Các câu hỏi tương tự
chứng minh các bất phương trình
\(A=\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)
\(B=\left(\frac{a+b}{c}\right)+\frac{b+c}{a}+\frac{c+a}{b}\ge6\left(a,b,c>0\right)\)
Cho a,b,c >0 CMR : \(\frac{c\left(ab+1\right)^2}{b^2\left(bc+1\right)}+\frac{a\left(bc+1\right)^2}{c^2\left(ac+1\right)}+\frac{b\left(ac+1\right)^2}{a^2\left(ab+1\right)}\ge6\)
chứng minh bất đẳng thức sau:
A=(a+b).\(\left(\frac{1}{a}+\frac{1}{b}\right)\)\(\ge4\)
B=\(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge6\)
1. Chứng minh bất đẳng thức sau
\(A=(a+b)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)
\(B=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge6;\left(a,b,c>0\right)\)
Bài 1. Cho a+b+c0. Đặt Pfrac{a-b}{b}+frac{b-c}{a}+frac{c-a}{b}; Qfrac{c}{a-b}+frac{a}{b-c}+frac{b}{c-a}.Tính P.Qb) Rút gọn rồi tính giá trị biểu thức Efrac{left(a-xright)^2}{aleft(b-aright)left(c-aright)}+frac{left(b-xright)^2}{bleft(a-bright)left(c-bright)}+frac{left(c-xright)^2}{cleft(a-cright)left(b-cright)}biết 1-frac{x^2}{abc}0
Đọc tiếp
Bài 1. Cho a+b+c=0. Đặt P=\(\frac{a-b}{b}+\frac{b-c}{a}+\frac{c-a}{b}\); Q=\(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\).Tính P.Q
b) Rút gọn rồi tính giá trị biểu thức E=\(\frac{\left(a-x\right)^2}{a\left(b-a\right)\left(c-a\right)}+\frac{\left(b-x\right)^2}{b\left(a-b\right)\left(c-b\right)}+\frac{\left(c-x\right)^2}{c\left(a-c\right)\left(b-c\right)}\)biết \(1-\frac{x^2}{abc}=0\)
\(Cho\)\(a;b;c>0\)\(TM\) \(a^2+b^2+c^2=3\)\(.\)\(CMR\)\(:\)
\(\frac{a^2b^2+7}{\left(a+b\right)^2}+\frac{b^2c^2+7}{\left(b+c\right)^2}+\frac{c^2a^2+7}{\left(c+a\right)^2}\ge6\)
Cho \(\frac{a-\left(c-b\right)}{b-c}+\frac{b-\left(a-c\right)}{c-a}+\frac{c-\left(b-a\right)}{a-b}=3\).
Chứng minh rằng \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-c}=0\)
Chứng minh\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-c\right)^2}=0\)
cho \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
cm \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)