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

c) \(\left(x^2+3^2\right)^2=x^4+18x+81\)

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

c) \(\left(3x-y^2\right)^2=9x^2-6xy^2+y^4\)

c) \(\left(x+2y^2\right)^2=x^2+4xy^2+4y^4\)

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

c) \(\left(2x+3y^2\right)^2=4x^2+12xy^2+9y^4\)

c) \(\left(4x-2y^2\right)^2=16x^2-16xy^2+4y^4\)

c) \(\left(4x^2-2y\right)^2=16x^4-16x^2y+4y^2\)

c) \(\left(\dfrac{1}{x}-5\right)\left(\dfrac{1}{x}+5\right)=\dfrac{1}{x^2}-25\)

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

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

c) \(\left(\dfrac{x}{y}-\dfrac{2}{3}\right)\left(\dfrac{x}{y}+\dfrac{2}{3}\right)=\dfrac{x^2}{y^2}-\dfrac{4}{9}\)

c) \(\left(\dfrac{x}{2}+\dfrac{y}{3}\right)\left(\dfrac{y}{3}-\dfrac{x}{2}\right)=\dfrac{y^2}{9}-\dfrac{x^2}{4}\)

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

c) \(\left(2x+\dfrac{3}{5}\right)\left(\dfrac{3}{5}-2x\right)=\dfrac{9}{25}-4x^2\)

c) \(\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)\left(\dfrac{4}{4}+\dfrac{1}{2}x\right)=\dfrac{1}{4}x^2-\dfrac{16}{9}\)

c) \(\left(\dfrac{2}{3}x^2-\dfrac{y}{2}\right)\left(\dfrac{2}{3}x^2+\dfrac{y}{2}\right)=\dfrac{4}{9}x^4-\dfrac{y^2}{4}\)

(x² + 3²)² 

^{=x^4 + 18x^2+81}

(x² + 3²)² 

= (x²)² + 2x² . 3² + (3²)² 

= x⁴ + 18x² + 81

( 2x - 3y )² = 4x² - 12xy + 9y²

( 3√x - y )² = 9x - 6y√x + y² ( x ≥ 0 )

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

\(\left(x+2y\right)^2=x^2+4xy+4y^2\)

\(36-12x+x^2=\left(6-x\right)^2\)

\(\left(x+5y\right)^2=x^2+10xy+25y^2\)

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

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

a/ \(36+x^2-12x=x^2-2x.6+6^2=\left(x+6\right)^2\)

b/ \(\left(x+2y\right)^2=x^2+2x.2y+\left(2y\right)^2=x^2+4xy+4y^2\)

c/ \(\left(\sqrt{x}-2\sqrt{y}\right)^2=\left(\sqrt{x}\right)^2-2\sqrt{x}.2\sqrt{y}+\left(2\sqrt{y}\right)^2=x-4\sqrt{xy}+4y\)

\(4x^2-20xy^2+25y^4=\left(2x\right)^2-2.2x.5y^2+\left(5y^2\right)^2=\left(2x-5y^2\right)^2\)

Áp dụng hằng đẳng thức: \(\left(A-B\right)^2=A^2-2AB+B^2\)

\(4x^2-20xy^2+25y^4\)

\(=\left(2x\right)^2-2\cdot2x\cdot5y^2+\left(5y\right)^2\)

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

\(4x^2-20xy^2+25y^4\)

\(=\left(2x\right)^2-2.2.5.xy^2+\left(5y\right)^2\)

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

Ta có ;

\(0.008-a^3b^6\)

\(=\left(0.2\right)^3-\left(ab^2\right)^3\)

\(=\left(0.2-ab^2\right)\left(0.04+0.2ab^2+a^2b^4\right)\)

\(0,008=0,2^3,a^6b^3=\left(a^2b\right)^3\)

=> \(0,2^3-\left(a^2b\right)^3=\left(0,2-a^2b\right)\left(0,04+0,2ab+a^4b^2\right)\)

Ta có :

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

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

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

\(=4y^2-9x^4\)

a, \(\left(\frac{x}{y}-\frac{2}{3}\right)\left(\frac{x}{y}+\frac{2}{3}\right)=\left(\frac{x}{y}\right)^2-\left(\frac{2}{3}\right)^2\)

b,\(\left(2\sqrt{x}-\frac{2}{3}\right)\left(\frac{2}{3}+2\sqrt{x}\right)=\left(2\sqrt{x}-\frac{2}{3}\right)\left(2\sqrt{x}+\frac{2}{3}\right)\)

\(=\left(2\sqrt{x}\right)^2-\left(\frac{2}{3}\right)^2\)

Trả lời:

a, \(\left(\frac{x}{y}-\frac{2}{3}\right)\left(\frac{x}{y}+\frac{2}{3}\right)\)\(=\left(\frac{x}{y}\right)^2-\left(\frac{2}{3}\right)^2=\frac{x^2}{y^2}-\frac{4}{9}\)

b, \(\left(2\sqrt{x}-\frac{2}{3}\right)\left(\frac{2}{3}+2\sqrt{x}\right)=\left(2\sqrt{x}-\frac{2}{3}\right)\left(2\sqrt{x}+\frac{2}{3}\right)=\left(2\sqrt{x}\right)^2-\left(\frac{2}{3}\right)^2=4x-\frac{4}{9}\)

\(\frac{1}{4}x^6-0,01y^2=\left(\frac{1}{2}x^3\right)^2-\left(0,1y\right)^2\)

                            \(=\left(\frac{1}{2}x^3-0,1y\right).\left(\frac{1}{2}x^3+0,1y\right)\)

Vậy \(\frac{1}{4}x^6-0,01y^2\)\(=\left(\frac{1}{2}x^3-0,1y\right).\left(\frac{1}{2}x^3+0,1y\right)\)

Tham khảo nhé ~

\(\frac{1}{4}x^6-0.01y^2\)

\(=\left(\frac{1}{2}x^3\right)^2-\left(0.1y\right)^2\)

\(=\left(\frac{1}{2}x^3-0.1y\right)\left(\frac{1}{2}x^3+0.1y\right)\)

Mong lần này không sai nữa ......