Violympic toán 8
Các câu hỏi tương tự
Cho \(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ac\right)\left(a-abc\right);abc\ne0;a\ne b\)
CMR:\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c\)
Cho abc=1. CMR \(\dfrac{a}{ab}+a+1+\dfrac{b}{bc}+b+1+\dfrac{c}{ac}+c+1=1\)
Cho abc=1.CMR:
\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+}=1\)
CMR nếu \(\left(a^2-bc\right).\left(b-abc\right)=\left(b^2-ac\right).\left(a-abc\right)\) và các số a, b, c, a-b khác 0 thì \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c\)
Cho a + b + c = 1 (a,b,c khác 1,2). Chứng minh
\(\dfrac{c+ab}{a^2+b^2+abc-1}+\dfrac{a+bc}{b^2+c^2+abc-1}+\dfrac{b+ac}{a^2+c^2+abc-1}=\dfrac{bc+ac+ab+8}{\left(a-2\right)\left(b-2\right)\left(a-2\right)}\)
1.Cho a,b,c > 0 cmr:\(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\)
Cho a, b, c >0 thỏa mãn: abc=1. CM: \(\dfrac{1}{a^2-ab+b^2}+\dfrac{1}{b^2-bc+c^2}+\dfrac{1}{c^2-ac+a^2}\le a+b+c\)
Cho a,b,c thỏa mãn abc=1
Tính B= \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
Cho 3 số a;b;c thỏa mãn a.b.c=1.CMR :\(\dfrac{a}{ab+a+1}\)+\(\dfrac{b}{bc+b+1}\)+\(\dfrac{c}{ac+c+1}\)=1