f

1. \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

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

\(\Rightarrow VT=VP\)

2. \(a^4-b^4=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\)

\(VP=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)=\left(a^2-b^2\right)\left(a^2+b^2\right)=a^4+a^2b^2-b^2a^2-b^4=a^4-b^4\)

\(\Rightarrow VT=VP\)

3. \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(bx+ay\right)^2\)

\(VT=\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(VP=\left(ax-by\right)^2+\left(bx+ay\right)^2=a^2x^2-2axby+b^2y^2+b^2x^2+2bxay+a^2y^2=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(\Rightarrow VT=VP\)

cảm ơn bạn nhiều nhé !!!!

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

2. \(x^2+5x+\dfrac{25}{4}=x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)

3. \(16x^2-8x+1=\left(4x-1\right)^2\)

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

5. \(\dfrac{1}{4}x^2-\dfrac{4}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)\left(\dfrac{1}{2}x+\dfrac{2}{3}y\right)\)

6. \(a^2-2ab+b^2-x^2=\left(a-b\right)^2-x^2=\left(a-b-x\right)\left(a-b+x\right)\)

7. \(4x^2-20x+25-y^2=\left(2x-5\right)^2-y^2=\left(2x-5-y\right)\left(2x-5+y\right)\)

bạn ơi bạn có thể giải thick cách lm đc ko

1.(x^2-1)^2=[(x-1).(x+1]^2

2). (x+5/2)^2

3). {4x-1)^2

a: \(A=2xy^2+x^2y^4+1\)

\(=\left(xy^2\right)^2+2\cdot xy^2\cdot1+1^2\)

\(=\left(xy^2+1\right)^2\)

\(=\left(2\cdot16^2+1\right)^2=\left(2\cdot256+1\right)^2=513^2\)

b: \(B=x^3+9x^2+27x+27\)

\(=x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3\)

\(=\left(x+3\right)^3=\left(97+3\right)^3=100^3=1000000\)

c: \(\left(2x+y^2-1\right)\left(2x+y^2+1\right)\)

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

\(=4x^2+4xy^2+y^4-1\)

1. \(\left(-a\right)^7\) : \(a^5\) = \(\left(-a\right)^2\) = a

2. 28 \(y^4z^3\) : 14 \(y^3z^2\) = 2yz

3. 25\(a^2bc^2\) : 5abc = 5ac

giúp mik với