bằng 99/100

99/100 chắc

bạn tham khảo ở đây nghe

https://hoc24.vn/cau-hoi/.1003659772279

Cô Hoài trả lời tin nhắn em với ạ!

a, \(\dfrac{15}{4}\) - 2,5 : |\(\dfrac{3}{4}\)\(x\) + \(\dfrac{1}{2}\)| = 3

    3,75 - 2,5:|\(\dfrac{3}{4}\)\(x\) +  \(\dfrac{1}{2}\)| = 3

             2,5:|\(\dfrac{3}{4}\)\(x\) + \(\dfrac{1}{2}\)|  = 3,75 - 3

            2,5 : |\(\dfrac{3}{4}\)\(x\) + \(\dfrac{1}{2}\)| = 0,75

                    |\(\dfrac{3}{4}\)\(x\) + \(\dfrac{1}{2}\)| = 2,5 : 0,75

                   |\(\dfrac{3}{4}\)\(x\) + \(\dfrac{1}{2}\)| = \(\dfrac{10}{3}\)

                   \(\left[{}\begin{matrix}\dfrac{3}{4}x+\dfrac{1}{2}=-\dfrac{10}{3}\\\dfrac{3}{4}x+\dfrac{1}{2}=\dfrac{10}{3}\end{matrix}\right.\)

                  \(\left[{}\begin{matrix}\dfrac{3}{4}x=-\dfrac{10}{3}-\dfrac{1}{2}\\\dfrac{3}{4}x=\dfrac{10}{3}-\dfrac{1}{2}\end{matrix}\right.\)

                    \(\left[{}\begin{matrix}\dfrac{3}{4}x=-\dfrac{26}{3}\\\dfrac{3}{4}x=\dfrac{17}{6}\end{matrix}\right.\)

                      \(\left[{}\begin{matrix}x=-\dfrac{46}{9}\\x=\dfrac{34}{9}\end{matrix}\right.\)

Bài 2: 

#include <bits/stdc++.h>

using namespace std;

long long a,b;

int main()

{

cin>>a>>b;

cout<<a+b;

return 0;

}

a. Ta có: $\sin x\in [-1;1]$ nên $|\sin x|\in [0;1]$

$\Rightarrow 1\leq 3-2|\sin x|\leq 3$

Vậy $y_{\min}=1; y_{\max}=3$

b.

$y=\frac{1-\cos 2x}{2}-\frac{3}{2}\sin 2x+1$

$2y=3-\cos 2x-3\sin 2x$
$3-2y=\cos 2x+3\sin x$

Áp dụng định lý Bunhiacopxky:

$(3-2y)^2\leq (\cos ^22x+\sin ^22x)(1+3^2)=10$

$\Rightarrow -\sqrt{10}\leq 3-2y\leq \sqrt{10}$

$\Rightarrow \frac{3-\sqrt{10}}{2}\leq y\leq \frac{3+\sqrt{10}}{2}$

Vậy $y_{\max}=\frac{1+\sqrt{10}}{2}; y_{\min}=\frac{1-\sqrt{10}}{2}$

c. 

\(y=\sqrt{5-\frac{1}{4}(2\sin x\cos x)^2}=\sqrt{5-\frac{1}{2}\sin ^22x}\)

Vì $\sin 2x\in [-1;1]$

$\Rightarrow \sin ^22x\in [0;1]$

$\Rightarrow \frac{3\sqrt{2}}{2}\leq \sqrt{5-\frac{1}{2}\sin ^22x}\leq \sqrt{5}$

d. 

$\cos (x+\frac{\pi}{3})\in [-1;1]$

$\Rightarrow 2(-1)+3\leq 2\cos (x+\frac{\pi}{3})+3\leq 2.1+3$

$\Rightarrow 1\leq y\leq 5$

$\Rightarrow y_{\min}=1; y_{\max}=5$

e.

\(y=2\sin ^2x-\cos 2x=1-\cos 2x-\cos 2x=1-2\cos 2x\)

Vì $\cos 2x\in [-1;1]$ nên:

$-1\leq 1-2\cos 2x\leq 3$

Vậy $y_{\min}=-1; y_{\max}=3$

f.

$y=(\sin ^2x+\cos ^2x)^2-2(\sin x\cos x)^2$

$=1-\frac{1}{2}\sin ^22x$
Vì $-1\leq \sin 2x\leq 1\Rightarrow 0\leq \sin ^22x\leq 1$

$\Rightarrow \frac{1}{2}\leq y\leq 1$

Vậy $y_{\min}=\frac{1}{2};y_{\max}=1$ 

1 D