`# \text {Kaizu DN}`

`a)`

`(3x + 6) + (7x - 14) = 0?`

\(\Rightarrow3x+6+7x-14=0\\ \Rightarrow\left(3x+7x\right)+\left(6-14\right)=0\\ \Rightarrow10x-8=0\\ \Rightarrow10x=8\Rightarrow x=\dfrac{8}{10}\\ \Rightarrow x=\dfrac{4}{5}\)

Vậy, \(x=\dfrac{4}{5}\) 

`b)`

`17y + 35 + 4x + 17 = 42`

\(\Rightarrow\left(17y+17\right)+\left(35+4x\right)=42\\ \Rightarrow17\left(y+1\right)+\left(35+4x\right)=42\)

Bạn xem lại đề ;-;.

b3:

a, <

b, <

c, <

d, >

c2, >

f, >

Bài 4:

0,4636363... < 0,463736< 0,4656365< 0,466< 7/15

b3:

a, <

b, <

c, <

d, >

c2, >

f, >

Bài 4:

0,4636363... < 0,463736< 0,4656365< 0,466< 7/15

c) và d) của Trí sai nên sửa lại

c) (2x - 4)/7 < 0

⇒ 2x - 4 < 0 (vì 7 > 0)

⇒ 2x < 4

⇒ x < 2

d) (5x - 8)/-10 < 0

⇒ 5x - 8 > 0 (vì -10 < 0)

⇒ 5x > 8

⇒ x > 8/5

a) \(\dfrac{x-2}{45}>0\Rightarrow x-2>0\Rightarrow x>2\)

b) \(\dfrac{x+3}{-2}>0\Rightarrow x+3< 0\Rightarrow x< -3\)

c) \(\dfrac{2x-4}{7}< 0\Rightarrow2x-4>0\Rightarrow2x>4\Rightarrow x>2\)

d) \(\dfrac{5x-8}{10}< 0\Rightarrow5x-8< 0\Rightarrow5x< 8\Rightarrow x< \dfrac{8}{5}\)

(7/2)² . 4³

= 49/4 . 4³

= 49.16

= 784

KQ = 784 nha

(7/2)^2•4^3

a) \(\dfrac{35}{101}=\dfrac{105}{303}< \dfrac{189}{303}\Rightarrow\dfrac{35}{101}< \dfrac{189}{303}\)

b) \(\dfrac{11}{13}< \dfrac{11+2}{13+2}=\dfrac{13}{15}< \dfrac{14}{15}\Rightarrow\dfrac{11}{-13}>\dfrac{-14}{15}\)

c) \(-\dfrac{32}{19}< 0< \dfrac{23}{32}\Rightarrow-\dfrac{32}{19}< \dfrac{23}{32}\)

d) \(1,561< 1,5661\Rightarrow-1,561>-1,5661\)

e) \(0,1=\dfrac{1}{10}=\dfrac{40}{400}< \dfrac{40+56}{400+56}=\dfrac{96}{456}< \dfrac{176}{456}\Rightarrow0,1< \dfrac{176}{456}\)

g) \(0,3=\dfrac{3}{10}=\dfrac{9}{30}< \dfrac{9+8}{30+8}=\dfrac{17}{38}< \dfrac{19}{38}\Rightarrow0,3< \dfrac{19}{38}\Rightarrow-0,3>\dfrac{-19}{38}\)