Đkiện: x\(\ge\)0 ; x\(\ne\)9
A=\(\frac{2\sqrt{x}}{\sqrt{x}+2}\)+\(\frac{7\sqrt{x}+4}{x-\sqrt{x}-6}\)-\(\frac{\sqrt{x}+2}{\sqrt{x}-3}\)=\(\frac{2\sqrt{x}}{\sqrt{x}+2}\)+\(\frac{7\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)-\(\frac{\sqrt{x}+2}{\sqrt{x}-3}\)=\(\frac{2\sqrt{x}.\left(\sqrt{x}-3\right)+7\sqrt{x}+4-\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
=\(\frac{2x-6\sqrt{x}+7\sqrt{x}+4-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)=\(\frac{x-3\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)=\(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)=\(\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)}\)=\(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)}\)-\(\frac{2}{\left(\sqrt{x}+2\right)}\)=1-\(\frac{2}{\left(\sqrt{x}+2\right)}\)
Để A có giá trị nguyên thì \(\sqrt{x}+2\) là ước của 2
Ư(2)=-1;1;2;-2 ( điều kiện \(\sqrt{x}+2\)\(\ge\)2)
nên \(\sqrt{x}+2\)=2
=>\(\sqrt{x}\)=0=> x=0(thỏa mãn điều kiện)
Mai Linh49 phút trước (13:04)Đkiện: x≥0 ; x≠9
A=2x√x√+2+7x√+4x−x√−6-x√+2x√−3=2x√x√+2+7x√+4(x√+2)(x√−3)-x√+2x√−3=2x√.(x√−3)+7
