1.Yes, they do

2..Yes, it is

3.People buy fruits and flowers from the market and decorate their house

4.People visit their family and friends

1 Yes, they do

2 Yes, it is

3 They often buy fruits and flowers from the market and decorate their houses

4 They often visit their family and friends

1, That
2, This
3, that
4, those
5, these - that
6, these
7, this
8, that
9, that
10, this
11, those
12, this
13, it
14, these
15, them

16, those

this/that dùng cho ng, vật số ít

this: dùng cho ng, vật ở gần

that: dùng cho ng, vật ở xa

these/those dùng cho ng, vật số nhiều

these: ở gần

those: ở xa

bạn có thể chụp rõ hơn đc ko ạ

nhỏ quá ko ai thấy đâu

1.

c, \(sin\left(\dfrac{\pi}{3}-x\right)=-\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{3}-x=arcsin\left(-\dfrac{1}{4}\right)+k.360^o\\\dfrac{\pi}{3}-x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k.360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}-arcsin\left(-\dfrac{1}{4}\right)+k.360^o\\x=-\dfrac{2\pi}{3}+arcsin\left(-\dfrac{1}{4}\right)+k.360^o\end{matrix}\right.\)

d, \(sin4x=\dfrac{2}{3}\)

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

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

1.

e, \(2sin2x+\sqrt{2}=0\)

\(\Leftrightarrow sin2x=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin2x=sin\left(-\dfrac{\pi}{4}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+k\pi\\x=\dfrac{5\pi}{8}+k\pi\end{matrix}\right.\)

Bài 1: 

a: \(36a^4-y^2=\left(6a^2-y\right)\left(6a^2+y\right)\)

n: \(6x^2+x-2\)

\(=6x^2+4x-3x-2\)

\(=\left(3x+2\right)\left(2x-1\right)\)

a: ΔADE vuông tại A

=>\(AD^2+AE^2=DE^2\)

=>\(DE^2=6^2+8^2=36+64=100=10^2\)

=>DE=10(cm)

b:

Sửa đề: Chứng minh \(DI\cdot DE=DA^2\)

Xét ΔDIA vuông tại I và ΔDAE vuông tại A có

\(\hat{IDA}\) chung

Do đó: ΔDIA~ΔDAE

=>\(\frac{DI}{DA}=\frac{DA}{DE}\)

=>\(DI\cdot DE=DA^2\)

c:Sửa đề: Tính DI,IE

Ta có: \(DI\cdot DE=DA^2\)

=>DI=6^2/10=3,6(cm)

Ta có: DI+IE=DE

=>IE=10-3,6=6,4(cm)

d: Xét ΔDAE có AM là phân giác

nên \(\frac{MD}{ME}=\frac{AD}{AE}=\frac68=\frac34\)

=>\(\frac{MD}{3}=\frac{ME}{4}\)

mà MD+ME=DE=10

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{MD}{3}=\frac{ME}{4}=\frac{MD+ME}{3+4}=\frac{10}{7}\)

=>\(MD=3\cdot\frac{10}{7}=\frac{30}{7}\) (cm)

mn ơi  giúp em

Bài 3:

\(a,=3x\left(y-4x+6y^2\right)\\ b,=5xy\left(x^2-6x+9\right)=5xy\left(x-3\right)^2\\ d,=\left(x+y\right)\left(x-12\right)\\ f,=2x\left(x-y\right)\left(5x-4y\right)\\ g,=\left(x-2\right)\left(x-2+3x\right)=\left(x-2\right)\left(4x-2\right)=2\left(x-2\right)\left(2x-1\right)\\ h,=x^2\left(1-5x\right)+3xy\left(5x-1\right)=x\left(1-5x\right)\left(x-3y\right)\\ i,=x\left(x-2\right)+4\left(x-2\right)=\left(x+4\right)\left(x-2\right)\\ j,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ k,=4x^2-12x+3x-9=\left(x-3\right)\left(4x+3\right)\\ l,=\left(x+5\right)^2-y^2=\left(x-y+5\right)\left(x+y+5\right)\\ m,=x^2-\left(2y-6\right)^2=\left(x-2y+6\right)\left(x+2y-6\right)\\ n,=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\\ =\left(x^2+5x+5\right)^2-1-24\\ =\left(x^2+5x+5\right)^2-25\\ =\left(x^2+5x\right)\left(x^2+5x+10\right)\\ =x\left(x+5\right)\left(x^2+5x+10\right)\)