Cho bất phương trình: $3(k-1)t+1>2k+t$
1. Giải bất phương trình với $k=1-2\sqrt{2}$
2. Tìm k để bất phương trình nhận mọi giá trị $t>1$ là nghiệm
1. Thay $k=1-2\sqrt{2}$, bất phương trình đã cho có dạng:
$-(6\sqrt{2}+1)t>1-4\sqrt{2}$, vậy $t<\frac{4\sqrt{2}-1}{6\sqrt{2}+1}$
2. Bất phương trình đã cho có thể viết : $(3k-4)t+(1-2k)>0          $                 (*)
Xét hàm số $f(t)=(3k-4)t+(1-2k)$ có đồ thị là một đường thẳng nên muốn cho bất phương trình (*) đúng với mọi t thì phải có $\begin{cases}3k-4>0 \\ f(1)=k-3\geq 0 \end{cases}$
Giải ra được $k\geq 3$

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