a) |x - 1,7| = 2,3

Xét 2 trường hợp:

TH1: x - 1,7 = -2,3

         x         = -2,3 +1,7

         x         = -0,6

TH2: x - 1,7 = 2,3

         x         = 2,3 + 1,7

         x         = 4

Vậy: Tự kl :<

c)

+)x<1=>/x-1/=1-x=2x-3=>1-x-(2x-3)=0=>4-3x=0=>x=4/3 (loại)

+)x>=1=>x-1=2x-3=>2x-x-3+1=0=>x-2=0=>x=2(t/m)

Vậy: x=2 haizz

x + 6 chia hết cho x + 3

=> x + 3 + 3 chia hết cho x + 3

=> 3 chia hết cho x + 3

=> (x + 3) \(\in\) Ư(3)

=> (x + 3) \(\in\) {-3; -1; 1; 3}

=> x \(\in\) {-6; -4; -2; 0}

1 B

2 A

3 D

4 C

5 B

6 A

7 C

8 A

9 B

10 A

19:14km. Câu 20 chịu:((((

Ồ hố thành niên này ko biết

Bài 1:

a. 

$545,26+117,3=662,56$

b.

$400,56-184,48=216,08$

c.

$4,21\times 3,2=13,472$

d.

$28,5:2,5=11,4$

Bài 2:

a. 2km 21 m = 2,021 km

b. 1020 kg = 1 tấn 20 kg 

c. 22 dam² 10 m² = 2210 m²

d. 90 giây = 1,5 phút

a: \(\frac{5x-5}{\left(x+1\right)^2}\cdot\frac{3+3x}{20-20x}\)

\(=\frac{5\left(x-1\right)}{\left(x+1\right)^2}\cdot\frac{3\left(x+1\right)}{-20\left(x-1\right)}\)

\(=\frac{5}{-20}\cdot\frac{3}{x+1}=\frac{-1}{4}\cdot\frac{3}{x+1}=\frac{-3}{4x+4}\)

b: \(\frac{x+4}{5x-25}\cdot\frac{x-5}{x^2+8x+16}\)

\(=\frac{x+4}{5\left(x-5\right)}\cdot\frac{x-5}{\left(x+4\right)^2}\)

\(=\frac{1}{5\left(x+4\right)}\)

c: \(\frac{x^2+xy}{5x^2-5y^2}\cdot\frac{3x^3-3y^3}{x^2-xy}\)

\(=\frac{x\left(x+y\right)}{5\left(x^2-y^2\right)}\cdot\frac{3\left(x^3-y^3\right)}{x\left(x-y\right)}\)

\(=\frac{x+y}{x-y}\cdot\frac{3\left(x-y\right)\left(x^2+xy+y^2\right)}{5\left(x-y\right)\left(x+y\right)}=\frac{3\left(x^2+xy+y^2\right)}{5\left(x-y\right)}\)

d: \(\frac{x^2y^2-4y^2}{4xy}\cdot\frac{x^2y}{2xy-x^2y}\)

\(=\frac{y^2\left(x^2-4\right)}{4xy}\cdot\frac{x^2y}{xy\left(2-x\right)}=\frac{-x^2y^3}{4xy\cdot xy}\cdot\frac{\left(x-2\right)\left(x+2\right)}{x-2}\)

\(=-\frac{y}{4}\cdot\left(x+2\right)=\frac{-y\left(x+2\right)}{4}\)

e: \(\frac{2x^3-2y^3}{3x+3y}\cdot\frac{15x^2-15y^2}{x^2+xy+y^2}\)

\(=\frac{2\left(x^3-y^3\right)}{3\left(x+y\right)}\cdot\frac{15\left(x^2-y^2\right)}{x^2+xy+y^2}\)

\(=\frac{30}{3}\cdot\frac{\left(x-y\right)\left(x^2+xy+y^2\right)\cdot\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(x^2+xy+y^2\right)}=10\left(x-y\right)^2\)

f: \(\frac{x^4-y^4}{x^2-2xy+y^2}\cdot\frac{x-y}{xy+x^2}\)

\(=\frac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x-y\right)^2}\cdot\frac{x-y}{x\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^2}\cdot\frac{\left(x^2+y^2\right)\left(x-y\right)}{x\left(x+y\right)}\)

\(=\frac{x^2+y^2}{x}\)

18.A
19.B
20.A
 Có sai sót gì mong bạn thông cảm
Thanks