Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{\left(a+b\right)\left(-x-y\right)-\left(a-y\right)\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{a\left(-x-y\right)+b\left(-x-y\right)-a\left(b-x\right)+y\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ax-ay-bx-by-ab+ax+by-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ay-bx-ab-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-xy+ay+ab+by}{abxy\left(xy+ay+ab+by\right)}=\dfrac{-1}{abxy}\)
Với \(a=\dfrac{1}{3};b=-2;x=\dfrac{3}{2};y=1\)
\(\Rightarrow A=\dfrac{-1}{\dfrac{1}{3}.\left(-2\right).\dfrac{3}{2}.1}=-1\)
\(C=\left(a+b\right)\left(a+1\right)\left(b+1\right)\)
\(\Leftrightarrow C=3\left(a+1\right)\left(b+1\right)\)
\(\Leftrightarrow C=\left(3a+1\right)\left(b+1\right)\)
\(\Leftrightarrow C=3a\left(b+1\right)+3\left(b+1\right)\)
\(\Leftrightarrow C=3ab+3a+3b+3\)
\(\Leftrightarrow C=3ab+3\left(a+b\right)+3\)
\(\Leftrightarrow C=3.\left(-5\right)+3.3+3\)
\(\Leftrightarrow C=\left(-15\right)+9+3\)
\(\Leftrightarrow C=\left(-3\right)\)
Vậy \(C=\left(-3\right)\)
- Chết cmnr :)) T làm nhầm 1 chỗ
Làm lại nè:
\(\Leftrightarrow C=3\left(a+1\right)\left(b+1\right)\)
\(\Leftrightarrow C=\left(3a+3\right)\left(b+1\right)\)
\(\Leftrightarrow C=3a\left(b+1\right)+3\left(b+1\right)\)
\(\Leftrightarrow C=3ab+3a+3b+3\)
\(\Leftrightarrow C=3.\left(-5\right)+3\left(a+b\right)+3\)
\(\Leftrightarrow C=\left(-15\right)+3.3+3\)
\(\Leftrightarrow C=\left(-15\right)+9+3\)
\(\Leftrightarrow C=\left(-3\right)\)
p/s : Không hiểu mắt tớ bị hỏng chỗ nào mà số 3 viết thành 1 nhưng đáp án vẫn đúng =))
Ta có:
\(a+b+c-abc=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+c\left(a+b\right)\right)-abc\)
\(=\left(a+b\right)ab+\left(a+b\right)^2c+abc+c^2\left(a+b\right)-abc\)
\(=\left(a+b\right)\left(ab+c^2+c\left(a+b\right)\right)\)
\(=\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Đồng thời:
\(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự:
\(b^2+1=\left(a+b\right)\left(b+c\right)\)
\(c^2+1=\left(a+c\right)\left(b+c\right)\)
Từ đó:
\(P=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}=1\)
Ta có: a - b - c = 0
=> \(\hept{\begin{cases}a-c=b\\a-b=c\\-b-c=-a\end{cases}}\Rightarrow\hept{\begin{cases}a-c=b\\-\left(a-b\right)=-c\\-\left(b+c\right)=-a\end{cases}}\Rightarrow\hept{\begin{cases}a-c=b\\-a+b=-c\\b+c=a\end{cases}}\)
Lại có: \(P=\left(1-\frac{c}{a}\right)\left(1-\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\)
\(\Rightarrow P=\frac{a-c}{a}.\frac{b-a}{b}.\frac{c+b}{c}=\frac{b}{a}.\frac{-c}{b}.\frac{a}{c}=-1\)
