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\(\frac{x}{4}=\frac{3}{2}\)
\(\Rightarrow\frac{x}{4}=\frac{6}{4}\)
\(\Rightarrow x=6\)
vậy_
b)\(\frac{2}{x}=\frac{x}{8}\)
\(\Rightarrow x^2=2\cdot8\)
\(x^2=16\Rightarrow x=4\)
c) \(\frac{x+3}{4}=\frac{5}{3}\)
\(3\left(x+3\right)=4\cdot5\)
\(3x+9=20\)
\(3x=11\)
\(x=\frac{11}{3}\)
a) \(\left(x+5\right)^3=64\)
\(\Leftrightarrow\left(x+5\right)^3=4^3\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)
Vậy x = - 1
b) \(x:\left(-\frac{3}{5}\right)^2=-\frac{3}{5}\)
\(\Leftrightarrow x=\left(-\frac{3}{5}\right)^2.\left(-\frac{3}{5}\right)\)
\(\Leftrightarrow x=\left(-\frac{3}{5}\right)^3\)
\(\Leftrightarrow x=-0,216\)
Vậy x = - 0, 216
c) \(\left(\frac{4}{7}\right)^4.x=\left(\frac{4}{7}\right)^6\)
\(\Leftrightarrow x=\left(\frac{4}{7}\right)^6:\left(\frac{4}{7}\right)^4\)
\(\Leftrightarrow x=\left(\frac{4}{7}\right)^2\)
\(\Leftrightarrow\text{x}=\frac{16}{49}\)
Vậy x = 16/49
d) \(\left(-\frac{1}{3}\right)^3x=\frac{1}{81}\)
\(\Leftrightarrow-\frac{1}{27}x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left(-\frac{1}{27}\right)\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy x = - 1/3
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
1: x=3/4-1/2=3/4-2/4=1/4
2: x-1/5=2/11
=>x=2/11+1/5=21/55
3: x-5/6=16/42-8/56
=>x-5/6=8/21-4/28=5/21
=>x=5/21+5/6=15/14
4: x/5=5/6-19/30
=>x/5=25/30-19/30=6/30=1/5
=>x=1
5: =>|x|=1/3+1/4=7/12
=>x=7/12 hoặc x=-7/12
6: x=-1/2+3/4
=>x=3/4-1/2=1/4
11: x-(-6/12)=9/48
=>x+1/2=3/16
=>x=3/16-1/2=-5/16
1)x= 1/4
2)x= 2/11+ 1/5
x= 21/55
3)x - 5/6 = 5/21
x = 5/21+5/6
x = 15/14
4)x/5 = 5/6 + -19/30
x:5 = 1/5
x = 1/5.5
x = 1
5) |x| - 1/4 = 6/18
|x| = 6/18 - 1/4
|x| =7/12
⇒x= 7/12 hoặc -7/12
6)x = -1/2 +3/4
x= 1/4
7) x/15 = 3/5 + -2/3
x:15 = -1/15
x = -1/15. 15
x = -1
8)11/8 + 13/6 = 85/x
85/24 = 85/x
⇒ x = 24
9) x - 7/8 = 13/12
x = 13/12 + 7/8
x = 47/24
10)x - -6/15 = 4/27
x = 4/27 + (-6/15)
x = -34/135
11) -(-6/12)+x = 9/48
x= 9/48 - 6/12
x = -5/16
12) x - 4/6 = 5/25 + -7/15
x -4/6 = -4/15
x = -4/15 + 4/6
x = 2/5
a, (-0,2)2 \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)
= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)
= 0,2 - \(\dfrac{2}{3}\)
= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)
= - \(\dfrac{7}{15}\)
b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)
= \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)
= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)
= \(\dfrac{1}{2}\)
a) |x|-7,4 = 2,6
|x| = 2,6+7,4
|x| = 10 <=> x= -10 hoặc x=10
b) |x| = 3,9 <=> x= -3,9 hoặc x=3,9
c) \(\left|x\right|+\dfrac{2}{3}+\dfrac{7}{6}=\dfrac{13}{6}\)
\(\left|x\right|+\dfrac{11}{6}=\dfrac{13}{6}\)
\(\left|x\right|=\dfrac{13}{6}-\dfrac{11}{6}\)
\(\left|x\right|=\dfrac{2}{6}=\dfrac{1}{3}\) <=> x = \(\dfrac{-1}{3}\) hoặc x= \(\dfrac{1}{3}\)
d) \(\left|4^x+2\right|=2^6.\dfrac{5}{6}\)
=> \(\left|4^x+2\right|=\dfrac{160}{3}\)
\(\Rightarrow\left[{}\begin{matrix}TH1:4^x+2=\dfrac{160}{3}\\TH2:4^x+2=\dfrac{-160}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4^x=\dfrac{154}{3}\\4^x=\dfrac{-166}{3}\end{matrix}\right.\) => x ϵ ∅
e) \(\left|x^9\right|=8^3\)
|x9| = (23)3
|x9| = 29
\(\Leftrightarrow\left\{{}\begin{matrix}x^9=2^9\\x^9=-2^9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy x=-2 hoặc x=2
a,\(\left|x\right|-7,4=2,6\)
\(\left|x\right|=10\)
Vậy \(x=10\) hoặc \(x=-10\)
b,\(\left|x\right|=3,9\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3,9\\x=-3,9\end{matrix}\right.\)
Vậy...
c,\(\left|x\right|+\dfrac{2}{3}+\dfrac{7}{6}=\dfrac{13}{6}\)
\(\left|x\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy...
d,Mình chưa hiểu rõ đề câu này \(2^6\dfrac{5}{6}\) là gì ạ?
e,Chưa hiểu đề luôn.
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