Giải bất phương trình sau : 
$(x^{2}-4)\sqrt{x+5}+(x+1)\sqrt{x+2}+x^{3}+2>x^{2}+6x$
hehe gõ đề sai :v mà vẫn ra nhỉ –  ☼SunShine❤️ 05-05-16 12:53 PM
-_- em ns r kết quả khéo còn hơn ... cái kiểu ns cách lm của a thì e chịu –  ☼SunShine❤️ 05-05-16 12:30 PM
Di tu, k lam, noi cach lam thoi –  ★★★★★★★★★★JOHNNN 509★★★★★★★★★★ 05-05-16 12:27 PM
:)) �i tu r?i. K l�m. �t ra c?ng n�i cach lam. Con hon thang jian chi noi mink ket qua =)))) –  ★★★★★★★★★★JOHNNN 509★★★★★★★★★★ 05-05-16 12:26 PM
ẹc -_- chỉ ns là giỏi mà chả thấy lm –  ☼SunShine❤️ 05-05-16 12:22 PM
Ak, cai nay tinh nham cung dk. Nhu may dua cap 1 =))) –  ★★★★★★★★★★JOHNNN 509★★★★★★★★★★ 05-05-16 12:20 PM
$DK: x\geq -2$
$(x^2-4)(\sqrt{x+5}-2)+(x+1)(\sqrt{x+2}-2)+(x^3+x^2-4x-4)>0$
$(x+1)(x-2)$ $(................)$ >0
Dể dàng cm dc cái đỏ >0 dựa vào DK.
$\Rightarrow (x+1)(x-2)>0\Rightarrow x<-1;x>2$
Vậy $-2\leq x<-1;x>2$

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