Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\frac{2000a}{ab+2000a+2000}+\frac{b}{bc+b+2000}+\frac{c}{ac+c+1}\)
\(=\frac{a\cdot abc}{ab+abc\cdot a+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)
\(=\frac{a^2bc}{ab+a^2bc+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)
\(=\frac{a^2bc}{ab\left(ac+c+1\right)}+\frac{b}{b\left(ac+c+1\right)}+\frac{c}{ac+c+1}\)
\(=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}\)
\(=\frac{ac+c+1}{ac+c+1}=1\)
Đặt bt là P ta có
P = 2000a/(ab + 2000a + 2000) + b/(bc + b + 2000) + c/(ac + c + 1)
= 2000ac/(abc + 2000ac + 2000c) + b/(bc + b + abc) + c/(ac + c + 1)
= 2000ac/(2000 + 2000ac + 2000c) + 1/(1 + c + ac) + c/(ac + c + 1)
= ac/(1 + ac + c) + 1/(ac + c + 1) + c/(ac + c + 1)
= (ac + c + 1)/(ac + c + 1) = 1
P=\(\dfrac{2000a}{ab+2000a+2000}\)
P=\(\dfrac{a^2bc}{ab+a^2bc+abc}\)
P=\(\dfrac{a^2bc}{ab\left(1+ac+c\right)}\)
P=\(\dfrac{ac}{1+ac+c}\)
1/
\(A=a^3+a^2-b^3+b^2+ab-3a^2b+3ab^2-3ab\)
\(A=\left(a^3-3a^2b+3ab^2-b^3\right)+\left(a^2-2ab+b^2\right)=\left(a-b\right)^3+\left(a-b\right)^2=7^3+7^2=392\)