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

cần chi tiết ko bạn

hay mỗi đáp án nhợ

Ta có:\(\sqrt{14-2\sqrt{33}}\)=1,584573983

\(\sqrt{33-4\sqrt{33}}\)=3,165714677

=>1,584573983\(-\)3,165714677=-1,581140694

nhé bạn

-21

-21

(-3 ) . 7 = 21

Chúc bn hc tốt nhá !

\(=\dfrac{35\cdot0.1\cdot154\cdot243\cdot50}{35\cdot154\cdot0.1\cdot121.5\cdot100}\)

\(=\dfrac{243}{121.5}\cdot\dfrac{1}{2}=1\)

(-5)³.\(x^2\) = - 1125

        \(x^2\) = (-1125) : (-5³)

        \(x^2\) = 9

         \(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-3; 3}

\(\left(-5\right)^3.x^2=-1125\)

\(x^2=-1125:\left(-5\right)^3\)

\(x^2=-1125:\left(-125\right)\)

\(x^2=9\)

\(x^2=3^2\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

⇒ Vậy \(x\in\left\{{}\begin{matrix}3\\-3\end{matrix}\right.\)

    2x+1 là ước của 3x+2
⇔3x+2 ⋮ 2x+1
⇒2(3x+2) ⋮ 2x+1
⇔6x+4 ⋮ 2x+1
⇔(2x+1)+(2x+1)+(2x+1)+1 ⋮ 2x+1
Để 3x+2 ⋮ 2x+1 thì 2x+1 ∈ Ư(1)
Ta có:
Ư(1)={±1}
⇒2x+1∈{±1}
⇒x∈{0;-1}
Vậy x={0;-1)

Thenkiu very much !!! 

Ta có : 2x + 1 là ước của 3x + 2

=> 3x + 2 ⋮ 2x + 1

=> 2(3x + 2) ⋮ 2x + 1

=> 6x + 4 ⋮ 2x + 1

=> (6x + 3) + 1 ⋮ 2x + 1

=> 3(2x + 1) + 1 ⋮ 2x + 1

Vì 3(2x + 1) ⋮ 2x + 1 nên 1 ⋮ 2x + 1

=> 2x + 1 ∈ Ư(1) ∈ {-1;1}

=> x ∈ {-1;0}

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

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

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

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

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

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

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