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15 tháng 1 2023

`a) 3-5+(-x+3)=6`

`=>5+(-x+3)=3-6`

`=>5+(-x+3)=-3`

`=>-x+3=-3-5`

`=>-x+3=-8`

`=>-x=-8-3`

`=>-x=-11`

`=>x=11`

__

`b)(-4-x)+(4-15)=-15`

`=>(-4-x)+-11=-15`

`=>-4-x=-15-(-11)`

`=>-4-x=-15+11`

`=>-4-x=-4`

`=>x=-4-(-4)`

`=>x=-4+4`

`=>x=0`

`c)(11+x)-(-11-9)=32`

`=>(11+x)-(-20)=32`

`=>(11+x)+20=32`

`=>11+x=32-20`

`=>11+x=12`

`=>x=12-11`

`=>x=1`

15 tháng 1 2023

`a)3-5+(-x+3)=6`

`5+(-x+3)=3-6`

`5+(-x+3)=-3`

`-x+3=-3-5`

`-x+3=-8`

`-x=-8-3`

`-x=-11`

`x=11`

`b,(-4-x)+(4-15)=-15`

`(-4-x)+(-11)=-15`

`-4-x=-15-(-11)`

`-4-x=-15+11`

`-4-x=-4`

`x=-4-(-4)`

`x=-4+4`

`x=0`

`c)(11+x)-(-11-9)=32`

`(11+x)-(-20)=32`

`(11+x)+20=32`

`11+x=32-20`

`11+x=12`

`x=12-11`

`x=1`

 

24 tháng 1 2022

Bài 2 : a, x = -36/9 = -4

b, đề sai 

c, <=> -2 =< x =< -3 => x = -1 

Bài 1: 

a: 2/8=9/36; 2/9=8/36; 8/2=36/9; 9/2=36/8

b: -2/4=9/-18; -2/9=4/-18; 4/-2=-18/9; 9/-2=-18/4

Bài 2: 

a: =>x/3=-4/3

hay x=-4

Câu b đề sai rồi bạn

a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)

=>x=-3; y=28; z=-60

b: 5/12=x/-72

=>x=-72*5/12=-6*5=-30

c: =>x+3=-5

=>x=-8

5 tháng 3 2020

-29-9(2x-1)\(^2\)= -110

(=) 9(2x-1)2 = (-29) +110

(=) 9(2x-1)2 = 81

(=) (2x-1)2 =81: 9

(=) (2x-1)2 =9

(=) (2x-1)2 = 32 =(-3)2

\(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)

\(\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)

\(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

vậy : ........

5 tháng 3 2020

a,\(-29-9\left(2x-1\right)^2=-110\)

\(=>-29+110=9.\left(2x-1\right)^2\)

\(=>81=9.\left(2x-1\right)^2\)

\(=>\left(2x-1\right)^2=9\)

\(=>\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}=>\orbr{\begin{cases}x=\frac{4}{2}=2\\x=\frac{-2}{2}=-1\end{cases}}}\)

25 tháng 11 2023

a)

\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+98\right)=0\)

\(x+x+2+x+4+...+x+98=0\)

\(50x+\left(98+2\right).\left[\left(98-2\right):2+1\right]:2=0\)

\(50x+100.49:2=0\)

\(50x+49.50=0\)

\(50x=0-49.50\)

\(50x=-2450\)

\(x=-2450:50\)

\(x=-49\)

b)

\(\left(x-5\right)+\left(x-4\right)+\left(x-3\right)+...+\left(x+11\right)+\left(x+12\right)=99\)

\(x+x+x+...+x-5-4-3-...+11+12=99\)

\(18x+6+7\text{+ 8 + 9 + 10 + 11 + 12 = 99}\)

\(18x+63=99\)

\(18x=99-63\)

\(18x=36\)

\(x=36:18\)

\(x=2\)

 

25 tháng 11 2023

giúp mình với, mình đang vội!

21 tháng 1 2022

\(a,\left(2x-5\right)+17=6\\ \Rightarrow2x-5=-11\\ \Rightarrow2x=-6\\ \Rightarrow x=-3\\ b,10-2\left(4-3x\right)=-4\\ \Rightarrow2\left(4-3x\right)=14\\ \Rightarrow4-3x=7\\ \Rightarrow3x=-3\\ \Rightarrow x=-1\\ c,24:\left(3x-2\right)=-3\\ \Rightarrow3x-2=-8\\ \Rightarrow3x=-6\\ \Rightarrow x=-2\\ d,5-2x=-17+12\\ \Rightarrow5-2x=-5\\ \Rightarrow2x=10\\ \Rightarrow x=5\)

a: =>2x-5=-11

=>2x=-6

hay x=-3

b: =>2(4-3x)=14

=>4-3x=7

=>3x=-3

hay x=-1

c: =>3x-2=-8

=>3x=-6

hay x=-2

18 tháng 1 2022

\(a,2x-5+17=6\\ \Rightarrow2x=-6\\ \Rightarrow x=-3\\ b,\Leftrightarrow10-8+6x=-4\\ \Leftrightarrow6x=-6\Leftrightarrow x=-1\\ d,\Rightarrow-2x=-10\\ \Rightarrow x=5\)

câu c giống câu b nhó

10 tháng 8 2023

a) \(x\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) \(\left(-7-x\right)\left(-x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)

c) \(\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3