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

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

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

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

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

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 ???????????????????

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

a) 3x(x - 1) - 3(x - 1) = 0

(x - 1)(3x - 3) = 0

3(x - 1)(x - 1) = 0

3(x - 1)² = 0

x - 1 = 0

x = 1

b) x² - x = 0

x(x - 1) = 0

x = 0 hoặc x - 1 = 0

*) x - 1 = 0

x = 1

Vậy x = 0; x = 1

c) 25x² - 100x = 0

25x(x - 4) = 0

25x = 0 hoặc x - 4 = 0

*) 25x = 0

x = 0

*) x - 4 = 0

x = 4

Vậy x = 0; x = 4

d) (2x - 1)² - 64 = 0

(2x - 1 - 8)(2x - 1 + 8) = 0

(2x - 9)(2x + 7) = 0

*) 2x - 9 = 0

2x = 9

x = 9/2

*) 2x + 7 = 0

2x = -7

x = -7/2

Vậy x = -7/2; x = 9/2

giúp tui với mng ơi tui đang cần gấp ạaaa

Câu 3: 

a: \(\left(x+2\right)^2=x^2+4x+4\)

b: \(\left(x+3\right)^2=x^2+6x+9\)

c: \(\left(x-3\right)^2=x^2-6x+9\)

d: \(\left(x-7\right)^2=x^2-14x+49\)

e: \(x^2-6x+9=\left(x-3\right)^2\)

f: \(x^2-8x+16=\left(x-4\right)^2\)

g: \(=\left(x-10\right)\left(x+10\right)\)

h: \(=\left(x-11\right)\left(x+11\right)\)