9dm2 20cm2 = .920..cm2

5 tạ 7 yến = .570..kg

3 giờ 5 phút = .185..phút

6 phút 32 giây = .392..giây

9dm2 20cm2 = .920..cm2

5 tạ 7 yến = .570..kg

3 giờ 5 phút = .185..phút

6 phút 32 giây = .392..giây

nhá

`(1+2cosx)(3-cosx)=0`

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{2}\\cosx=3\left(L\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{3}+k2\pi\\x=\dfrac{-2\pi}{3}+k2\pi\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{2\pi}{3}+k\pi\)

`(k \in ZZ)`

\(\Leftrightarrow\left[{}\begin{matrix}1+2\cos x=0\\3-\cos x=0\end{matrix}\right.\)

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

Mà \(-1\le\cos x\le1\)

\(\Rightarrow\cos x=-\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\pi+k2\pi\\x=\dfrac{4}{3}\pi+k2\pi\end{matrix}\right.\)

Vậy ...

5. 

ĐKXĐ: \(cos\left(x-30^0\right)\ne0\Leftrightarrow x\ne120^0+k180^0\)

Pt tương đương:

\(\left[{}\begin{matrix}tan\left(x-30^0\right)=0\\cos\left(2x-150^0\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-30^0=k180^0\\2x-150^0=90^0+k180^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=120^0+k90^0\end{matrix}\right.\)

Kết hợp ĐKXĐ: \(\Rightarrow x=30^0+k180^0\)

6.

\(\Leftrightarrow2\sqrt{2}sinx.cosx+2cosx=0\)

\(\Leftrightarrow2cosx\left(\sqrt{2}sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)

5. \(\Leftrightarrow\left[{}\begin{matrix}\tan\left(x-30\right)=0\\\cos\left(2x-150\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-30=360k\\\left[{}\begin{matrix}2x-150=90+360k\\2x-150=270+360k\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=30+360k\\x=120+180k\\x=210+180k\end{matrix}\right.\)

Vậy ...

6, \(\Leftrightarrow2\sqrt{2}\sin x.\cos x+2\cos x=0\)

\(\Leftrightarrow\cos x\left(1+\sqrt{2}\sin x\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{3}{2}\pi+k2\pi\end{matrix}\right.\\x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{7}{4}\pi+k2\pi\end{matrix}\right.\)

Vậy ...

Bài 5:

a: 2x-(3-5x)=4(x+3)

=>2x-3+5x=4x+12

=>7x-3=4x+12

=>3x=15

=>x=5

b: =>5/3x-2/3+x=1+5/2-3/2x

=>25/6x=25/6

=>x=1

c: 3x-2=2x-3

=>3x-2x=-3+2

=>x=-1

d: =>2u+27=4u+27

=>u=0

e: =>5-x+6=12-8x

=>-x+11=12-8x

=>7x=1

=>x=1/7

f: =>-90+12x=-45+6x

=>12x-90=6x-45

=>6x-45=0

=>x=9/2

152 + (-173) - (-18) - 127 

= 152 + (-173) + 18 - 127

= (152 + 18) + (-173) - 127

= 170 + (-300)

= -130

\(152+\left(-173\right)-\left(-18\right)-127\)

\(=\left(152+18\right)-\left(173+127\right)\)

\(=170-300=-130\)

152+(-173)-(-18)-127

=152-173+18-127

=(152+18)-(173+127)

=170-300

=-130

ăn cá bỏ lờ  (Lờ là đồ đan bằng tre dùng để đón cá mà bắt) Chê kẻ quên ơn: Hạng người ăn cá bỏ lờ ấy thực là đáng khinh.

nhìn hơi rối

bn tách ra đc k ặc

a,1/5+4/11+4/5+7/11

=(1/5+4/5)+(4/11+7/11)

=1+1

=2

Chọn B

1367.54+1367.45+1367

=1367.(54+45+1)
=1367.100

=136700

a) 1/5 + 4/11 + 4/5 + 7/11 = (1/5 + 4/5) + (4/11 + 7/11) = 1 + 1 = 2

=> B

b) 1367 x 54 + 1367 x 45 + 1367 = 1367 x (54 + 45 + 1) = 1367 x 100 = 136700

Đáp án là môn đua xe đạp.