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

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

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

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

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a,\(8y^2-\dfrac{1}{8}\)

=\(\dfrac{1}{8}\left(64y^2-1\right)\)

=\(\dfrac{1}{8}\left(8y-1\right)\left(8y+1\right)\)

b,\(\dfrac{1}{25}y^2-64m^2\)

=\(\left(\dfrac{1}{5}y-8m\right)\left(\dfrac{1}{5}y+8m\right)\)

(a+b)³-(a-b)³=a³+3a²b+3ab²+b³-(a³-3a²b+3ab²-b³)

                    =a³+3a²b+3ab²+b³-a³+3a²b-3ab²+b³

                        =6a²b+2b³

Áp dụng hđt a³-b³=(a-b)(a²+ab+b²) ấy

\(\left(a+b\right)^3-\left(a-b\right)^3=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

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

\(=2b\left(3a^2+b^2\right)\)

Ta có : \(\left(a+b\right)^3-\left(a-b\right)^3\) 

\(=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\)

\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3\)

\(=6a^2b+2b^3\)

\(\left(25-m^2\right)=\left(5-m\right)\left(5+m\right)\)

giải chi tiết hộ với em mới học

`25-m^2`

`=5^2-m^2`

`=(5-m)(5+m)`

Áp dụng HĐT : `a^2-b^2=(a-b)(a+b)`

\(10a-25-a^2=-\left(a^2-10a+25\right)=-\left(a^2-2.a.5+5^2\right)=-\left(a-5\right)^2\)

`10a-25-a^2`

`=-(a^2-10a+25)`

`=-(a^2-2.a.5+5^2)`

`=-(a-5)^2`

Áp dụng hằng đẳng thức : `(a-b)^2=a^2-2ab+b^2`

-a ^2 + 10 a - 25

- ( a^2 -10 a + 25 )

- ( a - 5 )^2

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

Chúc bạn học tốt và nhớ click cho mình với nhá!

= (5x-25) + (5x - x²)

= 5(x-5) + x(5-x)

= 5(x-5) - x(x-5)

= (5 - x)(x - 5)

\(10x-25-x^2=-\left(x^2-2\cdot x\cdot5+5^2\right)=-\left(x-5\right)^2\)

\(=x+10\sqrt{x}+25-20=\left(\sqrt{x}+5\right)^2-\left(2\sqrt{5}\right)^2\\ =\left(\sqrt{x}+5-2\sqrt{5}\right)\left(\sqrt{x}+5+2\sqrt{5}\right)\)

\(12-3\sqrt{12}=9-\sqrt{108}+3=9-2\sqrt{27}+3=\left(3-\sqrt{3}\right)^2\)

Bạn ei là hàng đẳng thức \(\left(a-b\right)^2\)??

\(12-3\sqrt{12}=12-3\sqrt{4.3}=12-3.2.\sqrt{3}\)

\(=9-2.3.\sqrt{3}+3=3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2\)

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

mik cảm ơn ạ ^^