\Rightarrow x=20+2\)
\(\Rightarrow x=22\)
k) \(20-\left(x+14\right)=5\)
\(\Rightarrow x+14=20-5\)
\(\Rightarrow x+14=15\)
\(\Rightarrow x=15-14\)
\(\Rightarrow x=1\)
l) \(14+3\left(5-x\right)=27\)
\(\Rightarrow3\left(5-x\right)=27-14\)
\(\Rightarrow3\left(5-x\right)=13\)
\(\Rightarrow5-x=\dfrac{13}{3}\)
\(\Rightarrow x=5-\dfrac{13}{3}\)
\(\Rightarrow x=\dfrac{2}{3}\)
28 tháng 10 2023
a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)
=>\(\left(x-2\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)
mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)
nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)
=>\(2^x\left(1+2+2^2+2^3\right)=120\)
=>\(2^x\cdot15=120\)
=>\(2^x=8\)
=>x=3
e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
