Bài 2:

Ta có:\(2\sqrt{48}< 2\sqrt{49}\) ; 

         \(3\sqrt{27}>3\sqrt{25}\)

mà \(2\sqrt{49}< 3\sqrt{25}\left(14< 15\right)\)

\(\Rightarrow3\sqrt{27}>3\sqrt{25}>2\sqrt{49}>2\sqrt{48}\)

\(\Rightarrow3\sqrt{27}>2\sqrt{48}\)

b)

Ta có:\(\sqrt{50}+\sqrt{2}>\sqrt{49}+\sqrt{1}\) 

        \(\sqrt{50+2}< \sqrt{64}\)

mà \(\sqrt{49}+\sqrt{1}=\sqrt{64}\left(8=8\right)\)

\(\Rightarrow\sqrt{50}+\sqrt{2}>8>\sqrt{50+2}\)

\(\Rightarrow\sqrt{50}+\sqrt{2}>\sqrt{50+2}\)

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

bài 1:

bn lấy giá trị của √(4^2-3,9^2) là dc

bài 2

AB+BC=2√(3^2+4^2)=??

23.17-23.14

=23.(17-14)

=23.3 = 69

23.(17-14)=23.3=69

2.

\(cosx+cos3x=1+\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow2cos2x.cosx=1+cos2x+sin2x\)

\(\Leftrightarrow2cos2x.cosx=2cos^2x+2sinx.cosx\)

\(\Leftrightarrow cosx\left(cos2x-cosx-sinx\right)=0\)

\(\Leftrightarrow cosx\left(cos^2x-sin^2x-cosx-sinx\right)=0\)

\(\Leftrightarrow cosx\left(cosx+sinx\right)\left(cosx-sinx-1\right)=0\)

\(\Leftrightarrow cosx.\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right).\left[\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin\left(x+\dfrac{\pi}{4}\right)=0\\cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\\x+\dfrac{\pi}{4}=\pm\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\\x=k2\pi\end{matrix}\right.\)

Bài 2

5 C

Bài 3

1 D

6 C

Còn lại ol r nhé

2) 5. C

3) 2. D

6. C

Còn lại ok nha

Bài 1: 

a: 726314:29=25045(dư 9)

b: 172864:315=548,7746032

3 A

4 D

5 A

6 C

7 C

8 A

9 A

10 C

11 C

12 A

13 A

14 B

15 A

16 D

17 D

18 C

19 D

20 D

21 C

22 C

23 C

24 C

25 C

Ex2

1 remember 

2 read

3 be repaired

4 focus

5 not enter

6 apoligize

7 attend

8 sit

9 think

10 drive

11 be taken

12 reduce

13 allow

14 regulated

15 swim

16 monitor

17 out

18 be invited

19 send

20 work 

mình làm những bài bn chưa lm nhé

9B

10A

bài 2

have repainted

bàii 3

ride - walikking

swimming

watch

2 have repainted

4 have done 

8 have stepped

9 have you ever played

7) 1 riding -walking

2 swimming

3 watching

8)

2 my mother doesn't like listening to music

5 doing more exercises makes you healthier

còn lại ok nha

chịu ???????????????????

Bài 5: 

a: BC=10cm

b: HA=4,8cm

HB=3,6(cm)

HC=6,4(cm)

Bài 6:

\(x^3=6+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow x^3=6+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=6\\ y^3=34+3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\right)\\ \Leftrightarrow y^3=34+3y\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=34\\ \Leftrightarrow P=x^3-3x+y^3-3y+1980=6+34+1980=2020\)

gfrưerrrrrrrrrrr

Bài 2:

a. $x^2=12y^2+1$ lẻ nên $x$ lẻ 

Ta biết một scp khi chia 8 dư $0,1,4$. Mà $x$ lẻ nên $x^2$ chia $8$ dư $1$

$\Rightarrow 12y^2+1\equiv 1\pmod 8$

$\Rightarrow 12y^2\equiv 0\pmod 8$

$\Rightarrow y^2\equiv 0\pmod 2$

$\Rightarrow y$ chẵn. Mà $y$ nguyên tố nên $y=2$.

Khi đó: $x^2=12y^2+1=12.2^2+1=49\Rightarrow x=7$ (tm)

Bài 2:

b.

$x^2=8y+1$ nên $x$ lẻ. Đặt $x=2k+1$ với $k$ tự nhiên.

Khi đó: $8y+1=x^2=(2k+1)^2=4k^2+4k+1$

$\Rightarrow 2y=k(k+1)$

Vì $(k,k+1)=1, k< k+1$ và $y$ nguyên tố nên xảy ra các TH sau:

TH1: $k=2, k+1=y\Rightarrow y=3\Rightarrow x=5$ (tm) 

TH2: $k=1, k+1=2y\Rightarrow y=1$ (vô lý) 

TH3: $k=y, k+1=2\Rightarrow y=1$ (vô lý)

Vậy $(x,y)=(5,3)$ là đáp án duy nhất thỏa mãn.

Bài 3:

a. $A=\frac{5-x}{x+2}=\frac{7-(x+2)}{x+2}=\frac{7}{x+2}-1$

Để $A$ min thì $\frac{7}{x+2}$ min

Điều này xảy ra khi $x+2$ là số nguyên âm lớn nhất

$\Rightarrow x+2=-1$

$\Rightarrow x=-3$. Khi đó: $A_{\min}=\frac{7}{-1}-1=-7-1=-8$

b.

$B=\frac{x^2-4x+10-8}{x^2-4x+10}=1-\frac{8}{x^2-4x+10}$

Để $B$ min thì $\frac{8}{x^2-4x+10}$ max

Điều này xảy ra khi $x^2-4x+10$ min 

Mà: $x^2-4x+10=(x-2)^2+6\geq 6$ với mọi $x$

$\Rightarrow x^2-4x+10$ max = 6 khi $x=2$

Khi đó: $B_{\min}=1-\frac{8}{x^2-4x+10}=1-\frac{8}{2^2-4.2+10}=\frac{-1}{3}$