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DT
Đỗ Thanh Hải
CTVVIP
15 tháng 10 2021
1 lived
2 wasn't hearing - was thinking
3 walked - was
4 were you doing - phoned
5 was reading - heard
6 was walking - saw
7 had watched - wrote
8 met
9 had you done - moved
10 was - didn't attend
II
1 when we were having lunch
2 had read the instruction, I started the machine
3 went out for a rest, we had finished our assignment
4 she left, it was raining
5 learning English
6 to help me
7 the man to open the briefcase
8 my friend break the bottle
9 him fall off the bike
10 a foreign language in a short time is not easy
16 tháng 6 2023
\(=\dfrac{35\cdot0.1\cdot154\cdot243\cdot50}{35\cdot154\cdot0.1\cdot121.5\cdot100}\)
\(=\dfrac{243}{121.5}\cdot\dfrac{1}{2}=1\)
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
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 $y_{\max}=\frac{1+\sqrt{10}}{2}; y_{\min}=\frac{1-\sqrt{10}}{2}$
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
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.
$\cos (x+\frac{\pi}{3})\in [-1;1]$
$\Rightarrow 2(-1)+3\leq 2\cos (x+\frac{\pi}{3})+3\leq 2.1+3$
$\Rightarrow 1\leq y\leq 5$
$\Rightarrow y_{\min}=1; y_{\max}=5$
1 D