\(pt\text{⇔}\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\text{⇔}x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\\ \text{⇔}17x=17\text{⇔}x=1\)

Vậy nghiệm của phương trình : \(S=\left\{1\right\}\)

Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2=27\)

\(\Leftrightarrow17x=17\)

hay x=1

\(d,\left(2x-1\right)\left(x+3\right)+2x\left(2-x\right)=10-8\left(x+4\right)\\ \Leftrightarrow2x^2+6x-x-3+4x-2x^2=10-8x-32\\ \Leftrightarrow17x=19\Leftrightarrow x=\dfrac{19}{17}\)

vậy phương trình đã cho có nhiệm \(x=\dfrac{19}{17}\)

Ta có: \(\left(2x-1\right)\left(x+3\right)+2x\left(2-x\right)=10-8\left(x+4\right)\)

\(\Leftrightarrow2x^2+6x-x-3+4x-2x^2=-8x-22\)

\(\Leftrightarrow18x=-19\)

hay \(x=-\dfrac{19}{18}\)

1 has never met

2 are playing

3 Have you ever sung

4 has worked

5 will help

6 will wait

7 will phone

8 will pay

9 will stay

10 will invite

11 is studying

12 goes - is going

13 is raining - is sleeping

14 will be

1, has never met

2,are playing

3,Have-ever sung

4,has worked

5, will help

6,will wait

7,will phone

8,will pay

9,will stay

10 will invite

11is studying

12, goes, has gone

13,is raining

14, is sleeping

15, will be

Teenagers become more independent and responsible to their parents.

Many teenagers in the 21st century are less selfish and ready to help other people in need

Teenagers are also more business-orientated and have to find their own source of opportunity

Most high school students believe they should be treated like adults

The primary role of young people is to get a good education to become better citizens of tomorrow.

Câu b

5x-8=2x+7

<=> 3x=7+8=15

<=>x=5

Câu c:

<=>4x=3+5

<=>4x=8

<=>x=2

Câu d

<=>6=4x

<=> x=3/2

Câu e

<=> 2x+8-6x+15=3

<=>4x=20

<=>x=5

b)5x-8=2x+7

⇔3x=15

⇔ x=5

c)4x²+2x-5=4x²-2x+3

⇔ 4x=8

⇔ x=2

d)2x³-3x+6=2x³+x

⇔ 4x=6

⇔ \(x=\dfrac{3}{2}\)

e)2(x+4)-3(2x-5)=3

⇔ 2x+4-6x+15=3

⇔ -4x+19=3

⇔ 4x=16

⇔ x=4

a: Ta có: \(\widehat{ABC}=\widehat{ACB}\)(ΔABC cân tại A)

\(\widehat{ACB}=\widehat{ECN}\)(hai góc đối đỉnh)

Do đó: \(\widehat{ABC}=\widehat{ECN}\)

Xét ΔMBD vuông tại D và ΔNCE vuông tại E có

BD=CE

\(\widehat{MBD}=\widehat{NCE}\)

Do đó: ΔMBD=ΔNCE

=>DM=EN

b: Ta có: DM\(\perp\)BC

EN\(\perp\)BC

Do đó: DM//EN

Xét ΔIDM vuông tại D và ΔIEN vuông tại E có

MD=EN

\(\widehat{MDI}=\widehat{ENC}\)(hai góc so le trong, DM//EN)

Do đó: ΔIDM=ΔIEN

=>IM=IN

=>I là trung điểm của MN

mấy cái  số mũ khó nhìn quá

Bài 5: 

a: Ta có: \(2^x=16\)

nên x=4

b: Ta có: \(3^x=243\)

nên x=5