Tìm x thuộc Z biết:
a) 2.(x-1)-(x-3)=4
b)2.(x+8)+(9-x)=(-10)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
-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 : ........
a,\(-29-9\left(2x-1\right)^2=-110\)
\(=>-29+110=9.\left(2x-1\right)^2\)
\(=>81=9.\left(2x-1\right)^2\)
\(=>\left(2x-1\right)^2=9\)
\(=>\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}=>\orbr{\begin{cases}x=\frac{4}{2}=2\\x=\frac{-2}{2}=-1\end{cases}}}\)
c) | x - 3 | + x - 3 = 0
| x - 3 | + x = 0 + 3
| x - 3 | + x = 3
| x - 3 | = 3 - x
=> x < 3
=> x = { 3 ; 2 ; 1 ; 0 ; -1 ; - 2 ; -3 ; .... }
2)
10 + 9 + 8 +.....+ x = 10
9 + 8 + ... + x = 10 - 10
9 + 8 + .... + x = 0
tổng : 9 + 8 + ... + x = \(\frac{\left(9+x\right).n}{2}\)trong đó n là số số hạng
ta có : \(\frac{\left(9+x\right).n}{2}=0\)
( 9 + x ) . n = 0 . 2
( 9 + x ) . n = 0
9 + x = 0 : n
9 + x = 0
x = 0 - 9
=> x = -9
a) | x | + 2 = 5
| x | = 5 - 2
| x | = 3
=> x = \(\orbr{\begin{cases}3\\-3\end{cases}}\)
b) | x + 2 | - x = 2
| x + 2 | = 2 + x
=> x + 2 \(\in\orbr{\begin{cases}Z^-\\Z^+\end{cases}}\)
=> x = { 0 ; 1 ; 2 ; 3 ; 4; .... }
hoặc x = { -1 ; -2 ; -3 ; -4 ; ... }
làm ko nổi nữa
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 = \(\dfrac{1}{4}\)
a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)
b: \(A=\dfrac{x^2-4-5+x+3}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}=\dfrac{x+2}{x-2}\)
c: Để A=3/4 thì 4x-8=3x+6
=>x=14
d: Để A nguyên thì \(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{3;1;4;0;6;-2\right\}\)
\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)
\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)
\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)
\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)
\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)
\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)
\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)
\(a,2\left(x-1\right)-\left(x-3\right)=4\)
\(2x-2-x+3=4\)
\(2x-x-2+3=4\)
\(x+1=4\)
\(x=4-1\)
\(x=3\)
Vậy \(x=3\).
\(b,2\left(x+8\right)+\left(9-x\right)=-10\)
\(2x+16+9-x=-10\)
\(2x-x+16+9=-10\)
\(x+25=-10\)
\(x=-10-25\)
\(x=-35\)
Vậy \(x=-35\).
a) \(2\left(x-1\right)-\left(x-3\right)=4\)
\(\Leftrightarrow2x-2-x+3=4\)
\(\Leftrightarrow2x-x-2+3-4=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
b) \(2\left(x+8\right)+\left(9-x\right)=-10\)
\(\Leftrightarrow2x+16+9-x=-10\)
\(\Leftrightarrow2x-x+16+9+10=0\)
\(\Leftrightarrow x+35=0\)
\(\Leftrightarrow x=-35\)