cho a,b>0a,\(\dfrac{1}{1+a}\)+\(\dfrac{1}{1+b}\)=\(\dfrac{2}{1+\sqrt{ab}}\)khi nàob,\(\dfrac{1}{1+a}\)+\(\dfrac{1}{1+b}\)>\(\dfrac{2}{1+\sqrt{ab}}\)khi nàoc,\(\dfrac{1}{1+a}\)+\(\dfrac{1}{1+b}\)>hoặc=\(\dfrac{2}{1+\sqrt{ab}}\)khi nàod,\(\dfrac{1}{1+a}\)+\(\dfrac{1}{1+b}\)<hoặc=\(\dfrac{2}{1+\sqrt{ab}}\)khi...
Đọc tiếp
cho a,b>0
a,\(\dfrac{1}{1+a}\)+\(\dfrac{1}{1+b}\)=\(\dfrac{2}{1+\sqrt{ab}}\)khi nào
b,\(\dfrac{1}{1+a}\)+\(\dfrac{1}{1+b}\)>\(\dfrac{2}{1+\sqrt{ab}}\)khi nào
c,\(\dfrac{1}{1+a}\)+\(\dfrac{1}{1+b}\)>hoặc=\(\dfrac{2}{1+\sqrt{ab}}\)khi nào
d,\(\dfrac{1}{1+a}\)+\(\dfrac{1}{1+b}\)<hoặc=\(\dfrac{2}{1+\sqrt{ab}}\)khi nào
