Chương I : Số hữu tỉ. Số thực

\(1,0,75-\dfrac{2}{3}-0,5=\dfrac{3}{4}-\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{9}{12}-\dfrac{8}{12}-\dfrac{1}{2}=\dfrac{1}{12}-\dfrac{1}{2}\)

\(=\dfrac{2}{24}-\dfrac{12}{24}=\dfrac{-10}{24}=\dfrac{-5}{12}\)

\(2,\dfrac{1}{5}-0,125-\dfrac{5}{4}=\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{5}{4}=\dfrac{8}{40}-\dfrac{5}{40}-\dfrac{5}{4}=\dfrac{3}{40}-\dfrac{5}{4}\)

\(=\dfrac{3}{40}-\dfrac{50}{40}=\dfrac{-47}{40}\)

\(3,1,25-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{5}{4}-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{2}{4}+\dfrac{4}{3}=\dfrac{6}{12}+\dfrac{16}{12}=\dfrac{22}{12}=\dfrac{11}{6}\)

\(4,0,15-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{5}{20}+\dfrac{2}{5}=\dfrac{-2}{20}+\dfrac{2}{5}\)

\(=\dfrac{-2}{20}+\dfrac{8}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)

\(5,5-3,4+\dfrac{1}{5}=\dfrac{5}{1}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25}{5}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25-17+1}{5}=\dfrac{9}{5}\)

\(6,\dfrac{1}{4}-0,3+\dfrac{4}{3}=\dfrac{1}{4}-\dfrac{3}{10}+\dfrac{4}{3}=\dfrac{10}{40}-\dfrac{12}{40}+\dfrac{4}{3}=\dfrac{-2}{40}+\dfrac{4}{3}\)

\(=\dfrac{-1}{20}+\dfrac{4}{3}=\dfrac{-3}{60}+\dfrac{80}{60}=\dfrac{77}{60}\)

\(7,0,2-3,25+4,7=\dfrac{1}{5}-\dfrac{13}{4}+\dfrac{47}{10}=\dfrac{4}{20}-\dfrac{65}{20}+\dfrac{47}{10}=\dfrac{-61}{20}+\dfrac{47}{10}\)

\(=\dfrac{-61}{20}+\dfrac{94}{20}=\dfrac{33}{20}=1,65\)

\(8,5,4+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{27}{5}+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{81}{15}+\dfrac{-35}{15}-\dfrac{-5}{7}\)

\(=\dfrac{46}{15}-\dfrac{-5}{7}=\dfrac{322}{105}-\dfrac{-75}{105}=\dfrac{397}{105}\)

\(9,\dfrac{-4}{2}+\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{-12}{6}+\dfrac{2}{6}-\dfrac{1}{4}=\dfrac{-10}{6}-\dfrac{1}{4}=\dfrac{-5}{3}-\dfrac{1}{4}\)

\(=\dfrac{-20}{12}-\dfrac{3}{12}=\text{ }\dfrac{-23}{12}\)

\(10,5,4-1,5-\left(7,2-1\right)=3,9-6,2=-2,3\)

\(11,4,9-\left(1,5-7,7+3\right)=4,9-\left(-3,2\right)=8,1\)

\(12,7,8-4,7+\left(5,3-1,4\right)=3,1+3,9=7\)

\(14,\dfrac{1}{2}-0,4+\dfrac{1}{5}\text{=}0,5-0,4+0,2=0,3\)

\(15,4,2-\dfrac{4}{5}+\dfrac{1}{2}=4,2-0,8+0,5=3,9\)

1: =1/8*9/4=9/32

2: =8/27*243/32=9/4

3: =(5/4*4/5)^5*(4/5)^2=16/25

4: \(=\left(-\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^2\cdot\left(\dfrac{6}{5}\right)^2=\dfrac{36}{25}\)

5: \(=\left(-\dfrac{4}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^{10}=\left(-1\right)\left(\dfrac{3}{4}\right)^7=-\left(\dfrac{3}{4}\right)^7\)

6: \(=\left(\dfrac{1}{3}\cdot\dfrac{-9}{2}\right)^4\left(-\dfrac{9}{2}\right)^2=\left(-\dfrac{3}{2}\right)^4\cdot\dfrac{81}{4}=\dfrac{9}{4}\cdot\dfrac{81}{4}=\dfrac{729}{16}\)

8: =(0,2*5)^4*5^2=25

10: =-0,5^5*2^10

=-0,5^5*2^5*2^5

=-32

13: =(0,5*2)^2*2^2=4

=>1/2-5/3+x-1/2=2-|5/3-1/4|

=>x-5/3=2-17/12=7/12

=>x=7/12+5/3=9/4

=>x-1/2=3+1/14-5-1/3-4/7+11/21

=>x-1/2=-97/42

=>x-21/42=-97/42

=>x=-97/42+21/42=-76/42=-38/21

(4x - 1)(2x² - 3x + 2) \(=8x^3-12x^2+8x-2x^2+3x-2\) \(=8x^3-14x^2+11x-2\)

=8x^3-12x^2+8x-2x^2+3x-2

=8x^3-14x^2+11x-2

Ta có : 

`x/4 = y/5 => x/ 16 = y/20`

`y/4=z/7 => y/20=z/35`

`-> x/16 =y/20 =z/35` và `-x-y+z=-2`

ADTC dãy tỉ số bằng nhau ta có :

`x/16 =y/20 =z/35= (-x-y+z)/(-16 -20+35)=(-2)/(-1)=2`

`=> x/16=2=>x=2.16=32`

`=> y/20=2=>y=2.20=40`

`=> z/35=2=>z=2.35= 70`

x/4 = y/5 ⇒ x/16 = y/20    (1)

y/4 = z/7 ⇒ y/20 = z/35    (2)

Từ (1) và (2) 

⇒ x/16 = y/20 = z/35

Áp dụng tính chất của dãy tỉ số bằng nhau ta được:

x/16 = y/20 = z/35 = (-x - y + z)/(-16 - 20 + 35) = -2/(-1) = 2

x/16 = 2 ⇒ x = 16.2 = 32

y/20 = 2 ⇒ y = 20.2 = 40

z/35 = 2 ⇒ z = 35.2 = 70

Vậy x = 32; y = 40; z = 70

a-b=6 nên a=b+6

\(A=\dfrac{3b+18-b}{2b+12+6}+\dfrac{3b-b-6}{2b-6}\)

\(=1+1=2\)

a: N=(-1)(-1)^2(-1)^3+(-1)^2(-1)^3(-1)^4+...+(-1)^2014*(-1)^2015*(-1)^2016

=1-1+1-1+...-1+1

=0

b: x^3-3x^2+2x-b chia hết cho x-3

=>2x-b chia hết cho x-3

=>2x-6+6-b chia hết cho x-3

=>6-b=0

=>b=6

=11/2(-1/5+3/7-4/5+4/7)

=11/2(1-1)

=0

\(\left(\dfrac{-1}{5}+\dfrac{3}{7}\right):\dfrac{2}{11}+\left(\dfrac{-4}{5}+\dfrac{4}{7}\right):\dfrac{2}{11}\) 

\(=\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\) 

\(=\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\) 

\(=\dfrac{2}{11}:0=0\) 

a: x>0

=>2a+5<0

=>a<-5/2

b: x<0

=>2a+5>0

=>a>-5/2

c: x=0

=>2a+5=0

=>a=-5/2

mik thank nha

mik thank nha