\(11-6\sqrt{2}=\left(3-\sqrt{2}\right)^2\)

\(6+4\sqrt{2}=\left(2+\sqrt{2}\right)^2\)

a. \(9x^2+30x+25=\left(3x+5\right)^2\)

b. \(\dfrac{4}{9}x^4-16x^2=\left(\dfrac{2}{3}x^2-4x\right)\left(\dfrac{2}{3}x^2+4x\right)=x^2\left(\dfrac{2}{3}x-4\right)\left(\dfrac{2}{3}x+4\right)\)

c. \(a^2y^2+b^2x^2-2axby=\left(ay-bx\right)^2\)

d. \(100-\left(3x-y\right)^2=\left(10-3x+y\right)\left(10+3x-y\right)\)

e. \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4=-\left(9x^4-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^4\right)=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)

f. \(64x^2-\left(8a+b\right)^2=\left(8x-8a-b\right)\left(8x+8a+b\right)\)

g. \(27x^3-a^3b^3=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)

a) \(x^2-9\)

= \(x^2-3^2\)

\(=\left(x-3\right)\left(x+3\right)\)

b) \(4x^2-1\)

\(=\left(2x\right)^2-1^2\)

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

c) \(9x^2-y^2\)

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

\(=\left(3x-y\right)\left(3x+y\right)\)

d) \(25x^2-9y^2\)

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

\(=\left(5x-3y\right)\left(5x+3y\right)\)

e) \(36x^2-25y^2\)

\(=\left(6x\right)^2-\left(5y\right)^2\)

\(=\left(6x-5y\right)\left(6x+5y\right)\)

a) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)

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

tương tự

a)\(-25+4x^2=\left(2x-5\right)\left(2x+5\right)\)

b)\(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

c)\(\frac{1}{9}x^2+\frac{2}{3}xy+y^2=\left(\frac{1}{3}x+y\right)^2\)

\(a,-25+4x^2=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)

\(b,-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

\(c,\frac{1}{9}x^2+\frac{2}{3}xy+y^2=\left(\frac{1}{3}x\right)^2+\frac{2.1}{3}xy+y^2=\left(\frac{1}{3}x+y\right)^2\)(sửa đề)

a) \(-25+4x=\left(2x-5\right)\left(2x+5\right)\)

b) \(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

c) \(\frac{1}{9}x^2+\frac{2}{3}x+y^2=\left(\frac{1}{3}x+y\right)^2\)

Good luck:3 (bài này dễ mà - _ -)

a) \(2x^2+2b^2=x^2+b^2+x^2+b^2=x^2+2xb+b^2+x^2-2xb+b^2=\left(x+b\right)^2+\left(x-b\right)^2\)

\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)

Bài 1 em dùng HĐT nha

Bài 2:

a. x² + 4x + 4

= x² + 2.2.x + 2²

= (x + 2)²

b. 9x² - 12x + 4

= (3x)² - 3x.2.2 + 2²

= (3x - 2)²

c. \(\dfrac{x^2}{4}+x+1\)

= \(\left(\dfrac{x}{2}\right)^2+2.\dfrac{x}{2}.1+1^2\)

= \(\left(\dfrac{x}{2}+1\right)^2\) 

a)\(x^2+2x+1=x^2+2x1+1^2=\left(x+1\right)^2\)

b)\(9x^2+y^2+6xy=3^2x^2+y^2+2.3x.y=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)

c)\(25a^2+4b^2-20ab=5^2a^2+2^2b^2-2.5a.2b=\left(5a\right)^2-2.5a.2b+\left(2b\right)^2=\left(5a-2b\right)^2\)

d)\(x^2-x+\frac{1}{4}=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)

a) Ta có: \(\left(x^2+9x+18\right)^2+2\left(x^2+9x\right)+37\)

\(=\left(x^2+9x+18\right)^2+2\cdot\left(x^2+9x+18\right)-36+37\)

\(=\left(x^2+9x+19\right)^2\)

b) Ta có: \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

\(=\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+2\left(x+1\right)\left(y+1\right)\)

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

c) Ta có: \(x^2-2x\left(y+2\right)+y^2+4y+4\)

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

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

d) Ta có: \(x^2+2x\left(y+1\right)+y^2+2y+1\)

\(=x^2+2\cdot x\cdot\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+y+1\right)^2\)