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a) \(25a^2-1=\left(5a-1\right)\left(5a+1\right)\)
b) \(a^2-9=\left(a-3\right)\left(a+3\right)\)
c) \(\dfrac{1}{4}a^2-\dfrac{9}{25}=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)\)
d) \(\dfrac{9}{4}a^4-\dfrac{16}{25}=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\)
e) \(\left(2a+b\right)^2-a^2=\left(2a+b-a\right)\left(2a+b+a\right)=\left(a+b\right)\left(3a+b\right)\)
f) \(16\left(x-1\right)^2-25\left(x+y\right)^2=\left(4x-4-5x-5y\right)\left(4x-4+5x+5y\right)=\left(-x-4-5y\right)\left(9x+5y-4\right)\)
a/ $25x^2-1\\=(5x)^2-1^2\\=(5x-1)(5x+1)$
b/ $a^2-9\\=a^2-3^2\\=(a-3)(a+3)$
c/ $\dfrac{1}{4}a^2-\dfrac{9}{25}\\=\left(\dfrac{1}{2}a\right)^2-\left(\dfrac{3}{5}\right)^2\\=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)$
d/ $\dfrac{9}{4}a^4-\dfrac{16}{25}\\=\left(\dfrac{3}{2}a^2\right)^2-\left(\dfrac{4}{5}\right)^2\\=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left[\left(\sqrt{\dfrac 3 2}a\right)^2-\left(\dfrac{2\sqrt 5}{5}\right)^2\right]\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left(\sqrt{\dfrac 3 2}a-\dfrac{2\sqrt 5}{5}\right)\left(\sqrt{\dfrac 3 2}a+\dfrac{2\sqrt 5}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)$
e/ $(2a+b)^2-a^2\\=(2a+b-a)(2a+b+a)\\=(a+b)(3a+b)$
f/ $16(x-1)^2-25(x+y)^2\\=[4(x-1)]^2-[5(x-y)]^2\\=[4(x-1)-5(x-y)][4(x-1)+5(x-y)]\\=[4x-4-5x+5y][4x-4+5x-5y]\\=(-x+5y-4)(9x-5y-4)$
a) \(=mp\left(m^2+mn-mp-np\right)=mp\left[m\left(m+n\right)-p\left(m+n\right)\right]=mp\left(m+n\right)\left(m-p\right)\)
b) \(=abm^2+abn^2+a^2mn+b^2mn=am\left(bm+an\right)+bn\left(bm+an\right)\)
\(=\left(bm+an\right)\left(am+bn\right)\)
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) x2-xy+5y-25
= x(2-y)+ 5(y-2)
= x(2-y)-5(2-y)
= (x-5)(2-y)
a: =x^3(x-y)+(x-y)
=(x-y)(x^3+1)
=(x-y)(x+1)(x^2-x+1)
b: =(a-1)^2-9b^2
=(a-1-3b)(a-1+3b)
TK
a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).
b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2
= (a + 4 – b)(a + 4 + b).
\(a,a^2\left(a-b\right)+ab\left(a-c\right)=a\left(a+b\right)\left(a-c\right)\\ c,=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ b,=\left(x-5\right)^2-9y^2=\left(x-5-3y\right)\left(x-5+3y\right)\\ d,=4\left(x^2-9x+14\right)=4\left(x-7\right)\left(x-2\right)\)
\(a,=\left(a-5\right)^2-4b^2=\left(a-2b-5\right)\left(a+2b-5\right)\\ b,=ax^2+a-a^2x-x=ax\left(a-x\right)+\left(a-x\right)=\left(ax+1\right)\left(a-x\right)\)
a: \(=\left(a-5-2b\right)\left(a-5+2b\right)\)
b: \(ax^2+a-a^2x-x\)
\(=ax\left(x-a\right)-\left(x-a\right)\)
\(=\left(x-a\right)\left(ax-1\right)\)