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4 tháng 9 2017

a,Ta có A=\(\dfrac{\left(54-23\right)\left(54+23\right)}{\left(36,5-25,5\right)\left(36,5+25,5\right)}=\dfrac{31.77}{11.62}=\dfrac{7}{2}\)

b,Ta có B=\(\dfrac{\left(82-34\right)\left(82+34\right)}{\left(30,5-1,5\right)\left(30,5+1,5\right)}=\dfrac{48.116}{29.32}=\dfrac{6.8.4.29}{29.8.4}=6\)

c,Ta có C=\(\dfrac{\left(86-54\right)\left(86^2+86.54+54^2\right)}{32}+86.54=86^2+86.54+54^2+86.54=\left(86+54\right)^2=19600\)

Đề sai rồi bạn

a: \(A=\dfrac{-\left(x+2\right)^2-2x\left(x-2\right)-4x^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)

\(=\dfrac{-x^2-4x-4-2x^2+4x-4x^2}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}\)

\(=\dfrac{-7x^2-4}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}=\dfrac{7x^2+4}{\left(x+2\right)\left(x-3\right)}\)

b: Khi x=1/3 thì \(A=\dfrac{7\cdot\dfrac{1}{9}+4}{\left(\dfrac{1}{3}-2\right)\left(\dfrac{1}{3}-3\right)}=\dfrac{43}{40}\)

15 tháng 12 2021

\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\\ A=\dfrac{2\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{2\left(-\dfrac{3}{2}\right)}{\dfrac{5}{2}}=\left(-3\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\)

\(B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)

17 tháng 12 2020

Ta có: \(A=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}-\dfrac{b^2}{\left(b-a\right)\left(c-b\right)}-\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\)

\(=\dfrac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\dfrac{b^2\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\dfrac{c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{\left(a-b\right)\left(ab+c^2\right)-c\left(a-b\right)\left(a+b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{\left(a-b\right)\left(ab+c^2-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{c^2+ab-c}{\left(a-c\right)\left(b-c\right)}\)