(8+12+16+...+100).(x-3)=2592. Tìm x
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2:
a: \(\Leftrightarrow x-1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)
b: \(\Leftrightarrow x+3\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
=>\(x\in\left\{-2;-4;0;-6;2;-8;12;-18\right\}\)
c: \(\Leftrightarrow x+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
=>\(x\in\left\{-2;-4;-1;-5;0;-6;1;-7;3;-9;9;-15\right\}\)
d: =>x+1+15 chia hết cho x+1
=>\(x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
=>\(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)
1/ Tính nhanh
1 - 4 + 7 - 10 + 13 - 16 + ... + 97 - 100
= ( - 3 ) + ( - 3 ) + ... + ( - 3 )
= ( - 3 ) x 17
= -51
a) X nhân 3/8 = 9/16
X= 9/16 chia 3/8
X= 9/16 nhân 8/3
X= 3/2
b) x chia 12/7=5/8
x= 5/8 nhân 12/7
x= 15/14
Lời giải:
a.
$x\times \frac{3}{8}=\frac{9}{16}$
$x=\frac{9}{16}: \frac{3}{8}$
$x=\frac{3}{2}$
b.
$x:\frac{12}{7}=\frac{5}{8}$
$x=\frac{5}{8}\times \frac{12}{7}$
$x=\frac{15}{14}$
\(a,x-\dfrac{7}{12}x=\dfrac{5}{24}-\dfrac{3}{8}x\)
\(\Leftrightarrow\dfrac{5}{12}x+\dfrac{3}{8}x=\dfrac{5}{24}\)
\(\Leftrightarrow\dfrac{19}{24}x=\dfrac{5}{24}\Leftrightarrow x=\dfrac{5}{19}\)
Vậy x = 5/19
\(b,\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\-3-\dfrac{x}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)
Vậy x = 1/2 hoặc x = -6
\(c,\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
Vậy x = 7 hoặc x = -1
1/ `|x|=10<=> x=\pm 10`
2/ `|x-8|=0<=>x-8=0<=>x=8`
3/ `7+|x|=12<=>|x|=5<=>x=\pm 5`
4/ `|x+1|=3`
$\Leftrightarrow\left[\begin{array}{1}x+1=3\\x+1=-3\end{array}\right.\\\Leftrightarrow\left[\begin{array}{1}x=3\\x=-4\end{array}\right.$
5/ `15-x=16-(14-42)`
`<=>15-x=16+28`
`<=>15-x=44`
`<=>x=-29`
6/ `210-(x-12)=168`
`<=>210-x+12=168`
`<=>222-x=168`
`<=>x=54`