\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow a+b=0\Leftrightarrow a=-b\)

\(\Rightarrow a^{23}+b^{23}=-b^{23}+b^{23}=0\)

Vậy \(\left(a^{23}+b^{23}\right)\left(a^{1995}+c^{1995}\right)=0\)

Câu 2: B

Câu 3: A

Câu 1: 

Ta có: \(5^{x+3}+5^x=126\cdot5^3\)

\(\Leftrightarrow5^x\left(5^3+1\right)=126\cdot5^3\)

\(\Leftrightarrow5^x=5^4\)

hay x=4

\(a+b+c=7\Rightarrow a+b+c-1=6\)

Ta có:\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Leftrightarrow49=23+2\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca=13\)

Lại có \(ab+c-6=ab+c-\left(a+b+c-1\right)=ab-a-b+1=\left(a-1\right)\left(b-1\right)\)

Tương tự \(bc+a-6=\left(b-1\right)\left(c-1\right)\)

                \(ca+b-6=\left(c-1\right)\left(a-1\right)\)

\(\Rightarrow A=\frac{1}{\left(a-1\right)\left(b-1\right)}+\frac{1}{\left(b-1\right)\left(c-1\right)}+\frac{1}{\left(c-1\right)\left(a-1\right)}\)

            \(=\frac{c-1+a-1+b-1}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=\frac{a+b+c-3}{abc-\left(ab+ac+bc\right)+\left(a+b+c\right)-1}\)

             \(=\frac{7-3}{3-13+7-1}=-1\)

Câu hỏi của hoàng thị huyền trang - Toán lớp 9 - Học toán với OnlineMath

Em tham khảo nhé!

a^3+b^3+c^3=3abc

=>(a+b)^3+c^3-3ab(a+b)-3bac=0

=>(a+b+c)(a^2+2ab+b^2-ac-bc+c^2)-3ab(a+b+c)=0

=>(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0

=>a^2+b^2+c^2-ab-bc-ac=0

=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0

=>(a-c)^2+(a-b)^2+(b-c)^2=0

=>a=b=c

=>A=(1+b/b)(1+b/b)(1+c/c)

=2*2*2=8