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Bài 2:
a: =>20-16-x+6=90
=>10-x=90
=>x=-80
b: =>|x+8|=24-3*(32-25)=24-3*7=3
=>x+8=3 hoặc x+8=-3
=>x=-11 hoặc x=-5
c: \(\Leftrightarrow30+2^x-6=1000:25=40\)
=>2^x=16
=>x=4
c: =>x+2+9 chia hết cho x+2
mà x>=0
nên \(x+2\in\left\{3;9\right\}\)
hay \(x\in\left\{1;7\right\}\)
a, 369 - (|-206| - 15 ) - ( - 206 + | -369|)
369-(206 - 15 ) + 206 - 369
= 369 - 206 + 15 + 206 - 369
= ( 369 - 369 ) + ( -206 + 206 ) + 15
= 0 + 0 + 15
= 15
b, 345 - 150 : [ ( \(3^3\)- 24 )\(^2\)- ( -21 ) ] + 2016\(^0\)
= 345 - 150 : [(27 - 24)\(^2\) + 21] + 1
= 345 - 150 : [3\(^2\)+ 21 ] + 1
= 345 - 150 : 30 + 1
= 345 - 5 + 1
= 341
c, - 2 + 6 - 12 + 16 -....-92 + 96
ta có : ( 6 + 16 + ... + 96 ) + ( -2 + -12 + ... + -92 )
mỗi bên đều có 10 số hạng
=> [ ( 6 + 96 ) . 10 : 2 ] + [ ( -2 + -92 ) . 10 : 2 ]
=> 510 + -470
= 40
\(a,369-\left(\left|-206\right|-15\right)-\left(-206+\left|-369\right|\right)\)
\(=369-\left(206-15\right)+206-369\)
\(=369-206+15+206-369\)
\(=\left(369-369\right)+\left(-206+206\right)+15\)
\(=0+0+15=15\)
\(b,345-150:\left[\left(3^3-24\right)^2-\left(-21\right)\right]+2016^0\)
\(=345-150:\left[\left(27-24\right)^2+21\right]+1\)
\(=345-150:\left[3^2+21\right]+1\)
\(=345-150:30+1\)
\(=345-5+1=341\)
\(c,-2+16-12+...+96\)
Ta có: \(\left[-2+...+\left(-92\right)\right]+\left[6+16+...+96\right]\)
Hai bên có \(10\) số hạng.
\(\Rightarrow\left\{\left[-2+\left(-12\right)\right].10:2\right\}+\left\{\left[6+16\right].10:2\right\}\)
\(\Rightarrow510+\left(-470\right)=40\)
\(2,\)
\(a,20-\left[4^2+\left(x-6\right)\right]=90\)
\(\Rightarrow20-16-x+6=90\)
\(\Rightarrow10-x=90\)
\(\Rightarrow x=-80\)
Vậy: \(x=-80\)
\(b,\left(x+3\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
\(c,1000:\left[30+\left(2^x-6\right)\right]=3^2+4^2\left(x\in N\right)\)
\(\Rightarrow1000:\left(30+2^x-6\right)=25\)
\(\Rightarrow24+2^x=40\)
\(\Rightarrow2^x=16\)
\(\Rightarrow x=4\)
Vậy: \(x=4\)
\(2,\)
\(a,20-\left[42+\left(x-6\right)\right]=90\)
\(\Rightarrow20-42-x+6-90=0\)
\(\Rightarrow x=-106\)
Vậy: \(x=-106\)
\(b,\left(x+3\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
\(c,1000:\left[30+\left(2x-6\right)\right]=32+42\left(x\in N\right)\)
\(\Rightarrow1000:\left(30+2x-6\right)=74\)
\(\Rightarrow1000:\left(24+2x\right)=74\)
\(\Rightarrow24+2x=\dfrac{500}{37}\)
\(\Rightarrow2x=-\dfrac{388}{37}\)
\(\Rightarrow x=-\dfrac{194}{37}\)
Mà \(x\in N\)
\(\Rightarrow x\in\varnothing\)
Vậy: \(x\in\varnothing\)
1: Ta có: \(20-2\left(x+4\right)=4\)
\(\Leftrightarrow2\left(x+4\right)=16\)
\(\Leftrightarrow x+4=8\)
hay x=4
5: Ta có: \(\left(x+1\right)^3=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
a) 359 - ( | -206 | - 15 ) - (- 206 + |- 369 | )
=359 - 206 + 15 - (- 206 ) - 369
=153 + 15 - ( - 206 ) - 369
=168 - ( - 206 ) - 369
=374 - 369
=5
sorry mình ko làm được câu b nhưng vẫn tk cho tớ nhé
dấu này là dấu gì vậy bạn: |