Toán \(\sqrt[3]{2x+3}+(x+2)\sqrt{x+3}=\sqrt[3]{6-x}-3\) 08/09/2021 By Hadley \(\sqrt[3]{2x+3}+(x+2)\sqrt{x+3}=\sqrt[3]{6-x}-3\)
Đáp án + giải thích các bước giải: `\root{3}{2x+3}+(x+2)\sqrt{x+3}=\root{3}{6-x}-3 (x>=-3)` `\to \root{3}{2x+3}+1+(x+2)\sqrt{x+3}-\root{3}{6-x}+2=0` `\to(2x+3+1)/(\root{3}{(2x+3)^2}-\root{3}{2x+3}+1)+(x+2)\sqrt{x+3}-(6-x-8)/(\root{3}{(6-x)^2}+2\root{3}{6-x}+4)=0` `\to (2(x+2))/(\root{3}{(2x+3)^2}-\root{3}{2x+3}+1)+(x+2)\sqrt{x+3}+(x+2)/(\root{3}{(6-x)^2}+2\root{3}{6-x}+4)=0` `\to (x+2) [2/(\root{3}{(2x+3)^2}-\root{3}{2x+3}+1)+\sqrt{x+3}+1/(\root{3}{(6-x)^2}+2\root{3}{6-x}+4)]=0` `\to x+2=0` `\to x=-2(TM)` Trả lời
Đáp án + giải thích các bước giải:
`\root{3}{2x+3}+(x+2)\sqrt{x+3}=\root{3}{6-x}-3 (x>=-3)`
`\to \root{3}{2x+3}+1+(x+2)\sqrt{x+3}-\root{3}{6-x}+2=0`
`\to(2x+3+1)/(\root{3}{(2x+3)^2}-\root{3}{2x+3}+1)+(x+2)\sqrt{x+3}-(6-x-8)/(\root{3}{(6-x)^2}+2\root{3}{6-x}+4)=0`
`\to (2(x+2))/(\root{3}{(2x+3)^2}-\root{3}{2x+3}+1)+(x+2)\sqrt{x+3}+(x+2)/(\root{3}{(6-x)^2}+2\root{3}{6-x}+4)=0`
`\to (x+2) [2/(\root{3}{(2x+3)^2}-\root{3}{2x+3}+1)+\sqrt{x+3}+1/(\root{3}{(6-x)^2}+2\root{3}{6-x}+4)]=0`
`\to x+2=0`
`\to x=-2(TM)`