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a) 7x – 14 = 7(x - 2)
b) 5x3 - 10x2y +5xy2 = 5x(x2 - 2xy - y2) = 5x(x - y)2
c) 25 – x2= (x - 5)(x + 5)
a, \(7\left(x-2\right)\)
c, \(\left(5-x\right)\left(5+x\right)\)
\(a,=\left(a-5\right)^2-4b^2=\left(a-2b-5\right)\left(a+2b-5\right)\\ b,=ax^2+a-a^2x-x=ax\left(a-x\right)+\left(a-x\right)=\left(ax+1\right)\left(a-x\right)\)
a: \(=\left(a-5-2b\right)\left(a-5+2b\right)\)
b: \(ax^2+a-a^2x-x\)
\(=ax\left(x-a\right)-\left(x-a\right)\)
\(=\left(x-a\right)\left(ax-1\right)\)
\(-9x^4+3x^2+2\\ =-9x^4+6x^2-3x^2+2\\ =-3x^2\left(3x^2-2\right)-\left(3x^2-2\right)\\ =-\left(3x^2-2\right)\left(3x^2+1\right)\)
\(-9x^4+3x^2+2\)
\(=-9x^4+6x^2-3x^2+2\)
\(=-3x^2\left(3x^2-2\right)-\left(3x^2-2\right)\)
\(=\left(3x^2-2\right)\left(-3x^2-1\right)\)
Hửm đề sai rồi phải là:
`a^2+2b^2-3ab`
`=a^2-ab-2ab+2b^2`
`=a(a-b)-2b(a-b)`
`=(a-b)(a-2b)`
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) \(4x^2\left(x+3\right)-8x\left(3+x\right)=4x\left(x+3\right)\left(x-2\right)\)
b) \(4x^2+y^2-25+4xy=\left(2x+y\right)^2-25=\left(2x+y-5\right)\left(2x+y+5\right)\)
c) \(\left(x-3\right)^2-\left(x+2\right)^2=\left(x-3-x-2\right)\left(x-3+x+2\right)=-5\left(2x-1\right)\)
\(-\left(3x^2-5\right)^2\)
\(- ( 3 x2 - 5)2\)
Phân tích xong ròi nhá
MAGICPENCIL
HÃY LUÔN :-)