Tìm x, biết:
a) x = 1 4 + 4 7
b) x − 1 = 7 2 + − 4 − 3
K
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GH
3
AH
Akai Haruma
Giáo viên
9 tháng 5 2021
Lời giải:
a)
$\frac{4}{7}x=\frac{2}{3}+\frac{1}{5}=\frac{13}{15}$
$x=\frac{13}{15}:\frac{4}{7}=\frac{91}{60}$
b)
$\frac{5}{7}:x=\frac{1}{6}-\frac{4}{5}$
$\frac{5}{7}:x=\frac{-19}{30}$
$x=\frac{5}{7}:\frac{-19}{30}=\frac{-150}{133}$
9 tháng 5 2021
a) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\)
\(\dfrac{4}{7}.x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{13}{15}\)
\(x=\dfrac{13}{15}:\dfrac{4}{7}\)
\(x=\dfrac{91}{60}\)
b) \(\dfrac{4}{5}+\dfrac{5}{7}:x=\dfrac{1}{6}\)
\(\dfrac{5}{7}:x=\dfrac{1}{6}-\dfrac{4}{5}\)
\(\dfrac{5}{7}:x=\dfrac{-19}{30}\)
\(x=\dfrac{5}{7}:\dfrac{-19}{30}\)
\(x=\dfrac{-150}{133}\)
VV
24 tháng 7 2021
a, \(x\) : \(\dfrac{13}{3}\) = -2,5
\(x\) = -2,5 . \(\dfrac{13}{3}\)
\(x\) = \(\dfrac{65}{6}\)
b,\(\dfrac{3}{5}\)\(x\) = \(\dfrac{1}{10}-\)\(\dfrac{1}{4}\)
\(\dfrac{3}{5}x\) = \(\dfrac{-3}{20}\)
\(x\) = \(\dfrac{-3}{20}\) : \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{-1}{4}\)
c, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)
\(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)
\(\dfrac{12}{13}x=2\)
\(x=2:\dfrac{12}{13}\)
\(x=\dfrac{13}{6}\)
HN
3
TG
18 tháng 6 2021
a) \(x+1^3=2^5-\left(-1^3\right)\)
\(\Rightarrow x+1=33\)
=> x = 32
b) \(3^7-x=1^4-\left(-3^5\right)\)
\(\Rightarrow2187-x=1+243=244\)
=> x = 1943
FH
2
QT
1
HQ
Hà Quang Minh
Giáo viên
7 tháng 10 2023
a) \(1\frac{2}{7} = 1 + \frac{2}{7} = \frac{9}{2}\)
\(\begin{array}{l}x:1\frac{2}{7} = - 3,5\\x:\frac{9}{7} = - \frac{7}{2}\\x = - \frac{7}{2}.\frac{9}{7}\\x = - \frac{9}{2}\end{array}\)
b) \(0,4.x - \frac{1}{5}.x = \frac{3}{4}\)
\(\begin{array}{l}\frac{2}{5}.x - \frac{1}{5}.x = \frac{3}{4}\\\left( {\frac{2}{5} - \frac{1}{5}} \right).x = \frac{3}{4}\\\frac{1}{5}.x = \frac{3}{4}\\x = \frac{3}{4}:\frac{1}{5}\\x = \frac{3}{4}.5\\x = \frac{{15}}{4}\end{array}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}x + \frac{1}{2} = - \frac{1}{3}\\x = - \frac{1}{3} - \frac{1}{2}\\x = - \frac{2}{6} - \frac{3}{6}\\x = \frac{{ - 5}}{6}\end{array}\)
Vậy \(x = \frac{{ - 5}}{6}\).
b)
\(\begin{array}{l}\left( { - \frac{2}{7}} \right) + x = - \frac{1}{4}\\x = - \frac{1}{4} - \left( { - \frac{2}{7}} \right)\\x = - \frac{1}{4} + \frac{2}{7}\\x = - \frac{7}{{28}} + \frac{8}{{28}}\\x = \frac{1}{{28}}\end{array}\)
Vậy \(x = \frac{1}{{28}}\).
KN
1
MN
2 tháng 5 2021
a, \(\left(\dfrac{1}{2}+\dfrac{4}{7}\right):x=\dfrac{-3}{4}\)
\(\dfrac{15}{14}:x=\dfrac{-3}{4}\)
=> x= \(\dfrac{-7}{10}\)
b, 0,5:x-\(1\dfrac{3}{4}\)= 25%
0,5:x-\(\dfrac{7}{4}=\dfrac{1}{4}\)
0,5:x = 2
=> x = \(\dfrac{1}{4}\)
L
1 tháng 8 2021
a) \(\left(x-4\right)^2-\left(x-4\right)=0\)
\(\left(x-4\right)\left(x-4-1\right)=0\)
\(\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
b) \(5x^2\left(x-7\right)+7\left(x-7\right)=0\)
\(\left(x-7\right)\left(5x^2+7\right)=0\)
\(\left[{}\begin{matrix}x-7=0\\5x^2+7=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x^2=\dfrac{-7}{5}\end{matrix}\right.\)
\(x=7\)
c) \(x^2\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right)\left(x^2-1\right)=0\)
\(\left[{}\begin{matrix}x=3\\x=\pm1\end{matrix}\right.\)
1 tháng 8 2021
a) (x - 4)^2=(x - 4)
(x - 4) (x -4)=(x -4 )
(x - 4) (x - 4)-(x - 4)=0
(x-4) (x-4-1)=0
(x-4) (x-5)=0
TH1:x-4=0 TH2:x-5=0
x=4 x=5
a) x = 1 4 + 4 7 ⇔ x = 1.7 + 4.4 28 = 23 28 ⇔ x = − 23 28 x = 23 28
b) x − 1 = 7 2 + − 4 − 3 ⇔ x − 1 = 29 6 ⇔ x − 1 = 29 6 x − 1 = − 29 6 ⇔ x = 35 6 x = − 23 6