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đk: x>= 5/3 $\sqrt{x+2}-2+\sqrt[3]{x-1}-1+\sqrt[4]{3x-5}-1+4-2\sqrt[5]{3x+26}=0\Leftrightarrow \frac{x-2}{\sqrt{x+2}+2}+\frac{x-2}{A=\sqrt[3]{(x-1)^{2}}+\sqrt[3]{x-1}+1}+\frac{3(x-2)}{B=\sqrt[4]{(3x-5)^{3}}+\sqrt[4]{(3x-5)^{2}}+\sqrt[4]{(3x-5)}+1}-\frac{6(x-2)}{C=\sqrt[5]{(3x+26)^{4}}+2.\sqrt[5]{(3x+26)^{3}}+4\sqrt[5]{(3x+26)^{2}}+8\sqrt[3]{3x+26}+16}=0. ta có x=2 là 1 No, vs x\neq 2\wedge x\geq \frac{5}{3} rút gọn x-2, CM: \frac{1}{\sqrt{x+2}+2}+\frac{3}{B}+\frac{1}{A}-\frac{6}{C}>0 hay \frac{1}{\sqrt{x+2}+2}-\frac{6}{C}>0 hay \frac{1}{\sqrt{x+2}+2}-\frac{6}{\sqrt[5]{(3x+26)^{4}}+16}>0$
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