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\(A=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1}\)
\(A=\frac{c}{abc+ac+c}+\frac{ac}{abc\cdot c+abc+ac}+\frac{1}{ac+c+1}\)
\(A=\frac{c}{ac+c+1}+\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}\)
\(A=\frac{ac+c+1}{ac+c+1}\)
\(A=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(A\left(x\right)=ax^2+bx+c\)
Thay \(A\left(-1\right)\) ta được:
\(A\left(-1\right)=a\left(-1\right)^2+b\left(-1\right)+c=a+c-b\)
\(=b-8-b=-8\)
b) \(\left\{{}\begin{matrix}A\left(0\right)=4\\A\left(1\right)=9\\A\left(2\right)=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b+c=9\\4a+2b+c=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b=5\\4a+2b=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b=5\\2a+b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a=0\\b=5\end{matrix}\right.\)
c)
Ta có: \(\left\{{}\begin{matrix}A\left(2\right)=4a+2b+c\\A\left(-1\right)=a-b+c\end{matrix}\right.\)
\(\Leftrightarrow A\left(2\right)+A\left(-1\right)=5a+b+2c=0\)
\(\Leftrightarrow A\left(2\right)=-A\left(-1\right)\)
\(\Leftrightarrow A\left(2\right)\times A\left(-1\right)=-\left[A\left(2\right)\right]^2\le0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left(a-\dfrac{1}{3}\right)\left(b+\dfrac{1}{2}\right)\left(c-3\right)=0\) (1)
Và: \(a+1=b+2=c+3\)
\(\Rightarrow a=b+2-1=b+1\)
Thay vào (1) ta có:
\(\left(b+1-\dfrac{1}{3}\right)\left(b+\dfrac{1}{2}\right)\left(c-3\right)=0\)
\(\Rightarrow\left(b+\dfrac{2}{3}\right)\left(b+\dfrac{1}{2}\right)\left(c-3\right)=0\) (2)
Mà: \(b+2=c+3\)
\(\Rightarrow c=b+2-3=b-1\)
Thay vào (2) ta có:
\(\left(b+\dfrac{2}{3}\right)\left(b+\dfrac{1}{2}\right)\left(b-1-3\right)=0\)
\(\Rightarrow\left(b+\dfrac{2}{3}\right)\left(b+\dfrac{1}{2}\right)\left(b-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}b=-\dfrac{2}{3}\\b=-\dfrac{1}{2}\\b=4\end{matrix}\right.\)
TH1 khi b=\(-\dfrac{2}{3}\)
\(\Rightarrow a=b+1=-\dfrac{2}{3}+1=\dfrac{1}{3}\)
\(\Rightarrow c=b-1=-\dfrac{2}{3}-1=-\dfrac{5}{3}\)
TH2 khi \(b=-\dfrac{1}{2}\)
\(\Rightarrow a=b+1=-\dfrac{1}{2}+1=\dfrac{1}{2}\)
\(\Rightarrow c=b-1=-\dfrac{1}{2}-1=-\dfrac{3}{2}\)
TH3 khi \(b=4\)
\(\Rightarrow a=b+1=4+1=5\)
\(\Rightarrow c=b-1=4-1=3\)
Vậy: ...
![](https://rs.olm.vn/images/avt/0.png?1311)
+)Ta có:\(\frac{a}{a^,}+\frac{b^,}{b}=1\) \(\iff\) \(ab+a^,b^,=a^,b\) \(\iff\) \(abc+a^,b^,c^,=a^,bc\) \(\left(1\right)\)
+)Ta có: \(\frac{b}{b^,}+\frac{c^,}{c}=1\)\(\iff\) \(bc+b^,c^,=b^,c\) \(\iff\) \(a^,bc+a^,b^,c^,=a^,b^,c\) \(\left(2\right)\)
Cộng (1) với (2) vế với vế ta được :
\(\implies\) \(abc+a^,b^,c^,+a^,bc+a^,b^,c^,=a^,bc+a^,b^,c^,\)
\(\implies\) \(abc+a^,b^,c^,=0\left(đpcm\right)\)
+)Ta có:\(\frac{a}{a^,}+\frac{b^,}{b}=1\) \(\iff \) \(ab+a^,b^,=a^,b\) \(\iff \) \(abc+a^,b^,c=a^,bc\left(1\right)\)
+)Ta có:\(\frac{b}{b^,}+\frac{c^,}{c}=1\) \(\iff \) \(bc+b^,c^,=b^,c\)\(\iff \) \(a^,bc+a^,b^,c^,=a^,b^,c\left(2\right)\)
Cộng \(\left(1\right)\) với \(\left(2\right)\) vế với vế ta được:\(abc+a^,b^,c+a^,bc+a^,b^,c^,=a^,bc+a^,b^,c\)
\(\implies\) \(abc+a^,b^,c^,=0\left(đpcm\right)\)