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a/ ( 35-x)-25=40-(15-18)
(35-x)-25=40-(-3)
(35-x)-25=43
35-x=43+25
35-x=68
x=35-68
x= -33
b/ x+42=-15+27: 3^2
x+42=-15+27 : 9
x+42=-15+3
x+42=-12
x=-12 -42
x= -54
\(\left|x+2\right|=3-\left(-8\right)\)
\(\Rightarrow \left|x+2\right|=11\)
\(\Rightarrow\hept{\begin{cases}x+2=-11\\x+2=11\end{cases}}\)
\(\text{Trường hợp : }x+2=-11\)
\(\Rightarrow x=-11-2\)
\(\Rightarrow x=-13\)
\(\text{Trường hợp : }x+2=11\)
\(\Rightarrow x=11-2\)
\(\Rightarrow x=9\)
\(\Rightarrow x\in\left\{-13;9\right\}\)
a) \(\left|x+2\right|=3-\left(-8\right)\)
\(\left|x+2\right|=3+8\)
\(\left|x+2\right|=11\)
\(\Rightarrow\orbr{\begin{cases}x+2=11\\x+2=-11\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=9\\x=-13\end{cases}}\)
Vậy.......
Bài 9:
Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)
Vậy: (x,y,z,t)=(-10;6;34;-18)
Bài 11:
Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)
\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)
Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)
\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)
Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)
\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)
Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)
\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)
Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)
\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)
Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)
\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)
Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)
Theo đề bài ta có:
268 - 18 ⋮ X = 250 ⋮ X
390 - 40 ⋮ X = 350 ⋮ X
Hay X là ƯC(250,350)
⇒ X ϵ ƯCLN(250,350)
Ta có: 250 = 2.53
350 = 2. 52. 7
⇒ ƯCLN(250,350) = 2. 52 = 50
⇒ 50 ⋮ X
⇒ X ϵ {1; 2; 5; 10; 25; 50}
268:x dư 18 => 250 chia hết cho x
390:x dư 40 => 350 chia hết cho x
\(250=5^3.2;350=5^2.2.7\\ ƯCLN\left(250;350\right)=5^2.2=50\\ x\inƯ\left(50\right)=\left\{1;2;5;10;25;50\right\}\)
Vì 390:x dư 40 => x>40
Vậy: x=50
1a) \(\frac{x-3}{x+7}=\frac{-5}{-6}\)
=> \(\frac{x-3}{x+7}=\frac{5}{6}\)
=> (x - 3).6 = 5.(x + 7)
=> 6x - 18 = 5x + 35
=> 6x - 5x = 35 + 18
=> x = 53
b) \(\frac{x-7}{x+3}=\frac{4}{3}\)
=> (x - 7). 3 = (x + 3). 4
=> 3x - 21 = 4x + 12
=> 3x - 4x = 12 + 21
=> -x = 33
=> x = -33
c) \(\frac{x-10}{6}=-\frac{5}{18}\)
=> (x - 10) . 18 = -5 . 6
=> 18x - 180 = -30
=> 18x = -30 + 180
=> 18x = 150
=> x = 150 : 18 = 25/3
d) \(\frac{x-2}{4}=\frac{25}{x-2}\)
=> (x - 2)(x - 2) = 25 . 4
=> (x - 2)2 = 100
=> (x - 2)2 = 102
=> \(\orbr{\begin{cases}x-2=10\\x-2=-10\end{cases}}\)
=> \(\orbr{\begin{cases}x=12\\x=-8\end{cases}}\)
e) \(\frac{7}{x}=\frac{x}{28}\)
=> 7 . 28 = x . x
=> 196 = x2
=> x2 = 142
=> \(\orbr{\begin{cases}x=14\\x=-14\end{cases}}\)
f) \(\frac{40+x}{77-x}=\frac{6}{7}\)
=> (40 + x) . 7 = (77 - x).6
=> 280 + 7x = 462 - 6x
=> 280 - 462 = -6x + 7x
=> -182 = x
=> x = -182
\(2x+3=8\)
\(\Rightarrow2x=8-3\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=\dfrac{5}{2}\)
\(x:5-2=3\)
\(\Rightarrow x:5=3+2\)
\(\Rightarrow x:5=5\)
\(\Rightarrow x=5\cdot5\)
\(\Rightarrow x=25\)
\(x:7-2=19\)
\(\Rightarrow x:7=19+2\)
\(\Rightarrow x:7=21\)
\(\Rightarrow x=21\cdot7\)
\(\Rightarrow x=147\)
Mình chưa rõ đề
\(20-\left(x+3\right)=5\)
\(\Rightarrow-x-3=5-20\)
\(\Rightarrow-x-3=-15\)
\(\Rightarrow-x=-15+3\)
\(\Rightarrow-x=-12\)
\(\Rightarrow x=12\)
18-(x+3)=40.5
18-(x+3)=200
x+3=18-200
x+3=-182
x=-182-3
x=-185
Vậy x=-185
\(18-\left(x+3\right):5=40\)
\(\left(x+3\right):5=18-40\)
\(\left(x+3\right):5=-22\)
\(x+3=-110\)
\(x=-113\)
vay \(x=-113\)