Tìm x:
a) 2\(\dfrac{1}{2}\). x + x = 2\(\dfrac{1}{17}\)
b) x - 3\(\dfrac{1}{2}\). x = -2\(\dfrac{6}{7}\)
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DT
Dang Tung
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26 tháng 2
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23 tháng 1 2022
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{y+2}+\dfrac{1}{y+2}=\dfrac{1}{2}-\dfrac{1}{18}\\\dfrac{2}{x-1}-\dfrac{1}{y+2}=\dfrac{1}{18}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=9\\\dfrac{2}{x-1}=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=9\\x-1=12\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(13;7\right)\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 12 2023
a, 12 - (2\(x^2\) - 3) = 7
2\(x^2\) - 3 = 12 - 7
2\(x^2\) - 3 = 5
2\(x^2\) = 8
\(x^2\) = 4
\(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
15 tháng 5 2022
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 8 2023
b, \(x+\dfrac{2}{3}\) = \(\dfrac{3}{5}\) - \(\dfrac{-1}{6}\)
\(x+\dfrac{2}{3}\) = \(\dfrac{23}{30}\)
\(x\) = \(\dfrac{23}{30}\) - \(\dfrac{2}{3}\)
\(x\) = \(\dfrac{1}{10}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 8 2023
a, \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x\) = \(\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}-x\) = \(\dfrac{1}{4}\)
x = \(\dfrac{2}{3}-\dfrac{1}{4}\)
\(x=\dfrac{5}{12}\)
TN
1
KV
26 tháng 12 2023
a) (x - 2).3⁵ = 3⁷
x - 2 = 3⁷ : 3⁵
x - 2 = 3²
x - 2 = 9
x = 9 + 2
x = 11
b) x² - 2x = 0
x(x - 2) = 0
⇒ x = 0 hoặc x - 2 = 0
*) x - 2 = 0
x = 2
Vậy x = 0; x = 2
c) (2x - 1)² = 49
⇒ 2x - 1 = 7 hoặc 2x - 1 = -7
*) 2x - 1 = 7
2x = 7 + 1
2x = 8
x = 8 : 2
x = 4
*) 2x - 1 = -7
2x = -7 + 1
2x = -6
x = -6 : 2
x = -3
Vậy x = -3; x = 4
VN
Võ Ngọc Phương
CTVHS
VIP
11 tháng 9 2023
Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 2 2023
a, \(\dfrac{x-1}{21}\) = \(\dfrac{3}{x+1}\)
( x-1)(x+1) = 21.3
x2 + x - x -1 = 63
x2 = 63 + 1
x2 = 64
x = + - 8
b, 2\(\dfrac{1}{2}\)x + x = 2\(\dfrac{1}{17}\)
x( \(\dfrac{5}{2}\) + 1) = \(\dfrac{35}{17}\)
x = \(\dfrac{35}{17}\) : ( \(\dfrac{5}{2}\)+1)
x = \(\dfrac{35}{17}\) x \(\dfrac{2}{7}\)
x = \(\dfrac{10}{17}\)
c, (x + \(\dfrac{1}{4}\) - \(\dfrac{2}{3}\) ) : ( 2 + \(\dfrac{1}{6}\) - \(\dfrac{1}{4}\)) = \(\dfrac{7}{46}\)
(x - \(\dfrac{5}{12}\)): \(\dfrac{23}{12}\) = \(\dfrac{7}{46}\)
(x - \(\dfrac{5}{12}\)) = \(\dfrac{7}{46}\) x \(\dfrac{23}{12}\)
x - \(\dfrac{5}{12}\) = \(\dfrac{7}{12}\)
x = \(\dfrac{7}{12}\) + \(\dfrac{5}{12}\)
x = 1
d, 2\(\dfrac{1}{3}\)x - 1\(\dfrac{3}{4}\)x + \(2\dfrac{2}{3}\) = 3\(\dfrac{3}{5}\)
x( \(\dfrac{7}{3}\) - \(\dfrac{7}{4}\)) + \(\dfrac{8}{3}\) = \(\dfrac{18}{5}\)
x\(\dfrac{7}{12}\) = \(\dfrac{18}{5}\) - \(\dfrac{8}{3}\)
x\(\dfrac{7}{12}\) = \(\dfrac{14}{15}\)
x = \(\dfrac{14}{15}\) : \(\dfrac{7}{12}\)
x = \(\dfrac{8}{5}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 12 2023
a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 12 2023
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
KV
26 tháng 12 2023
a) (x - 2)(x + 3) < 0 (1)
Do x là số nguyên nên x - 2 < x + 3
(1) x - 2 < 0 và x + 3 > 0
*) x - 2 < 0
x < 0 + 2
x < 2
*) x + 3 > 0
x > 0 - 3
x > -3
Vậy -3 < x < 2
AH
Akai Haruma
Giáo viên
10 tháng 3 2023
Lời giải:
a.
$\frac{2}{3}x-\frac{7}{6}=\frac{12}{7}-\frac{1}{2}=\frac{17}{14}$
$\frac{2}{3}x=\frac{17}{14}+\frac{7}{6}=\frac{50}{21}$
$x=\frac{50}{21}: \frac{2}{3}=\frac{25}{7}$
b.
$(1\frac{1}{2}+\frac{5}{3}-\frac{1}{6}):x=\frac{3}{4}-\frac{1}{2}$
$3:x=\frac{1}{4}$
$x=3: \frac{1}{4}=12$
