bn thiếu đề

x:2 +x:4x6 +x :2x4 = ?

=> x:2+x:24+x:8 = ?

=> x:(2+24+8)  =?

=> x:34  = ?

(đề bài thiếu hả bạn)

\(\frac{3}{2}\times\frac{4}{7}-x=\frac{2}{3}\)

\(\frac{6}{7}-x=\frac{2}{3}\)

\(x=\frac{6}{7}-\frac{2}{3}\)

\(x=\frac{18}{21}-\frac{14}{21}\)

\(x=\frac{4}{21}\)

\(\frac{3}{2}\chi\frac{4}{7}-x=\frac{2}{3}\)

\(\frac{6}{7}-x=\frac{2}{3}\)

\(x=\frac{6}{7}-\frac{2}{3}\)

\(x=\frac{4}{21}\)

\(A=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2011.2013}\right)\)

\(A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{4048144}{2011.2013}\)

\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2012.2012}{2011.2013}\)

\(A=\frac{2.3.4...2012}{1.2.3...2011}.\frac{2.3.4...2012}{3.4.5...2013}\)

\(A=2012.\frac{2}{2013}=\frac{4024}{2013}\)

Ta có :

\(A=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right).................\left(1+\dfrac{1}{2011.2013}\right)\)

\(A=\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\left(\dfrac{8}{8}+\dfrac{1}{8}\right)................\left(\dfrac{9999}{9999}+\dfrac{1}{9999}\right)\)

\(A=\dfrac{4}{3}.\dfrac{9}{8}...............\dfrac{10000}{9999}\)

\(A=\dfrac{4.9.................10000}{3.8.............9999}\)

\(A=\dfrac{2.2.3.3................100.100}{1.3.2.4...............99.101}\)

\(A=\dfrac{2.100}{101}=\dfrac{200}{101}\)

~ Chúc bn học tốt ~

2 cách thì cách 1 là tính theo cột dọc ( đặt tính ak ) , cách 2 là tính như bth ( cậu tự làm nha ) 

đáp án   : -2x^3 + x^2 -x + 6

(x + 40) . 15 = 5² . 3² . 4

(x + 40) . 15 = 25 . 9 . 4

(x + 40) . 15 = 900

x + 40 = 900 : 15

x + 40 = 60

x = 60 - 40

x = 20

( x + 40 ) x 15 = 5² x 3² x 4

⇒ ( x + 40 ) x 15 = 25 x 9 x 4

⇒ ( x + 40 ) x 15 = 900

⇒ x + 40 = 60

⇒ x = 20.

Vậy x = 20.

Xx0,5+Xx0,25x6+Xx0,5x4=8

Xx0,5+Xx1,5+Xx2=8

Xx(0,5+1,5+2)=8

Xx4               =8

X                   =8:4

X                   =2

A= (1+1/1 x 3)x(1+1/2x4)x(1+1/3x5)x............x(1+1/2011x2013)

\(=\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{8}{8}+\frac{1}{8}\right)....\left(\frac{4048143}{4048143}+\frac{1}{4048143}\right)\)

\(=\frac{4}{3}\cdot\frac{9}{8}\cdot...\cdot\frac{4048144}{4048143}\)

\(=\frac{4\cdot9\cdot....\cdot4048144}{3\cdot8\cdot....\cdot4048143}\)

\(=\frac{2\cdot2\cdot3\cdot3\cdot....\cdot2012\cdot2012}{1\cdot3\cdot2\cdot4\cdot....\cdot2011\cdot2013}\)

\(=\frac{2\cdot2012}{2013}=\frac{4024}{2013}\)