![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x-36\right):18=12\)
\(\Leftrightarrow x-36=216\)
\(\Leftrightarrow x=252\)
Các câu còn lại lm tương tự nhé
Hok tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5.\left(x+35\right)=515\)
\(x+35=515:5\)
\(x+35=103\)
\(x=103-35\)
\(\Rightarrow x=68\)
\(\left(x+40\right).15=75.12\)
\(\Rightarrow\left(x+40\right)=5.12\)[ rút gọn theo t/c đẳng thức ]
\(x+40=60\)
\(x=20\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (x+15) - (-x+1) = 12 b, (x-2) - (2x-2) = -2
x+15+x-1 = 12 x-2- 2x+2 = -2
2x +14 =12 -x = -2
2x = 2 x = 2
x = 1
c, x+2- (5x-16) = 4x-6 d, 15- 3x+4 = 85-x
x+2 - 5x+ 16 = 4x-6 19-3x = 85-x
-4x +18 = 4x-6 19-85 = -x+3x
-4x-4x = -6-18 -66 = 2x
-8x = -24 x = 33
x = 3
e, (-11)2 - 5(x-1) = 76+x
121 - 5x+5 = 76+x
126 -5x = 76+x
126- 76 = x+5x
50 = 6x
x = 25/3
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(24 - (-15) - 2\)
\(=39-2\)
\(=37\)
b) \((-85) + 10 - (-85) - 50\)
\(=[(-85)-(-85)]+10-50\)
\(=0+10-50\)
\(=10-50\)
\(=-40\)
c) \(71 - (-30) - (+18) + (-30)\)
\(=[(-30)-(-30)]+71-(+18)\)
\(=0+71-18\)
\(=71-18\)
\(=53\)
d) \(-(30) - (+37) + (+37) + (-85)\)
\(=[-(+37)+(+37)]-(30)+(-85)\)
\(=0-(30)+(-85)\)
\(=(-30)+(-85)\)
\(=-115\)
e) \((35-815) - (795-65)\)
\(=(-780)-730\)
\(=-1510\)
g) \((2002-79+15) + (-79+15)\)
\(=1938+(-64)\)
\(=1874\)
Bài 2:
a) \(25 - (30+x) = x - (27-8)\)
\(25-30-x=x-27+8\)
\(x+x=25-30+27-8\)
\(2x=14\)
\(x=14\div2\)
\(x=7\)
b) \((x-12) - 15 = (20-17) - (18+x) \)
\(x-12-15=13-18-x\)
\(x-27=-5-x\)
\(x+x=-5+27\)
\(2x=22\)
\(x=22\div2\)
\(x=11\)
c) \(15 - x = 7 - (-2)\)
\(15-x=9\)
\(x=15-9\)
\(x=6\)
d) \(x - 35 = (-12) - 3\)
\(x-35=-15\)
\(x=-15+35\)
\(x=20\)
e) \(\left|5-x\right|-26=-15\)
\(\left|5-x\right|=-15+26\)
\(\left|5-x\right|=11\)
Từ đây ta có:
*Nếu \(5-x=11\)
\(x=5-11\)
\(x=-6\)
*Nếu \(5-x=-11\)
\(x=5-(-11)\)
\(x=16\)
Vậy \(x=-6;x=16\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 125 - 5(x + 10) = 12
=> 5(x + 10) = 125 - 12 = 113
=> x + 10 = 113 : 5
=> x + 10 = 113/5
=> x = 113/5 - 10 = 63/5
b) (x - 1)2 = 4 => (x - 1)2 = (\(\pm\)2)2
=> x - 1 = 2 hoặc x - 1 = -2
=> x = 3 hoặc x = -1
c) 5x+1 - 52 . 17 = 52.8
=> 5x . 5 - 52 . 17 = 52 . 8
=> 5x . 5 = 52.8 + 52 . 17
=> 5x . 5 = 52(8 + 17)
=> 5x . 5 = 52 . 52
=> 5x = \(\frac{5^2\cdot5^2}{5}=5^3\)
=> x = 3
d) 84 + (12 - 3.x) : 6 = 85
=> 84 + (12 - 3.x) = 510
=> 12 - 3.x = 426
=> 3.x = -414
=> x = -138
e) (x + 1) + (x + 2) + (x + 3) + ... + (x + 99) + (x + 100) = 5150
=> x + 1 + x + 2 + x + 3 + ... + x + 99 + x + 100 = 5150
=> (x + x + .... + x) + (1 + 2 + 3+ ... + 100) = 5150
=> 100x + 5050 = 5150
=> 100x = 5150 - 5050 = 100
=> x = 1
Bài làm:
\(a,125-5.\left(x+10\right)=12\)
\(\Rightarrow5.\left(x+10\right)=125-12\)
\(\Rightarrow5.\left(x+10\right)=113\)
\(\Rightarrow x+10=\frac{113}{5}\)
\(\Rightarrow x=\frac{63}{5}\)
\(b,\left(x-1\right)^2=4\)
\(\Rightarrow\left(x-1\right)^2=2^2\)
\(\Rightarrow\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
\(c,5^{x+1}-5^2.17=5^2.8\)
\(\Rightarrow5^{x+1}-425=200\)
\(\Rightarrow5^{x+1}=200+425\)
\(\Rightarrow5^{x+1}=625\)
\(\Rightarrow5^{x+1}=5^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=3\)
\(d,84+\left(12-3.x\right):6=85\)
\(\left(12-3x\right):6=85-84\)
\(\left(12-3x\right):6=1\)
\(12-3x=6\)
\(3x=6\)
\(x=2\)
\(e,\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+99\right)+\left(x+100\right)=5150\)
\(100.x+\left(1+2+3+...+99+100\right)=5150\)
\(100.x+5050=5150\)
\(100.x=5150-5050\)
\(100.x=100\)
\(x=1\)
Học tốt nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2012 x 24 + 77 x 2012 - 2012
= 2012 x (24 + 77 - 1)
= 2012 x 100
= 201 200
b, 33 x 76 + 84 x 27 + 32 x 3 x 40
= 27 x 76 + 84 x 27 + 27 x 40
= 27 x (76 + 84 + 40)
= 27 x 200
= 5400
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\left(x-2\right)^{10}=\left(x-2\right)^{12}\)
\(\Leftrightarrow\left(x-2\right)^{12}-\left(x-2\right)^{10}=0\)
\(\Leftrightarrow\left(x-2\right)^{10}\left[\left(x-2\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^{10}=0\\\left(x-2\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x\in\left\{3;1\right\}\end{cases}}\)
Vậy \(x\in\left\{-1,2,3\right\}\)
b) Ta có: \(4^{x+2}+4^{x+3}+4^{x+4}+4^{x+5}=85.\left(2^{2016}:2^{2012}\right)\)
\(\Leftrightarrow4^{x+2}\left(1+4+4^2+4^3\right)=85.2^4\)
\(\Leftrightarrow4^{x+2}.85=1360\)
\(\Leftrightarrow4^x=16\)
\(\Leftrightarrow4^x=4^2\)
\(\Leftrightarrow x=2\)
Vậy x=2
`#3107`
`2^2 \times 85 + 40 \times 2^2 - 2^3 \times 12`
`= 2^2 \times (85 + 40 - 2 \times 12)`
`= 2^2 \times (85 + 40 - 24)`
`= 4 \times 101`
`= 404`
Help