= x^2 + 2.x.5 +5^2 +2(x^2 - 7^2 ) - 10x -25

= x^2 + 10x +25 +2(x^2- 49) -10x-25

= x^2 +10x + 25 + 2x^2 - 98 -10x-25

= (x^2 +2x^2) +(10x-10x) +(25-25-98)

= 3x^2 -98 

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=6a^2+2ab+15ab+5b^2

=2a(3a+b)+5b(3a+b)

=(2a+5b)(3a+b)

áp dụng định lí bézout lên tìm đc số dư trong phép chia là a*1^5+4*4^4-9

=a+64-9

=a+55

để chia hết thì a+55=0

<=>a=-55

\(N=x^2+2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2+\dfrac{39}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{39}{4}\)

\(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}\forall x\)

\(\Rightarrow N_{min}=\dfrac{39}{4}\)

a)\(x^2-\dfrac{3}{2}x-\dfrac{5}{2}x+15\)

(\(x-\dfrac{5}{2}\))(\(x-\dfrac{3}{2}\))

\(\text{b)30x^2-5xy-12xy+2y^2}\)

\(\left(5x+2y\left(6x-y\right)\right)\)

c)\(15x^2+3xy-10xy-2y^2\)

\(\left(3x-2y\right)\left(5x+y\right)\)

8x^2 + 6x - 35

= 2 × (4x^2 + 3x - 35/2)

= 2 × [(2x)^2 + 2 × 2x × 3/4 +(3/4)^2 - (9/16+35/2) ]

= 2 × [(2x+3/4)^2 -(289/16)

a)8x^2+6x-35=8x^2+20x-14x-35=4x(2x+5)-7(2x+5)=(4x-7)(2x+5)

b)2x^2-13x+20=2x^2-5x-8x+20=x(2x-5)-4(2x-5)=(x-4)(2x-5)

c)3x^2+13xy-30y^2=3x^2+18xy-5xy-30y^2=(3x-5y)(x+6y)