OIMEOI

may lam dai the thi cho no giai ho a 

1)  (3x+4)(x+1) = 3x²+7x+4 đặt là a

(6x+7)²= 36x²+84x+49 = 12a+1

=> a(12a+1)- 6 = 12a² -a -6 = (3a+2)(4a-3) = (9x²+21x+14)(12x²+28x+13)

2) (x-2)²=x²-4x+4 đặt là a

(2x-5)(2x-3)= 4x²-16x+15 =4a-1

=> a(4a-1)-5 = 4a²-a-5 = (4a-5)(a+1) = ( 4x²-16x+11)(x²-4x+5)

3) đặt (x+3)² =a ta làm tương tự

4) (x-2)(x-10)(x-4)(x-5) = (x²-12x+20)(x²-9x+20)

đặt x²+20=a => (a-12x)(a-9x)-54x² = a²-21ax+54x² = (a-18x)(a-3x) = (x²-18x+20)(x²-3x+20)

a: \(=-\left(x^2+10x-11\right)\)

\(=-\left(x^2+10x+25-36\right)\)

\(=-\left(x+5\right)^2+36< =36\)

Dấu '=' xảy ra khi x=-5

b: \(=-\left(x^2-6x+5\right)\)

\(=-\left(x^2-6x+9-4\right)\)

\(=-\left(x-3\right)^2+4< =4\)

Dấu '=' xảy ra khi x=3

c: \(=-2\left(x^2-x+\dfrac{5}{2}\right)\)

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

\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)

Dấu '=' xảy ra khi x=1/2

d: \(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

\(=-\left(x^2+2x-8\right)\)

\(=-\left(x^2+2x+1-9\right)\)

\(=-\left(x+1\right)^2+9< =9\)

Dấu '=' xảy ra khi x=-1

?1 . Có . Mẫu thức chung : 12x²y³z đơn giản hơn

?2 . \(\dfrac{3}{x^2-5x}=\dfrac{3}{x\left(x-5\right)}=\dfrac{6}{2x\left(x-5\right)}\)

\(\dfrac{5}{2x-10}=\dfrac{5}{2\left(x-5\right)}=\dfrac{5x}{2x\left(x-5\right)}\)

?3 . \(\dfrac{3}{x^2-5x}=\dfrac{3}{x\left(x-5\right)}=\dfrac{6}{2x\left(x-5\right)}\)

\(\dfrac{-5}{10-2x}=\dfrac{5}{2x-10}=\dfrac{5}{2\left(x-5\right)}=\dfrac{5x}{2x\left(x-5\right)}\)