toán lớp 2 à

toán lớp 2 ????

b: \(\left(a-b\right)^2-c^2=\left(a-b-c\right)\left(a-b+c\right)\)

c: \(4x^2+12x+9=\left(2x+3\right)^2\)

d: \(25x^2-20xy+4y^2=\left(5x-2y\right)^2\)

e: \(8x^6-27y^3=\left(2x^2-3y\right)\left(4x^2+6x^2y+9y^2\right)\)

ĐK: \(\hept{\begin{cases}x\ge2\\y\ge-\frac{1}{3}\end{cases}}\)

\(\sqrt{x-2}+x^3-6x^2+12x=\sqrt{3y+1}+27y^3+27y^2+9y+9\)

<=> \(\sqrt{x-2}+x^3-6x^2+12x-8=\sqrt{3y+1}+27y^3+27y^2+9y+1\)

<=> \(\sqrt{x-2}+\left(x-2\right)^3=\sqrt{3y+1}+\left(3y+1\right)^3\)

<=> \(\left(\sqrt{x-2}-\sqrt{3y+1}\right)+\left[\left(x-2\right)^3-\left(3y+1\right)^3\right]=0\)

<=> \(\frac{x-3y-3}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-3y-3\right)\left[\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2\right]=0\)

<=> \(\left(x-3y-3\right)\left(\frac{1}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2\right)=0\)

<=> \(x-3y-3=0\)

vì \(\frac{1}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2>0\)

<=> x = 3y + 3

Thế vào phương trình trên ta có: 

\(2+2\left(3y+3\right)^2-2y^2+3\left(3y+3\right)y-4\left(3y+3\right)-3y=0\)

<=> \(25y^2+30y+8=0\Leftrightarrow\orbr{\begin{cases}y=-\frac{2}{5}\\y=-\frac{4}{5}\end{cases}}\)không thỏa mãn đk 

Vậy hệ vô nghiệm.

B=x^3+27y^3+9x^2y+27xy^2 

  =x^3+9x^2y+27xy^2+27y^3

 =(x+3y)^3 ---> sử dụng hằng đẳng thức số 4 bạn nhé 

b) \(-y^8+10y^4x^3-25x^6\)

\(=-\left(y^8-10y^4x^3+25x^6\right)\)

\(=-\left[\left(y^4\right)^2-2.y^4.5x^3+\left(5x^3\right)^2\right]\)

\(=-\left(y^4-5x^3\right)^2\)

c) \(8x^3+36x^2y+54xy^2+27y^3\)

\(=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)

\(=\left(2x+3y\right)^3\)

d) \(-y^3+12y^2x-48yx^2+64x^3\)

\(=-\left(y^3-12y^2x+48yx^2-64x^3\right)\)

\(=-\left[y^3-3.y^2.4x+3.y.\left(4x\right)^2-\left(4x\right)^3\right]\)

\(=-\left(y-4x\right)^3\)

e) \(64x^6y^4-81x^2y^2\)

\(=\left(8x^3y^2\right)^2-\left(9xy\right)^2\)

\(=\left(8x^3y^2-9xy\right)\left(8x^3y^2+9xy\right)\)

f) \(64x^6-27y^6\)

\(=\left(4x^2\right)^3-\left(3y^2\right)^3\)

\(=\left(4x^2-3y^2\right)\left[\left(4x^2\right)^2+4x^2.3y^2+\left(3y^2\right)^2\right]\)

\(=\left(4x^2-3y^2\right)\left(16x^4+12x^2y^2+9x^4\right)\)

x²(y²-3) - 2y²(x-1) + 6x =7 

(xy)² -3x² - 2y²x +2y² +6x=7

x²y² -2.xy²+y² -3(x² -2x+1)+y² =4

y²(x²-2x+1) -3(x²-2x+1) +y²-3=1

(y²-3)(x²-2x+1+1)=1

(y²-3)(x²-2x+2)=1

còn lại bn tự làm nha!

chúc bạn làm bài tốt!