a,15+(3-x)=-10
b, 11-(53+x) = 97
c, 5x - (-25)=35
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: =>3x=10+14=24
=>x=8
b: =>25-30-x=-19
=>-5-x=-19
=>x+5=19
=>x=14
c: =>15-x=12+14=26
=>x=-11
d: =>4x+11-x-34=67
=>3x-23=67
=>3x=90
=>x=30
e: =>-x-14=289-36-289=-36
=>x+14=36
=>x=22
a) 3x=24
x=8
b) 25-30-x=-19
25-30+19=x
14=x
c) 15-x=26
x=-11
d) 4x+11-x-34=67
3x=90
x=30
e) -14-x= -36
x=-14 -36
x= 22
bài 1
a)\(15+\left(3-x\right)=-10\)
\(3-x=-10-15\)
\(3-x=-25\)
\(x=3-\left(-25\right)\)
\(x=28\)
b)\(11-\left(-53+x\right)=97\)
\(-53+x=11-97\)
\(-53+x=-86\)
\(x=-86-\left(-53\right)\)
\(x=-33\)
c)\(5x-\left(-25\right)=35\)
\(5x+25=35\)
\(5x=35-25\)
\(5x=10\)
\(x=10:5\)
\(x=2\)
bài 2
\(-50\le x< 51\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{-50;-49;...;50\right\}\)
Tổng các số nguyên x là:
\(-50+\left(-49\right)+...+50\)
\(=\left(-50+50\right)+\left(-49+49\right)+...+\left(-1+1\right)+0\)
\(=0+0+...+0+0\)
\(=0\)
bài 3:
Ta có:\(n-7\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau:
n - 7 | 1 | -1 | 5 | -5 |
n | 8 | 6 | 12 | 2 |
Vậy\(n\in\left\{8;6;12;2\right\}\)
dẫu . là dấu nhân nha mn.
mong mn giải giúp mình mình đang cần gấp
b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)
a)(x-1)(5x+3)=(3x-8)(x-1)
\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0
\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)
a) `123 - 27 + 27 + 13 - 123`
`= (123 - 123) + (27 - 27) + 13`
`= 0 + 0 + 13`
`= 13`
`---`
b) `175 + 25 + 13 + 15 - 175 - 25`
`= (175-175)+(25-25)+25+13`
`=0+0+25+13`
`=28`
`---`
c) \(53\times39+47\times39+53\times21+47\times21\)
\(=\text{[}\left(53+47\right)\times39]+\left[\left(53+47\right)\times21\right]\)
\(=\left(100\times39\right)+\left(100\times21\right)\)
\(=3900+2100\)
`=6000`
`---`
d) \(49\times35+49\times16-49\times61\)
`=` \(49\times\left(35+16-61\right)\)
`=` \(49\times\left(-10\right)\)
`=-490`
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
A/ 15+(3-x)=-10
15+3-x=-10
18-x=-10
-x=-10-18=-8
Suy ra x= 8
a, 15 + (3-x) = -10
3-x = -10 - 15
3-x = -25
x = 3 - (-25)
x = -28
b, 11 - ( 53 + x ) = 97
53 + x = 11 - 97
53 + x = -86
x = -86 - 53
x = -139
c, 5x - (-25) = 35
5x =35 + ( -25)
5x = 10
x = 10 : 5
x = 2