\((4.x-15)^3=5^3=4.x-15=5=4.x=20=>x=5\)

\(\dfrac{2\cdot3\cdot8}{4\cdot5\cdot6\cdot7}=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{8}{35}=\dfrac{2}{35}\)

Bằng 0

=0

nha

cac

bn

tk

và

kb

mk

nhé

(2.7.8.9.6 20.78.120) . (3-3)

(2.7.8.9.620.78.120) . 0

0

dấu " \(.\) " là dấu nhân

\(\left\{\left[27.11-\left(x+12\right).8\right].3-79\right\}:5=100\)

=>\(\left[297-8.x+12.8\right].3-79=100.5\)

=> \(\left[297-8.x+96\right].3-79=500\)

=> \(\left[297-8.x+96\right].3=500+79\)

=> \(\left[\left(297+96\right)-8.x\right].3=579\)

=> \(393-8.x=579:3\)

=> \(393-8.x=193\)

=> \(8.x=393-193\)

=> \(8.x=200=>x=200:8=25\)

Vậy \(x=25\)

Chúc bạn Hk tốt!!!!

`x^8+36x^4=0`

`<=>x^4(x^4+36)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x^4=0\\x^4+36=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^4=-36\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x\in\varnothing\end{matrix}\right.\)

__

`(x-5)^3 -x+5=0`

`<=> (x-5)^3 -(x-5)=0`

`<=> (x-5) [(x-5)^2 -1]=0`

`<=> (x-5)(x-5-1)(x-5+1)=0`

`<=>(x-5)(x-6)(x-4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-6=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

__

`5(x-2)-x^2+4=0`

`<=>5(x-2)-(x^2-4)=0`

`<=>5(x-2)-(x-2)(x+2)=0`

`<=>(x-2)(5-x-2)=0`

`<=>(x-2)(-x-3)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\-x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

a) 

x 8 = − 1 4 x .4 = − 1.8 x .4 = − 8 x = − 2

b)

6 x − 3 = 9 2 x − 7 6. 2 x − 7 = 9. x − 3 12 x − 42 = 9 x − 27 3 x = 15 x = 5

c)

x − 1 2 2 = 4 x − 1 2 = ± 2

TH1:

x − 1 2 = 2 x = 5 2

TH2:

x − 1 2 = − 2 x = − 3 2

 Vậy  x = 5 2 hoặc  x = − 3 2

\(\left(x^2+1-x^3\right)^8=\sum\limits^8_{k=0}C^k_8.\left(x^2-x^3\right)^k\)

\(=\sum\limits^8_{k=0}C^k_8\sum\limits^k_{i=0}C^i_k.\left(x^2\right)^{k-i}\left(x^3\right)^i\)

\(=\sum\limits^8_{k=0}\sum\limits^k_{i=0}C^k_8C^i_k.x^{2k+i}\)

\(\Rightarrow2k+i=8\)

Ta có: \(\left\{{}\begin{matrix}2k+i=8\\i\in N\\k\in N\\0\le i\le k\le8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}i=2\\k=3\end{matrix}\right.\)

\(\Rightarrow\) Hệ số của \(x^8\) trong khai triển là \(C^3_8C^2_3=168\).

3x8/10+8/10x4+3x8/10

=8/10(3+4+3)

=8/10x 10

=8

( 1 +8+ 2+3+..... +8+ 9+10 ) x( 6x8-48)

=(1+2+3+.......+8+9+10)x 48-48

=(1+2+3+.........+8+9+10)x0

= 0
^    ^ chúc bạn học tốt k cho mình nha

Bằng 0 nhé

what the hell?