rút gọn A
A=(-a+b-c) - (-a-b-c)
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C = a 2 − a a + a + 1 − a 2 + a a − a + 1 + a + 1 ( D K : a ≥ 0 ) C = a ( a ) 3 − 1 a + a + 1 − a ( a ) 3 + 1 a − a + 1 + a + 1 = a ( a − 1 ) − a ( a + 1 ) + a + 1 = a − a − a − a + a + 1 = a - 1 2
Ta có:
aa = 10 a + a =11a
bbb = 100b + 10b + b =111b
abab = 1000a+10a+100b+b=1010a + 101b
aabb=1000a+100a+10b+b=1100a+11b
Đáp số:
a: =b-c-a+c+1-a-b+c
=-2a+1
b: =a-b-c-b+c+a+c-b-a
=c-3b+a
c: =2(a-b-b+c-c+a)
=2(2a-2b)
=4a-4b
a) \(\left(b-c\right)-\left(a-c-1\right)-\left(a+b-c\right)\)
\(=b-c-a+c+1-a-b+c\)
\(=c-2a+1\)
b) \(\left(a-b-c\right)-\left(b-c-a\right)+\left(c-b-a\right)\)
\(=a-b-c-b+c+a+c-b-a\)
\(=a-3b+c\)
c) \(2\cdot\left(a-b\right)-2\cdot\left(b-c\right)-2\cdot\left(c-a\right)\)
\(=2\cdot\left(a-b-b+c-c+a\right)\)
\(=2\cdot\left(2a-2b\right)\)
\(=4a-4b\)
=(b+c)(ac-a2+bc-ab)+(b+c)(ac-bc+a2-ab)+(c+a)(a+b)(b-c)
=(b+c)(ac-a2+bc-ab+ac-bc+a2-ab)+(a+c)(a+b)(b-c)
=(b+c)(2ac-2ab)-(a+c)(a+b)(c-b)
=(b+c).2a.(c-b)-(a2+ab+ac+bc)(c-b)
=(c-b)(2ab+2ac-a2-ab-ac-bc)
=(c-b)(-a2+ab+ac-bc)=(c-b)[a(b-a)-c(b-a)]
=(c-b)(b-a)(a-c)
Ta có: \(\left(b-c\right)^3+\left(c-a\right)^3-\left(a-b\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(b-c+c-a\right)\left[\left(b-c\right)^2-\left(b-c\right)\left(c-a\right)+\left(c-a\right)^2\right]-\left(a-b\right)\left[1+3\left(b-c\right)\left(c-a\right)\right]\)
\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2\right)-\left(a-b\right)\left(1+3bc-3ab-3c^2+3ac\right)\)
\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2+1+3bc-3ab-3c^2+3ac\right)\)
\(=\left(b-a\right)\left(b^2-2ab+a^2+1\right)\)
\(=\left(b-a\right)^3+\left(b-a\right)\)
\(=b^3-3b^2a+3ba^2-a^3+b-a\)
`#3107.101107`
`3(a + b)(a + c)(b + c)`
`= 3(a^2 + 2ab + bc)(b + c)`
`= 3(a^2b + a^2c + 2ab^2 + 2abc + b^2c + bc^2)`
`= 3a^2b + 3a^2c + 6ab^2 + 6abc + 3b^2c + 3bc^2`
\(A=\left(-a+b-c\right)-\left(-a-b-c\right)\)
\(=-a+b-c+a+b+c\)
Âm với dương triệt ha
\(=b+b=2b\)
A = -a + b - c + a + b + c
A = 2b