tu hoc moi gioi

(81-1^2)(81-2^2)(81-3^2)....(81-9^2)

= (81-1^2)(81-2^2)(81-3^2)....(81-81)

= (81-1^2)(81-2^2)(81-3^2)....0

= 0

Bài 2: 

x=13 nên x+1=14

\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)

\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)

=14-x=1

x=13 nên x+1=14

f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14

=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14

=14-x=1

a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)

\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)

\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)

\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)

\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)

b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)

\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)

\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)

\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)

\(\Leftrightarrow x=\frac{1}{-3}\)

c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)

\(\Leftrightarrow x=\frac{5}{6}\)

d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)

\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{6}\)

ta có : 

\(\frac{x+1}{3}+\frac{x+1}{9}+\frac{x+1}{27}+\frac{x+1}{81}=\left(x+1\right)\times\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)\)

\(=\left(x+1\right)\times\frac{40}{81}=\frac{56}{81}\text{ nên }x+1=\frac{56}{40}=\frac{7}{5}\)

vậy \(x=\frac{7}{5}-1=\frac{2}{5}\)

16/81

a)= ( 81 , 6 x 27 , 3 – 17 , 3 x 81 , 6 ) x 0

=0 ( vì 0 nhân với số nào cũng bằng 0)

b)= ( 13 , 75 – 0 , 48 x 5 ) x ( 42 , 75 : 3 + 2 , 9 ) x ( 9-9)

= ( 13 , 75 – 0 , 48 x 5 ) x ( 42 , 75 : 3 + 2 , 9 ) x 0

= 0 ( vì 0 nhân với số nào cũng bằng 0)

c)=( 792 , 81 x 0 , 25 + 792 , 81 x 0 , 75 ) x 0

=0 ( vì 0 nhân với số nào cũng bằng 0 )

a ) ( 81 , 6 x 27 , 3 – 17 , 3 x 81 , 6 ) x ( 32 x 11 - 3200 x 0 , 1 - 32 )

= ( 81 , 6 x 27 , 3 – 17 , 3 x 81 , 6 ) x ( 352 - 320 - 32 )

= ( 81 , 6 x 27 , 3 – 17 , 3 x 81 , 6 ) x 0

= 0.

b ) ( 13 , 75 – 0 , 48 x 5 ) x ( 42 , 75 : 3 + 2 , 9 ) x ( 1 , 8 x 5 – 0 , 9 x 10 )

= ( 13 , 75 – 0 , 48 x 5 ) x ( 42 , 75 : 3 + 2 , 9 ) x ( 9 - 9 )

= ( 13 , 75 – 0 , 48 x 5 ) x ( 42 , 75 : 3 + 2 , 9 ) x 0

= 0

c ) ( 792 , 81 x 0 , 25 + 792 , 81 x 0 , 75 ) x ( 11 x 9 – 900 x 0 , 1 – 9 )

= ( 792 , 81 x 0 , 25 + 792 , 81 x 0 , 75 ) x ( 99 - 90 - 9 )

= ( 792 , 81 x 0 , 25 + 792 , 81 x 0 , 75 ) x 0

= 0.

Hok tốt !

a)32 x 11 -3200x 0,1 -32

=32x11-32x100x0,1-32

=32x(11-10-1)=32x0=0

nên(81,6x27,3-17,3x81,6)x(32x11-3200x0,1-32)

=(81,6x27,3-17,3x81,6)x0

=0

b)1,8x5 -0,9x10=1,8x5-0,9x2x5

=1,8x5-1,8x5

=0

nên (13,75-0,48 x 5)x(42,75:3+2,9)x(1,8x5-0,9x10)=0

c)(792,81x0,25+792,81x0,75)x(11x9-900x0,1-9)

=792,81x(0,25+0,75)x(99-90-9)

=792,81x1x0

=0

Vũ Hồng Linh bạn check lại bài đầu dùm =_=" 

\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)

\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)

\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)

\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)

=> Làm nốt 

Mấy bài kia cũng làm tương tự

(- \(\dfrac{1}{3}\))³.\(x\) = \(\dfrac{1}{81}\)

          \(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))³

          \(x\) =  - (\(\dfrac{1}{3}\))⁴ :(\(\dfrac{1}{3}\))³

           \(x=-\dfrac{1}{3}\)

Vậy \(x=-\dfrac{1}{3}\)

(3\(x\) + 1)³ =  - 27

(3\(x\) + 1)³ = (-3)³

3\(x\) + 1 = -3

 3\(x\)       = - 3 - 1

 3\(x\) = -4

    \(x=-\dfrac{4}{3}\)

Vậy \(x=-\dfrac{4}{3}\)

a: =>1/3:x=3/5-2/3=9/15-10/15=-1/15

=>x=-1/3:1/15=5

b: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)

=>x*2/3=-1

=>x=-3/2

c: =>2x+1=4 hoặc 2x+1=-4

=>x=3/2 hoặc x=-5/2

h: =>x-3=4

=>x=7

g: =>2x-1=3

=>2x=4

=>x=2

f: \(\Leftrightarrow x\cdot\left(\dfrac{3}{2}-\dfrac{7}{3}\right)=\dfrac{3}{2}-\dfrac{2}{3}\)

=>x*-5/6=5/6

=>x=-1

d: =>|2x-1|=3

=>2x-1=3 hoặc 2x-1=-3

=>x=-1 hoặc x=2