My best friend is Luong Anh. She is 13 years old. She is in the same class as me. We have been best friends for 7 years when we were in class 1A1. Because her house is near my house, she often goes to my house to play with me.

She is tall and thin. She has an oval face and a long hair. She is very beautiful.

She is a good student. She always gets good marks. However, she is very friendly. She often helps everybody. When I can't do my homeworks, she help me complete them. When I feel sad, she supports me. I'm very happy because I have her.

We are best friends forever!

S₂ = 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰

Tổng S₂ có 100 số, nhóm 5 số vào 1 nhóm thì vừa hết. 

Ta có:

S₂= (2 + 2² + 2³ + 2⁴ + 2⁵) + ... + (2⁹⁶ + 2⁹⁷ + 2⁹⁸ + 2⁹⁹ + 2¹⁰⁰)

= 2.(1 + 2 + 2² + 2³ + 2⁴) + ... + 2⁹⁶. (1 + 2 + 2² + 2³ + 2⁴)

= 2. 31 + ... + 2⁹⁶ . 31

= 31 (2 + ...+ 2⁹⁶)

1,   |x-3|=5

TH1: x-3\(\ge\)0 => x\(\ge\)3

x-3=5

x=5+3

x=8 (thỏa mãn )

TH2: x-3<0 => x<3

-(x-3)=5

-x+3=5

-x=5-2

-x=3

x=-3 (thỏa mãn )

Vậy x={8;-3}

2,    |x| - |-5|=14

|x|-5=15

|x|=15+5

|x|=20

TH1: x\(\ge\)0

x=20 (thỏa mãn )

TH2 : x<0

x=-20 (thỏa mãn )

Vây x={20;-20}

ơ sao em gửi bài rồi mà ko thấy hiện lên vậy ạ?

1. Do you like drinking coffe ? Why / Why not ?

=>Yes, I like drinking coffee because it's very delicious.

2. What do you often eat for breakfast ?

=>I often eat steamed sticky rice with roasted sesame and salt for breakfast.

3. What do you often eat for lunch ?

=>I often eat rice, meat and vegetable for lunch.

4. What do you often eat for dinner ?

=>I often eat rice, fry eggs and vegetable for dinner.

ông già noel không có thật mà

Ta có : 2/3 số vở 5a = 3/5 số vở 5b = 4/7 số vở 5c hay

           24/36 số vở 5a = 24/40 số vở 5b = 24/42 số vở 5c hay

            1/36 số vở 5a = 1/40 số vở 5b = 1/42 số vở 5c

Ta có sơ đồ :

5A : 36 phần

5B : 40 phần 

5C : 42 phần 

Số vở lớp 5A là : 177 : ( 36 + 40 + 42 ) x 36 = 54 ( quyển )

Số vở lớp 5B là : 177 : ( 36 + 40 + 42 ) x 40 = 60 ( quyển )

Số vở lớp 5C là : 177 - 54 - 60 = 63 ( quyển )

d)\(6x^2+7x-3=0\Leftrightarrow6x^2-2x+9x-3=0\Leftrightarrow2x\left(3x-1\right)+3\left(3x-1\right)=0\)

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

e)\(10x^2-14x-12=0\Leftrightarrow2\left(5x^2-7x-6\right)=0\Leftrightarrow5x^2-7x-6=0\)

\(\Leftrightarrow5x^2+3x-10x-6=0\Leftrightarrow x\left(5x+3\right)-2\left(5x+3\right)=0\Leftrightarrow\left(x-2\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\5x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-\frac{3}{5}\end{array}\right.\)

a)\(4x^2-7x-2=0\Leftrightarrow4x^2+x-8x-2=0\Leftrightarrow x\left(4x+1\right)-2\left(4x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-\frac{1}{4}\end{array}\right.\)

b)\(3x^2+10x+3=0\Leftrightarrow3x^2+9x+x+3=0\Leftrightarrow3x\left(x+3\right)+\left(x+3\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x+3\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}3x+1=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{3}\\x=-3\end{array}\right.\)

c)\(x^2-x-20=0\Leftrightarrow x^2+4x-5x-20=0\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-5=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=5\\x=-4\end{array}\right.\)

Mình nghĩ cái hình của bạn như này :)

\(\Delta ABC\) là tam giác cân <=> \(\widehat{BAC}=\widehat{ABC}=\widehat{ACB}=60^o\)

a)BD là tia phân giác của \(\widehat{ABC}\) => \(\widehat{ABD}=\widehat{DBC}=\frac{1}{2}.\widehat{ABC}=\frac{1}{2}.60^o=30^o\)

\(\Delta BDC\) có: \(\widehat{DBC}+\widehat{BDC}+\widehat{BCD}=180^o\) (tổng 3 góc trong 1 tam giác)

=>\(30^o+\widehat{BDC}+60^o=180^o\Rightarrow\widehat{BDC}=180^o-60^o-30^o=90^o\)

=>Góc BDC là góc vuông => \(BD\perp AC\) (đpcm)

b)CE là tia phân giác của góc ACB => \(\widehat{ACE}=\widehat{ECB}=\frac{1}{2}.\widehat{ACB}=\frac{1}{2}.60^o=30^o\)

\(\Delta EBC\) có: \(\widehat{ECB}+\widehat{BEC}+\widehat{EBC}1=180^o\) (tổng 3 góc trong 1 tam giác)

=>\(30^o+\widehat{BEC}+60^o=180^o\Rightarrow\widehat{BEC}=180^o-60^o-30^o=90^o\)

=>Góc BEC là góc vuông => \(CE\perp AB\) (đpcm)

c) Câu này có vấn đề nhé bạn, mình tính ra góc ACB=60^o, nhưng góc BOC=120^o cơ :v