{[(3² + 1).10 - (8:2 + 6)]:2} + 55

= {[( 9 + 1 ) . 10 - ( 4 + 6 )] : 2 } + 55

= [( 10 . 10 - 10 ) : 2 ] + 55

= [( 100 - 10 ) : 2 ] + 55

= ( 90 : 2 ) + 55

= 45 + 55 = 100

{[(3² + 1).10 - (8:2 + 6)]:2} + 55

= {[(3² + 1).10 - (4 + 6)]:2} + 55

=[(3² + 1).10 - 10):2] + 55

= [(9 +1) x 10 - 10) : 2 + 55

= ( 10 x 10 - 10) : 2 + 55

= -90 : 2 + 55

= -45 + 55

= 10

{ [ (3² + 1) .10 - (8 : 2 + 6) ] : 2 } + 55

= { [ 10 . 10 - 10 ] : 2 } + 55

= { 90 : 2 } + 55

= 45 + 55

= 100

\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)

a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)

ĐKXĐ: x ≠ -1

⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)

⇔ 65 + 52 = -3(x + 1)

⇔ 117 = -3x - 3

⇔ 117 + 3 = -3x

⇔ 120 = -3x 

⇔ x = \(\dfrac{120}{-3}=-40\) (TM)

b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)

⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)

⇔ 4x = -2,75

⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)

c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)

⇔  \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)

⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x² + 96x + 144x + 48

⇔ 936x + 1560x - 624x - 96x - 144x - 288x² = 48 - 312 - 520 + 312

⇔ 1632x - 288x² = -472

⇔ -288x² + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)

⇔ x = 5,942459684 \(\approx\) 6

c: Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Leftrightarrow2x+1=13\)

hay x=6

1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
= 1 – 1/2 + 1/2- 1/4 + 1/4 – 1/8 + 1/8 – 1/16 + 1/16 – 1/32 + 1/32 – 1/64 + 1/64 – 1/128 + 1/128 – 1/256 – 1/256 – 1/512
= 1 – 1/512
= 511/512

hok tot

chỗ k ở đâu để mình k cho

1) \(\frac{45^{20}\cdot20^{10}}{3^{15}\cdot6^3}=\frac{3^{40}\cdot5^{20}\cdot5^{10}\cdot2^{20}}{3^{15}\cdot2^3\cdot3^3}\)

\(=\frac{2^{20}\cdot3^{40}\cdot5^{30}}{2^3\cdot3^{18}}=2^{17}\cdot3^{22}\cdot5^{30}\)

2) Ta có: \(2^{2009}+2^{2008}+...+2^1+2^0\)

\(=2^{2010}-1\) đã CM ở rất nhiều bài rồi

=> \(2^{2010}-2^{2010}+1=1\)

ý a bạn nhé

`->D` 

vì : `4/16=(4:4)/(16:4)=1/4`

      `2/8=(2:2)/(8:2)=1/4`

      `3/12=(3:3)/(12:3)=1/4`

32 . {160 : [ 300 - (175 + 21 . 5) ] }

= 32 . {160 : [ 300 - 280 ] }

= 32 . {160 : 20}

= 32 . 8

= 256

Trả lời 

184

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