a) Giải PT

$ \frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}} $

b)Tìm $m \in (0,\frac{5}{6})$ sao cho

$\int_0^2\left| \frac{1}{3}x^{3} + mx^{2} -2x -2m - \frac{1}{3} \right|dx=4$

c) Chứng minh rằng $x^4 - k^2x^2 + 2kx - 1 = 0$ (k là tham số) có 4 nghiệm phân biệt

d)  $2^{x^{2}-2x+2} + 2^{x^{2}+3} > 2^{2x^{2}-2x}+32$

c) Bạn thao khảo tại đây
http://toan.hoctainha.vn/Hoi-Dap/Cau-Hoi/113626/bien-luan-phuong-trinh/13020#13020
Đặt $a=x-{1\over x}, b=x-{4\over x}$. PT
$\Leftrightarrow \sqrt{a+b}=\sqrt{a}-b\implies b(b-2\sqrt{a}-1)=0$

$b=0$, thì  $x^2=4\implies x=\pm 2$, ta chọn $x=2$.

 $b=2\sqrt{a}+1$, thif  $\sqrt{a+b}=\sqrt{a}-b\iff |\sqrt{a}+1|=-\sqrt{a}-1$ vô nghiệm vì vế phải $>0>$  vế trái
Vậy $x=2$.

Bạn cần đăng nhập để có thể gửi đáp án

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