Rút gọn biểu thức P = a b − 2 a b + 1 : b − a 2 .
A. 1 b
B. 1 a
C. b
D. 1 b
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a) Ta có:
\(A=\left(-a+b-c\right)-\left(-a-b-c\right)\)
\(=-a+b-c+a+b+c\)
\(=\left(-a+a\right)+\left(b+b\right)+\left(-c+c\right)\)
\(=0+2b+0\)
\(=2b\)
b) \(A=2b=2.\left(-1\right)=-2\)
Bài 1 :
\(A=\left(-a+b-c\right)-\left(-a-b-c\right)\)
\(=-a+b-c+a+b+c=2b\)
Ta có b = -1 ta được : \(2b=2\left(-1\right)=-2\)
Vậy \(A=-2\)
\(B=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)=-2a+3b-4c+2a+3b+4c\)
\(=6b\)
Ta có : b = -1 khi đó: \(B=6b=6\left(-1\right)=-6\)
Vậy B = -6
\(H=\dfrac{a^2\left(a^{-2}b^3\right)^2\cdot b^{-1}}{\left(a^{-1}\cdot b\right)\cdot a^{-5}\cdot b^{-2}}\)
\(=\dfrac{a^2\cdot a^{-4}\cdot b^6\cdot b^{-1}}{a^{-1-5}\cdot b^{1-2}}\)
\(=\dfrac{a^{-2}\cdot b^5}{a^{-4}\cdot b^{-1}}=a^{-2+4}\cdot b^{5+1}=a^2b^6\)
\(H=\dfrac{a^2.a^{-4}.b^6.b^{-1}}{a^{-1}.b.a^{-5}.b^{-2}}=\dfrac{a^{2-4}.b^{6-1}}{a^{-1-5}.b^{1-2}}=\dfrac{a^{-2}.b^5}{a^{-6}.b^{-1}}=a^{-2-\left(-6\right)}.b^{5-\left(-1\right)}=a^4b^6\)
\(a,\left(a+b\right)^2-\left(a-b\right)^2\)
\(=a^2+2ab+b^2-a^2+2ab-b^2\)
\(=4ab\)
\(b,\left(a+b\right)^3-\left(a-b\right)-\left(2b\right)^3\)
\(=a^3+3a^2b+3ab^2+b^3-a+b-8b^3\)
a) \(\left(a+b\right)^2-\left(a-b\right)^2\)
\(\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(\left(2b\right)\left(2a\right)\)
\(4ab\)
b) \(\left(a+b\right)^3-\left(a-b\right)-\left(2b\right)^3\)
\(a^3+3a^2b+3ab^2+b^3-a+b-8b^3\)
\(a\left(a^2-1\right)+3\left(a^2b+ab^2\right)+b\left(b^2+1-8b^2\right)\)
\(a\left(a-1\right)\left(a+1\right)+3\left[ab\left(a+b\right)\right]+b\left(-7b^2+1\right)\)
`T=sqrt{1/(a-b)^2+1/(b-c)^2+1/(c-a)^2}`
`=sqrt{1/(a-b)^2+1/(b-c)^2+1/(c-a)^2+2/((a-b)(b-c))+2/((b-c)(c-a))+2/((c-a)(a-b))-2/((a-b)(b-c))-2/((b-c)(c-a))-2/((c-a)(a-b))}`
`=sqrt{(1/(a-b)+1/(b-c)+1/(c-a))^2-(2(a-b+b-c+c-a))/((a-b)(b-c)(c-a))}`
`=sqrt{(1/(a-b)+1/(b-c)+1/(c-a))^2-0}`
`=sqrt{1/(a-b)+1/(b-c)+1/(c-a))^2}`
`=|1/(a-b)+1/(b-c)+1/(c-a)|`
\(a,=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2b^3=6a^2b\\ b,=\left(6x+1-6x+1\right)^2=2^2=4\)