2.
a) 4x(x-1)-6x+6
= 4x(x-1)-6(x-1)
= (4x-6)(x-1)
3.
a) 6x²-24x=0
    6x(x-4)=0
TH1: 6x=0         TH2: x-4=0
           x=0                     x=4
Vậy x\(\in\){0;4}

2. a. \(4x\left(x-1\right)-6x+6\)

\(=4x\left(x-1\right)-6\left(x-1\right)\)

\(=\left(4x-6\right)\left(x-1\right)\)

3. a. \(6x^2-24x=0\)

\(\Leftrightarrow6x\left(x-4\right)=0\)

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

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

Bài 3: 

a: \(\Leftrightarrow6x\left(x-4\right)=0\)

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

\(\left(3x-1\right)\times\left(4x+7\right)-6x\times\left(2x-3\right)=63\)

\(\Leftrightarrow12x^2+17x-7-12x^2+18x=63\)

\(\Leftrightarrow35x=63+7\)

\(\Leftrightarrow35x=70\)

\(\Rightarrow x=2\)

P/s Các bạn tham khảo nha

a: =>7x=63

hay x=9

b: =>3x=-15

hay x=-5

d: =>-6x=-16

hay x=8/3

a) \(7x=63\Leftrightarrow x=9\)

b) \(3x=-15\Leftrightarrow x=-5\)

c) \(2x-5=0\Leftrightarrow2x=5\Leftrightarrow x=\dfrac{5}{2}\)

d) \(-6x=-16\Leftrightarrow x=\dfrac{8}{3}\)

| x -1 | + | 2x - 2 | + | 6 - 6x | = 63

TH1: x-1+2(x-1)+6(x-1)=63

<=>9(x-1)=63

<=>x-1=7

<=> x=8

TH2: 1-x+2(1-x)+6(1-x)=63

<=>9(1-x)=63

<=>1-x=7

<=>x=-6

Vậy \(x\in\left\{8;-6\right\}\)

tr1

=>9(x-1)=63

=> x= 8

th2 

=>1-x = 7

=>x=-6

Hướng dẫn giải:

625 x 10 = 6250

540 : 10 = 54

3456 x 100 = 345 600

63 000 : 100 = 630

302 761 x 1000 = 302 761 000

1 011 000 : 1000 = 1011

Hướng dẫn giải:

625 x 10 = 6250

540 : 10 = 54

3456 x 100 = 345 600

63 000 : 100 = 630

302 761 x 1000 = 302 761 000

1 011 000 : 1000 = 1011

\(2,ĐK:x\ge4;y\ge1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-4}=a\\\sqrt{y-1}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}a+b=4\\a^2+b^2=58\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2ab+58=16\\a^2+b^2=58\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}ab=-21\\a+b=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=4-b\\b^2-4b-21=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}b=7\Rightarrow a=-3\\b=-3\Rightarrow a=7\end{matrix}\right.\left(loại\right)\)

Vậy hệ vô nghiệm

\(1,\\ \forall x=0\\ HPT\Leftrightarrow1=19\left(\text{vô lí}\right)\\ \forall x\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x^3}+y^3=19\\\dfrac{y}{x^2}+\dfrac{y^2}{x}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{1}{x}+y\right)^3-3\cdot\dfrac{y}{x}\left(\dfrac{1}{x}+y\right)=19\\\dfrac{y}{x}\left(\dfrac{1}{x}+y\right)=-6\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}+y=a\\\dfrac{y}{x}=b\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}a^3-3ab=19\\ab=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+y=1\\\dfrac{y}{x}=-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}1+xy=x\\y=-6x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3};y=-2\\x=-\dfrac{1}{2};y=3\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(\dfrac{1}{3};-2\right);\left(-\dfrac{1}{2};3\right)\)

C

C?

\(C.\dfrac{63}{200}m^2\)

mik làm 1 câu thôi các cau khác 1 chang lun chỉ khác số 

A = 1 + (-2) + 3 + (-4) + 5 +(-6) + ... + 99 + (-100)

A=(1+(-2))+(3+(-4))+.....+(99+(-100))

A=(-1)+(-1)+......+(-1) CÓ 50 SỐ 

A= -50