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22 tháng 6 2016
\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)
=\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)
= \(-\frac{35}{27}+\frac{47}{21}\)
= \(\frac{178}{189}\)
c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)
= \(\frac{117}{13}-\frac{311}{65}\)
= \(\frac{274}{65}\)
d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
NT
1
30 tháng 8 2019
a) 2x.3/5 - 5x = 3/2 - 7x
6x/5 - 5x = 3/2 - 7x
-19x/5 = 3/2 - 7x
-19x = 15/2 - 35x
-19x + 35x = 15/2
16x = 15/2
x = 15/2 : 16
x = 15/32
b) (x - 1/5)2 = 4/25
(x - 1/5)2 = (+-2/5)2
x - 1/5 = +-2/5
x - 1/5 = 2/5 hoặc x - 1/5 = -2/5
x = 2/5 + 1/5 x = -2/5 + 1/5
x = 3/5 x = -1/5
NN
26 tháng 12 2017
\(a,\left(\dfrac{2}{3}\right)^x=\left(\dfrac{4}{9}\right)^4.\)
\(\left(\dfrac{2}{3}\right)^x=\left[\left(\dfrac{2}{3}\right)^2\right]^4.\)
\(\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^8\Rightarrow x=8.\)
Vậy.....
\(b,\left(2x-1\right)^2=25.\)
\(\left(2x-1\right)^2=\left(\pm5\right)^2.\)
\(\Rightarrow\left(2x-1\right)=\pm5.\)
+) Xét \(2x-1=5\), ta có:
\(2x-1=5.\)
\(\Rightarrow2x=6.\)
\(\Rightarrow x=3.\)
+) Xét \(2x-1=-5\), ta có:
\(2x-1=-5.\)
\(\Rightarrow2x=-4.\)
\(\Rightarrow x=-2.\)
Vậy.....
NL
6
18 tháng 7 2019
\(a,\left(x-3\right)^2=1\)
=> \(\sqrt{\left(x-3\right)^2}=\sqrt{1}\)
=> \(\left|x-3\right|=1\)
=> \(\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1+3=4\\x=-1+3=2\end{matrix}\right.\)
Vậy \(x\in\left\{4;2\right\}\)
\(\left(2x+1\right)^3=-8\)
=> \(\sqrt[3]{\left(2x+1\right)^3}=\sqrt[3]{-8}\)
=> \(2x+1=-2\)
=> \(2x=-2-1=-3\)
=> \(x=-3:2=-\frac{3}{2}\)
Vậy \(x\in\left\{-\frac{3}{2}\right\}\)
\(c,\left(x-\frac{1}{4}\right)^2=\frac{1}{25}\)
=> \(\sqrt{\left(x-\frac{1}{4}\right)^2}=\sqrt{\frac{1}{25}}\)
=> \(\left|x-\frac{1}{4}\right|=\frac{1}{5}\)
=> \(\left[{}\begin{matrix}x-\frac{1}{4}=\frac{1}{5}\\x-\frac{1}{4}=-\frac{1}{5}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{1}{5}+\frac{1}{4}=\frac{9}{20}\\x=-\frac{1}{5}+\frac{1}{4}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{9}{20};\frac{1}{20}\right\}\)
7 tháng 8 2017
\(2x-15=21\Rightarrow2x=36\Rightarrow x=18\)
\(2\left|x+2\right|+7=25\Rightarrow2\left|x+2\right|=18\Rightarrow\left|x+2\right|=9\)
\(\Rightarrow\left[{}\begin{matrix}x+2=9\Rightarrow x=7\\x+2=-9\Rightarrow x=-11\end{matrix}\right.\)
\(3x+12=2x-4\)
\(\Rightarrow3x=2x-16\Rightarrow-x=16\Rightarrow x=-16\)
\(\left|2x-5\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=1\Rightarrow2x=6\Rightarrow x=3\\2x-5=-1\Rightarrow2x=4\Rightarrow x=2\end{matrix}\right.\)
\(2\left|x-5\right|+3=4\Rightarrow2\left|x-5\right|=1\Rightarrow\left|x-5\right|=0,5\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0,5\Rightarrow x=5,5\\x-5=-0,5\Rightarrow x=4,5\end{matrix}\right.\)
7 tháng 8 2017
a. \(2x-15=21\\ \Leftrightarrow2x=36\\ \Leftrightarrow x=18\)
c. \(3x+12=2x-4\Leftrightarrow x=-16\\ \)
ND
2
8 tháng 7 2016
a, (2x-3)3 = -64
=> (2x-3)3 = -43
=> 2x-3=-4
=> 2x = -1
=> x = -1 : 2
=> x = -1/2
b, (2x-3)2 =25
=> (2x-3)2 =5^2
=> 2x-3 = 5
=> 2x = 8
=> x = 4
c, (3x-4)2 =36
=> (3x-4)2 =62
=> 3x-4 = 6
=> 3x = 10
=> x = 3.(3)
d, 2x+1 = 64
=> 2x+1 = 26
=> x+1 = 6
=> x = 5
NT
0
S
15 tháng 8 2018
a, \(2x\left(x-5\right)-x\left(2x+3\right)=25\)
\(\Rightarrow2x^2-10x-2x^2-3x=25\)
\(\Rightarrow-13x=25\Rightarrow x=\dfrac{-25}{13}\)
b, \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\dfrac{5}{2}\)
\(\Rightarrow3y^3-y^2+y-3y^2+y-1+4y^2-3y^3=\dfrac{5}{2}\)
\(\Rightarrow2y-1=\dfrac{5}{2}\Rightarrow2y=\dfrac{7}{2}\Rightarrow y=\dfrac{7}{4}\)
c, \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Rightarrow2x^2+3\left(x^2+x-x-1\right)=5x^2+5x\)
\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Rightarrow-5x=3\Rightarrow x=\dfrac{-3}{5}\)
12 tháng 1 2023
1: \(=\dfrac{1}{4}:\dfrac{-1}{4}-2\cdot\dfrac{-1}{8}+5-4\)
\(=-1+1+\dfrac{1}{4}=\dfrac{1}{4}\)
2: \(=5^{20}\cdot\dfrac{1}{5^{20}}+\left(\dfrac{3}{8}\cdot\dfrac{4}{3}\right)^8-1=1-1+\dfrac{1}{2}^8=\dfrac{1}{2^8}\)
\(\left(2x-3\right)^2=\dfrac{4}{25}\\ \Rightarrow\left(2x-3\right)^2=\left(\pm\dfrac{2}{5}\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{2}{5}\\2x-3=-\dfrac{2}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=\dfrac{17}{5}\\2x=\dfrac{13}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
\(\left(2x-3\right)^2=\dfrac{4}{25}\\ \Leftrightarrow\left(2x-3\right)^2=\left(\dfrac{2}{5}\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{2}{5}\\-2x+3=\dfrac{2}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{5}\\-2x=-\dfrac{13}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
Vậy...