A=68,9+8,75+1,25
=68,9+(8,75+1,25)

=6,89+10=689
B=4,7 nhân 6,1+ 4,7 nhân 3,9

=4,7 nhân (6,1+3,9)

=4,7 nhân 10=47

ta có : A= 68,9+8,75+1,25

=(8,75+1,25)+68,9

=10+68,9

=78,9

11

a , x+6=49-3-13

x+6=a

x=a-6

x=n

b,231-(x-6)=n

x-6=231-n

x-6=a

x=a+6

x=m

B1:

a)49-3.(x+6)=13

3.(x+6)=49-13

3.(x+6)=36

x+6=36/3

x+6=12

x=12-6

x=6

bài sai đề thì phải

Lời giải:

$x^3+x^2-12x=0$

$\Leftrightarrow x(x^2+x-12)=0$

$\Leftrightarrow x(x^2+4x-3x-12)=0$

$\Leftrightarrow x[x(x+4)-3(x+4)]=0$

$\Leftrightarrow x(x-3)(x+4)=0$

$\Rightarrow x=0$ hoặc $x-3=0$ hoặc $x+4=0$

$\Lefotrightarrow x=0; x=3$ hoặc $x=-4$

A, \(0,489\times360+4,89\times63+4,89\times0,1\)

\(=0,489\times10\times36+4,89\times63+4,89\times0,1\)

\(=4,89\times36+4,89\times63+4,89\times0,1\)

\(=4,89\times\left(36+63+0,1\right)\)

\(=4,89\times99,1\)

\(=484,599\)

B, \(0,256\times3900+25,6\times60+256\times0,1\)

\(=0,256\times100\times39+25,6\times40+25,6\)

\(=25,6\times39+25,6\times40+25,6\)

\(=25,6\times\left(39+40+1\right)\)

\(=25,6\times100\)

\(=2560\)

Bài 1: 

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^3-10x^2-6x\)

Bài 4: 

a: =>3x+10-2x=0

=>x=-10

c: =>3x²-3x²+6x=36

=>6x=36

hay x=6

Bài 1:

\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)

Bài 4:

\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)

Bài 1:

\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)

\(\left(x-3\right)\left(2x+6\right)=0\)

<=> \(\hept{\begin{cases}x-3=0\\2x+6=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=3\\x=-3\end{cases}}\)

học tốt

a. (x - 3) . (2x + 6) = 0

<=> \(\orbr{\begin{cases}x-3=0\\2x+6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\2x=-6\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy x = + 3

(x-3)(2x+6)=0

x(2+6-3)=0

x.5=0

x=0:5=0

mk ko bt cộng trừ chi nhân

và mk nghĩ A chia hết cho 40

Trl :

\(A=3+3^2+3^3+3^4+...+3^{25}\)

\(=3+\left(3^2+3^3+3^4+3^5\right)+...+\left(3^{22}+3^{23}+3^{24}+3^{25}\right)\)

\(=3+3^2.\left(1+3+3^2+3^3\right)+...+3^{22}.\left(1+3+3^2+3^3\right)\)

\(=3+3^2.40+...+3^{22}.40\)

\(=3+40.\left(3^2+...3^{22}\right)\)

Vì \(40.\left(3^2+...+3^{22}\right)⋮40\)và \(3:40\)(dư 3)

Nên \(3+40.\left(3^2+...+3^{22}\right):40\)dư 3

Vậy ....

=> 3A=3^2+3^3+3^4+3^5+......+3^26+3^27

=> 3A=3(3+3^2+.........+3^26

ta sử dụng cách khử liên tiếp

còn 3 và 3^27

3A-A

2A=3^27-3

A= 3^27 - 3 phần 2