Question

Cho 3 số a;b;c thỏa mãn a.b.c=1

Chứng minh

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

2)

CHứng tỏ rằng

\(a=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-....-\frac{1}{2^{10}}>\frac{1}{2^{11}}\)

\(b=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}>\frac{1}{100}\)

\(c=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)

Answer 1

1.\(VT=\frac{c}{abc+ac+c}+\frac{b}{bc+b+abc}+\frac{abc}{abc+bc+b}=\frac{c}{ac+c+1}+\frac{1}{ac+c+1}+\frac{ac}{ac+c+1}=\frac{ac+c+1}{ac+c+1}=1=VP\)

Answer 2

ac+c+1\ac+c+1 =1