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a, 2 . (x - 5) = -24 b, 2 . (x - 5) - 7 = 13
2.x - 2.5 = -24 2 . x - 10 - 7 = 13
2.x - 10 = -24 2 . x - 17 = 13
2.x = -24 + 10 = -14 2.x = 13 + 17 = 30
x = -14 : 2 = -7 x = 30 : 2 = 15
c, 5 . ( 3 - x) + 2 . (x - 7) = -14 d, 12 . (x - 5) + 7 . (3 - x) = 5
15 - 5.x + 2.x - 14 = -14 12.x - 60 + 21 - 7.x = 5
- 3.x + 1 = -14 5.x - 39 = 5
-3 . x = -14 - 1 = -15 5.x = 5 + 39 = 44
x = -15 : ( -3) = 5 x = 44 : 5 = 8,8
e, 30 . ( x + 2) - 6 . ( x - 5 ) - 24 . x = 160
30 . x + 60 - 6x + 30 - 24.x = 160
30. x - 30 .x +90 = 160
0 .x = 160 - 90 = 70
Vậy không có x
a) 2 . (x - 5) = -24
(x - 5) = -24 : 2 = -12
x = -12 + 5
x = -7
b) 2. (x - 5) - 7 = 13
2 . (x - 5) = 13 + 7
2 . (x - 5) = 20
x - 5 = 20 : 2 = 10
x = 10 + 5 = 15
nhiều quá :((
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(-x=45\)
\(x=45\)
\(b,5\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36==12\)
\(3x=48\)
\(x=16\)
\(c,3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=0\)
\(4x-20=0\)
\(4x=20\)
\(x=5\)
Cố nốt nha bn !
cảm ơn, bn nha:)))
mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???
a, => -12x+60+21-7x=5
=> 71-19x=5
=> 19x = 71-5 = 66
=> x = 66 : 19 = 4
b, => 30x+60-6x+30-24=100
=> 24x+76 = 100
=> 24x = 100-76 = 24
=> x = 24 : 24 = 1
Tk mk nha
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
a) \(-12\left(x-5\right)+7\left(3-x\right)=5\)
\(-12x+60+21-7x=5\)
\(-19x=5-60-21\)
\(-19x=-76\)
\(x=4\)
vậy \(x=4\)
tương tự
a,-12.(x-5)+7.(3-x)=5
=>-12x-(-60)+21+7x=5
=>-12x+60+21+7x=5
=>-12x+7x=5-60-21
=>-5x=-76
=>x=-76:-5
=>x=15,2
b,30.(x+2)-6.(x-5)-24.x=100
=>30.x+60-(6.x-30)-24.x=100
=>30.x+60-6.x+30-24x=100
=>30.x-6.x-24.x=100-60-30
=>0x=10(vô lý)
Vậy không có x thỏa mãn
