VD ve goc xOy va ve tia Ot nam giua goc xOy 

lay diem A nam trong goc  xOt  ta thay A ko nam trong goc tOy

=> xOt va tOy ko co diem trong chung

Ban cu hieu nom na la neu 1 diem nam trong goc nay ma ko nay trong goc kia thi 2 goc do ko co diem trong chung

Minh hoc kem Van nen dien dat hoi kem mong bn thong cam

=> x+5=-4 => x=-9 

=> x=-4

vậy x thuộc ( -9; -4)

<=>  \(x^2+5x+4=0\)

   \(x^2+4x+x+4=0\)

\(x\left(x+4\right)+\left(x+4\right)=0\)

\(\left(x+4\right)\left(x+1\right)=0\)

=> x=-4  hoặc x=-1

1. Sam isn't good at soccer. He loses every game.

2. Nina is very  kind. Everyone loves her

3. Don't try to use this cell phone. It doesn't work

4. What is the matter with you? You look so sad

5. I live quite close to the school, so I walk to there

6. I always go to school by car. I own a car

7. I hate cooking. I just eat

8. He has got three dogs and a cat. He loves them very much

9. I want to buy this house. How much does it cost ?

10. I love English. I learn very well

\(2S=2^2+2^3+2^4+.....+2^{2020}\)

\(S=2+2^2+2^3+.....+2^{2019}\)

Trừ 2 vế của 2 đẳng thức ta có:

\(\Rightarrow2S-S=S=2^{2020}-2\)

S=2+2²+2³+.....+2²⁰¹⁹

2S=2.(2+2²+2³+......+2²⁰¹⁹)

2S=2²+2³+2⁴+......+2²⁰²⁰

2S-S=(2²+2³+2⁴+.....+2²⁰²⁰)-(2+2²+2³+.......+2²⁰¹⁹⁾

S=2²⁰²⁰-2

\(2x-\frac{5}{9}=\frac{1}{3}+\text{ }[\frac{11}{3}-4+\frac{2}{3}]\)

\(2x-\frac{5}{9}=\frac{1}{3}+\text{ }[-\frac{1}{3}+\frac{2}{3}]\)
\(2x-\frac{5}{9}=\frac{1}{3}+\frac{1}{3}\)
\(2x-\frac{5}{9}=\frac{2}{3}\)
\(2x=\frac{11}{9}\)
   \(x=\frac{11}{18}\)
Vậy \(x=\frac{11}{18}\)

a) ta có -3x+5y=33

=> 5y=33+3x

=> y=(33+3x)/5

thay y=(33+3x)/5 vào x/y=3/4 ta đc

x/y=3/4

x/(33+3x)/5=3/4

5x/(33+3x)=3/4

x=9

thay x=9 vào x/y=3/4 ta đc

x/y=3/4

9/y=3/4

y=12

Ta có:\(\frac{x}{y}=\frac{3}{4}\Leftrightarrow\frac{x}{3}=\frac{y}{4}=\frac{-3x}{-9}=\frac{5y}{20}=\frac{-3x+5y}{-9+20}=\frac{33}{11}=3\)

(Áp dụng tính chất dãy tỉ số bằng nhau)

\(\Rightarrow\hept{\begin{cases}x=3\cdot3=9\\y=3\cdot4=12\end{cases}}\)

a) bài 1

để \(x\in Z\)thì \(3x-1⋮x-1\)

mà \(x-1⋮x-1\)

\(\Rightarrow3\left(x-1\right)⋮x-1\)

\(\Rightarrow\left(3x-1\right)-\left[3x-3\right]⋮x-1\)

\(\Rightarrow2⋮x-1\)

\(\Rightarrow x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

ta có bảng

vậy \(x\in\left\{2;0;3;-1\right\}\)

còn nữa mà bạn