Ta có: \(a^2+1=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)

Tương tự: \(\left\{{}\begin{matrix}b^2+1=\left(a+b\right)\left(b+c\right)\\c^2+1=\left(c+a\right)\left(b+c\right)\end{matrix}\right.\)

=> \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)

Mặt khác: \(a+b+c-abc=a\left(1-bc\right)+b+c\)

                \(=a\left(ab+ca\right)+b+c\)     (Vì ab+bc+ca=1)

               \(=\left(a^2+1\right)\left(b+c\right)\)

               \(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)    (Vì \(a^2+1=\left(a+b\right)\left(c+a\right)\))

\(T=1\)

giúp mh nhanh vs

\(A=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1}\)

\(A=\frac{c}{abc+ac+c}+\frac{ac}{abc\cdot c+abc+ac}+\frac{1}{ac+c+1}\)

\(A=\frac{c}{ac+c+1}+\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}\)

\(A=\frac{ac+c+1}{ac+c+1}\)

\(A=1\)

1:

=>x+2xy=8y

=>x+2xy-8y=0

=>x(2y+1)-8y-4=-4

=>x(2y+1)-4(2y+1)=-4

=>(2y+1)(x-4)=-4

mà x,y là số nguyên

nên (x-4;2y+1) thuộc {(-4;1); (4;-1)}

=>(x,y) thuộc {(0;0); (8;-1)}

\(S=\frac{105}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+105}\)

\(S=\frac{abc}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\)

\(S=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\)

\(S=\frac{bc+b+1}{bc+b+1}=1\)

1 nha bn

1 nhé bn