=3(x^2-y^2-4x+4y)

=3[(x-y)(x+y)-4(x-y)]

=3(x-y)(x+y-4)

\(D=\left(\dfrac{3}{4}xy+3y\right)\left(x-12y\right)\\ =x\left(\dfrac{3}{4}xy+3y\right)-12y\left(\dfrac{3}{4}xy+3y\right)\\ =\dfrac{3}{4}x^2y+3xy-9xy^2-36y^2\)

Cảm ơn ạ

2(x - 2)²⁰²⁴ + 5|y - 1| = 0 

Do (x - 2)²⁰²⁴ ≥ 0 với mọi x

|y - 1| ≥ 0 với mọi y

2(x - 2)²⁰²⁴ + 5|y - 1| = 0

⇔ x - 2 = 0 và y - 1 = 0

*) x - 2 = 0

x = 2

*) y - 1 = 0

y = 1

Vậy x = 2; y = 1

1. 

$=153^2+2.47.153+47^2=(153+47)^2=200^2=40000$

2.

$=1,24^2-2.1,24.0,24+0,24^2=(1,24-0,24)^2=1^2=1$

3. Không phù hợp để tính nhanh 

4. 

$=15^8-(15^8-1)=1$

5.

$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+...+(2019^2-2020^2)$

$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+(2019-2020)(2019+2020)$

$=(-1)(1+2)+(-1)(3+4)+(-1)(5+6)+....+(-1)(2019+2020)$

$=(-1)(1+2+3+4+....+2019+2020)=(-1).2020(2020+1):2=-2041210$

6:

\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^4-1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^8-1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^{2020}-1\right)\left(2^{2020}+1\right)+1\\ =2^{4040}-1+1=2^{4040}\)

\(1,\left(3x+2\right)\left(5-x^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\5-x^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\-x^2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{2}{3};-\sqrt{5};\sqrt{5}\right\}\)

\(2,-2x-\dfrac{2}{3}\left(\dfrac{3}{4}-\dfrac{1}{8}x\right)=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow-2x-\dfrac{1}{2}+\dfrac{1}{12}x=-\dfrac{1}{8}\)

\(\Leftrightarrow-2x+\dfrac{1}{12}x=-\dfrac{1}{8}+\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{23}{12}=\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{9}{46}\)

Vậy \(S=\left\{-\dfrac{9}{46}\right\}\)

\(3,\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\)

\(\Leftrightarrow\dfrac{1}{12}.\dfrac{21}{4}=\dfrac{7}{2}.\dfrac{1}{3x-2}\)

\(\Leftrightarrow\dfrac{7}{16}=\dfrac{7}{6x-4}\)

\(\Leftrightarrow6x-4=7:\dfrac{7}{16}\)

\(\Leftrightarrow6x-4=16\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

Vậy \(S=\left\{\dfrac{10}{3}\right\}\)

\(4,\dfrac{x-1}{x+2}=\dfrac{4}{5}\left(dk:x\ne-2\right)\)

\(\Rightarrow5\left(x-1\right)=4\left(x+2\right)\)

\(\Rightarrow5x-5=4x+8\)

\(\Rightarrow x=13\left(tmdk\right)\)

Vậy \(S=\left\{13\right\}\)

mk c.ơn bn

=>x^2=0 và (y-1)^4=0

=>x=0 và y=1

này mình có vài câu không làm được, xin lỗi bạn nha

\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)

TH1: x<-5

=>-x-2-x-3-x-5=12x

=>12x=-3x-10

=>15x=-10

=>x=-2/3(loại)

TH2: -5<=x<-3

=>-x-2-x-3+x+5=12x

=>12x=-x

=>x=0(loại)

Th3: -3<=x<-2

=>-x-2+x+3+x+5=12x

=>12x=x+6

=>11x=6

=>x=6/11(loại)

TH4: x>=-2

=>12x=x+2+x+3+x+5=3x+10

=>9x=10

=>x=10/9(nhận)

\(\left|x-2\right|=x-2\)

\(TH_1:x\ge2\)
\(x-2=x-2\) ( luôn đúng )

\(TH_2:x< 2\)

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

\(\Leftrightarrow-x-x=-2-2\)

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\left(ktm\right)\)

Vậy pt có vô số nghiệm hay \(x\in R\)

4:

a: =>x-3=2

=>x=5

b: =>(2x+6+3x-9)/(x^2-9)=(3x+5)/(x^2-9)

=>5x-3=3x+5

=>2x=8

=>x=4

c: =>(4x-3x)/12x^2=12/12x^2

=>4x-3x=12

=>x=12

a: =>2x=3-x

=>3x=3

=>x=1

b: =>(3x-1)/(x^2-9)=(2x+6-x+3)/(x^2-9)

=>3x-1=x+9

=>2x=10

=>x=5

c: =>3/8x-4/8x=1/x^2

=>1/x^2=-1/8x

=>8/8x^2=-x/8x^2

=>-x=8

=>x=-8(nhận)

a: =>x=3x-15

=>-2x=-15

=>x=15/2

b: =>(x-2-2x-4)/(x^2-4)=(2x-3)/(x^2-4)

=>2x-3=-x-6

=>3x=-3

=>x=-1