\(8x-2x+8=48\)

\(6x=48-8\)

\(6x=40\)

\(x=\frac{20}{3}\)

vậy \(x=\frac{20}{3}\)

x8 - x * 2 + 8 = 48

x.(8-2)=48-8

x.6=40

x=40/6=20/3

Chỗ x8 là sao??

( 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?

\(37\times46+37\times51+111\)

\(=37\times46+37\times51+37\times3\)

\(=37\times\left(46+51+3\right)\)

\(=37\times100\)

\(=3700\)

------------

\(172\times47+86\times8-34\times4\)

\(=172\times47+172\times4-34\times4\)

\(=172\times\left(47+4\right)-34\times4\)

\(=172\times51-34\times4\)

\(=172\times3\times17-17\times2\times4\)

\(=17\times\left(172\times3-2\times4\right)\)

\(=17\times\left(516-8\right)\)

\(=17\times508\)

\(=8636\)

------------

\(35\times215-35\times200-175\)

\(=35\times215-35\times200-35\times5\)

\(=35\times\left(215-200-5\right)\)

\(=35\times10\)

\(=350\)

------------

\(48\times56+28\times104\)

\(=48\times2\times28+28\times104\)

\(=28\times\left(48\times2+104\right)\)

\(=28\times200\)

\(=5600\)

88 800

261 304

không phải nhé

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

`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`

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

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

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

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

\(\Rightarrow2k+i=8\)

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 \(C^3_8C^2_3=168\).

X8 LÀ X MŨ 8 HAY X*8 VẬY BẠN