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a. (5.x + 3) : 4 – 16 = 6
(5.x + 3) : 4 = 22
(5.x + 3) = 88
5.x = 85
x = 13
b. 7(23 – 3.x) : 2 – 35 = 35
7(23 – 3.x) : 2 = 70
7(23 – 3.x) = 35
7 . 3.x = 12
7.x = 4
x = 4/7
a.) (5.x + 3) : 4 – 16 = 6
=> (5.x + 3) : 4 = 22
=> 5.x + 3 = 88
=> 5x = 85
=> x = 17
b) 7(23 – 3.x) : 2 – 35 = 35
=> 7(23 – 3.x) : 2 = 70
=>7(23 – 3.x) = 140
=> 23 – 3.x = 20
=> 3x = 3
=> x = 1
a, \(x\) \(\times\) \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\) = \(\dfrac{5}{6}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) + \(\dfrac{3}{4}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{19}{12}\)
\(x\) = \(\dfrac{19}{12}\) : \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{19}{6}\)
b, \(x\) : \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\) = \(\dfrac{5}{6}\)
\(x\): \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) + \(\dfrac{3}{4}\)
\(x\) : \(\dfrac{1}{2}\) = \(\dfrac{19}{12}\)
\(x\) = \(\dfrac{19}{12}\) \(\times\) \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{19}{24}\)
c, \(x\) \(\times\) \(\dfrac{3}{4}\) + \(x\) \(\times\) \(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
\(x\) \(\times\) ( \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\)) = \(\dfrac{7}{8}\)
\(x\) \(\times\) 1 = \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{7}{8}\)
d, \(x\times\) \(\dfrac{3}{4}\) - \(x\) \(\times\) \(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
\(x\) \(\times\) ( \(\dfrac{3}{4}\) - \(\dfrac{1}{4}\)) = \(\dfrac{7}{8}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{7}{8}\) : \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{7}{4}\)
a, \(390-\left(x-7\right)=13^2:12\)
\(390-\left(x-7\right)=\) \(\dfrac{169}{12}\)
\(x-7=390-\dfrac{169}{12}\)
\(x-7=\dfrac{4511}{12}\)
\(x=\dfrac{4511}{12}+7\)
\(x=\dfrac{4595}{12}\)
Vậy ...
b, \(\left(x-35.2^2\right):7=3^3-24\)
\(\left(x-35.4\right):7=27-24\)
\(\left(x-140\right):7=3\)
\(\Leftrightarrow\left(x-140\right)=3.7\)
\(\Leftrightarrow x-140=21\)
\(\Leftrightarrow x=161\)
Vậy .....
c) \(x-6:2-\left(4^2.3-24\right):2:6=3\)
\(x-3-\left(16.3-24\right):2:6=3\)
\(x-3-\left(48-24\right):2:6=3\)
\(x-3-24:2:6=3\)
\(x-3-2=3\)
\(x=3+2+3\)
\(x=8\)
Vậy ......
d) \(4x-5=5+5^2+5^3+.....+5^{99}\)
Đặt :
\(A=5+5^2+.........+5^{99}\)
\(\Leftrightarrow5A=5^2+5^3+..........+5^{100}\)
\(\Leftrightarrow5A-A=\left(5^2+5^3+......+5^{100}\right)-\left(5+5^2+....+5^{99}\right)\)
\(\Leftrightarrow4A=5^{100}-5\)
\(\Leftrightarrow A=\dfrac{5^{100}-5}{4}\)
\(\Leftrightarrow4x+5=\dfrac{5^{100}-5}{4}\)
Đến đây thì sao nữa nhỉ ?
e) \(\left(2x-1\right)^4=625\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=5\\\left(2x-1\right)^4=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy ....
tốt nhất để tớ chả lời cho
a,4/7.x=1/5+2/3
4/7.x=13/15
13/15:4/7=91/60
b
=-7/8.x=1/3-2/9
-7/8.x=1/9
x=1/9:7/8
x=8/63
c
5/7:x=1/6-4/5
5/7:x=-19/30
x=-116/133
\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
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\(a,\frac{3}{35}-\left[\frac{3}{5}+x\right]=\frac{2}{7}\)
\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)
\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{2\cdot5}{35}\)
\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{10}{35}=\frac{-7}{35}\)
\(\Rightarrow x=\frac{-7}{35}-\frac{3}{5}=\frac{-7}{35}-\frac{3\cdot7}{35}=\frac{-7}{35}-\frac{21}{35}=\frac{-28}{35}=\frac{-4}{5}\)
\(b,\frac{5}{6}+\frac{1}{6}:x=\frac{3}{4}\)
\(\Rightarrow\frac{1}{6}:x=\frac{3}{4}-\frac{5}{6}\)
\(\Rightarrow\frac{1}{6}:x=\frac{18}{24}-\frac{20}{24}\)
\(\Rightarrow\frac{1}{6}:x=\frac{-2}{24}\)
\(\Rightarrow x=\frac{1}{6}:\frac{-2}{24}=\frac{1}{6}:\frac{-1}{12}=\frac{12}{-6}=\frac{-12}{6}=-2\)
\(c,\left|x-3\frac{4}{7}\right|=-0,5\)
Vì \(\left|x-3\frac{4}{7}\right|\ge0\)mà \(-0,5< 0\)nên \(x\in\varnothing\)
Vậy không có giá trị x thỏa mãn
a) 3/35-(3/5+x)=2/7
<=> 3/5+x=3/35-2/7
<=> 3/5+x=3/35-10/35
<=> 3/5+x=-7/35
<=> 3/5+x=-1/5
<=> x=-1/5-3/5
<=> x=-4/5
Vậy x=-4/5