Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
C1
B = (a + b - c)-(b + c - a)-(a - b - c)
=a+b-c-b-c+a-a+b+c
=a+b-c
C2
ta có:
a) (a-b)+(c-d)
=a-b+c-d
=a+c-b-d
=(a+c)-(b+d)
vậy .....
b)ta có:
(a-b)-(c-d)
=a-b-c+d
=a+d-b-c
=a+d-b-c
=(a+d)-(b+c)
a ) a + b = -1 => a = -1 - b
b + c = 1 => c = 1 - b
Thay vào a + c = 6.
Ta được : -1 - b + 1 - b = 6
=> -2b = 6
=> b = -3
=> a = -1 - - 3 = 2
=> c = 1 - - 3 = 4
b ) ab = - 35
=> \(a=\dfrac{-35}{b}\)
bc = 7
\(\Rightarrow c=\dfrac{7}{b}\)
Thay vào abc = 35, ta được :
\(\dfrac{-245}{b}=35\Leftrightarrow35b=-245\Rightarrow b=-7\)
=> \(a=-\dfrac{35}{-7}=\dfrac{35}{7}\)
=> \(c=\dfrac{7}{-7}=-1\)
c ) Đặt a + b + c = -6 (1)
b + c + d = -9 (2)
c + d + a = -8 (3)
d + a + b = -7 (4)
Cứ thay từ từ rồi sẽ ra .
Bài 2:
a) Ta có: \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=3^{n+1}\cdot10+2^{n+3}\cdot3⋮6\)
b) Ta có: \(4^{13}+32^5-8^8\)
\(=2^{26}+2^{25}-2^{24}\)
\(=2^{24}\left(2^2+2-1\right)\)
\(=2^{24}\cdot5⋮5\)
c) Ta có: \(2014^{100}+2014^{99}\)
\(=2014^{99}\left(2014+1\right)\)
\(=2014^{99}\cdot2015⋮2015\)
a) Ta có: \(\left|x+7\right|-\left(-8\right)=-25+73\)
\(\Leftrightarrow\left|x+7\right|+8=48\)
\(\Leftrightarrow\left|x+7\right|=40\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=40\\x+7=-40\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=33\\x=-47\end{matrix}\right.\)
Vậy: \(x\in\left\{33;-47\right\}\)
c) Ta có: \(-\left(a-b\right)+\left(b-c\right)-\left(a-c\right)=2b-2a\)
\(\Leftrightarrow-a+b+b-c-a+c=2b-2a\)
\(\Leftrightarrow-2a+2b-2b+2a=0\)
\(\Leftrightarrow0a+0b=0\)(luôn đúng)
Vậy: \(\left\{{}\begin{matrix}a\in Z\\b\in Z\end{matrix}\right.\)
d) Ta có: \(-\left(-a+b+c\right)+\left(b+c-1\right)=-\left(b-a\right)-\left(1-b\right)\)
\(\Leftrightarrow a-b-c+b+c-1=-b+a-1+b\)
\(\Leftrightarrow a-1=a-1\)(luôn đúng)
Vậy: \(\left\{{}\begin{matrix}a\in Z\\b\in Z\\c\in Z\end{matrix}\right.\)
e) Ta có: \(-\left(-a+b+c\right)+\left(b-c+6\right)=a+6\)
\(\Leftrightarrow a-b-c+b-c+6=a+6\)
\(\Leftrightarrow a+6-2c-a-6=0\)
\(\Leftrightarrow-2c=0\)
hay c=0
Vậy: \(\left\{{}\begin{matrix}a\in Z\\b\in Z\\c=0\end{matrix}\right.\)
a) (b - c + 6) - (7 - a + b) + c
= b - c + 6 - 7 + a - b + c
= (b-b)+(-c+c)+(6-7)+a
= -1 + a = a - 1
b) -(a-b-c) + (-c+b+a) - (a+b)
= -a+b+c-c+b+a-a-b
= (-a+a-a) + (b+b-b)+(c-c)
= -a + b + 0
= -a + b = b - a
a, \(\left(a-b\right)+\left(c-d\right)=\left(a+c\right)-\left(b+d\right)\)
\(a-b+c-d=a+c-b-d\)
\(\Rightarrow VT=VP\left(đpcm\right)\)
b, \(\left(a-b\right)-\left(c-d\right)=\left(a+d\right)-\left(b+c\right)\)
\(a-b-c+d=a+d-b-c\)
\(\Rightarrow VT=VP\left(đpcm\right)\)
c, \(a-\left(b-c\right)=\left(a-b\right)+c=\left(a+c\right)-b\)
\(a-b+c=a-b+c=a+c-b\)
\(\Rightarrowđpcm\)
d, \(\left(a-b\right)-\left(b+c\right)+\left(c-a\right)-\left(a-b-c\right)=-\left(a+b-c\right)\)
\(a-b-b-c+c-a-a+b+c=-a-b+c\)
\(-a-b+c=-a-b+c\)
\(\Rightarrow VT=VP\left(đpcm\right)\)
e, \(-\left(-a+b+c\right)+\left(b+c-1\right)=\left(b-c+6\right)-\left(7-a+b\right)+c\)
\(a-b-c+b+c-1=b-c+6-7+a-b+c\)
\(a-1=-1+a\Rightarrow a-1=a+\left(-1\right)\Rightarrow a-1=a-1\)
\(\Rightarrow VT=VP\left(đpcm\right)\)