toán 10

A

angleofdarkness

$\sqrt{x-1} (3x^2+x+1)+3x^2+2x-1=0$

\Leftrightarrow $\sqrt{x-1} (3x^2+x+1)+(3x^2+x+1)=1-x+1$

\Leftrightarrow $(3x^2+x+1)(\sqrt{x-1}+1)=1-(\sqrt{x-1})^2=(\sqrt{x-1}+1)(1-(\sqrt{x-1}).$

\Leftrightarrow $3x^2+x+1=1-\sqrt{x-1}$

\Leftrightarrow $3x^2+x=-\sqrt{x-1}$

Đến đây bạn tự làm tiếp
 
S

sasuke_156

dat can bac 2 cua x-1=a,3x^2+2x-1=b thi phuong trinh viet lai la ab+a^2+b-1=0 <=>
(a+1)*(a+b-1)=0
den day ban tu giai tiep nhe
 
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