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

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

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

b) Ta có: 5x-8=2x+7

\(\Leftrightarrow5x-2x=7+8\)

\(\Leftrightarrow3x=15\)

hay x=5

c) Ta có: \(4x^2+2x-5=4x^2-2x+3\)

\(\Leftrightarrow2x-5+2x-3=0\)

\(\Leftrightarrow4x=8\)

hay x=2

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

\(a,9^2.3^5=3^4.3^5=3^9\)

\(b,2^5.4^2.16=2^5.2^4.2^4=2^{13}\)

\(c,9^2.27.3^5=3^4.3^3.3^5=3^{12}\)

\(d,5^2.25^2.125=5^2.5^4.5^3=5^9\)

\(e,7^4.343.49^2=7^4.7^3.7^4=7^{11}\)

\(f,100^2.10^5.1000=10^{13}\)