=−5. .

a) 2 y − 1 y − 2 x + 1 x y 2y−1 − x 2x+1 = x ( 2 y − 1 ) x y − y ( 2 x + 1 ) x y = xy x(2y−1) − xy y(2x+1) = 2 x y − x − 2 x y − y x y = − x − y x y = xy 2xy−x−2xy−y = xy −x−y . b) 2 x 3 : 5 6 x 2 3 2x : 6x 2 5 2 x 3 : 5 6 x 2 = 2 x 3 . 6 x 2 5 3 2x : 6x 2 5 = 3 2x . 5 6x 2 = 4 x 3 5 = 5 4x 3 .

Ta có x 2 − 4 x + 9 = ( x − 2 ) 2 + 5 ⩾ 5 x 2 −4x+9=(x−2) 2 +5⩾5. Suy ra B = 1 x 2 − 4 x + 9 = 1 ( x − 2 ) 2 + 5 ⩽ 1 5 B= x 2 −4x+9 1 = (x−2) 2 +5 1 ⩽ 5 1 . Dấu bằng xảy ra khi x = 2 x=2.

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) Rút gọn A = ( x − 1 ) 2 ( x − 1 ) ( x + 1 ) = x − 1 x + 1 A= (x−1)(x+1) (x−1) 2 = x+1 x−1 . b) Với x = 3 x=3 thì A = 3 − 1 3 + 1 = 1 2 A= 3+1 3−1 = 2 1 Với x = 3 2 x= 2 3 thì A = − 3 2 − 1 − 3 2 + 1 = 5 A= − 2 3 +1 − 2 3 −1 =5 c) Ta có biến đối: A = x − 1 x + 1 = 1 + − 2 x + 1 A= x+1 x−1 =1+ x+1 −2 . Để biểu thức A A nguyên khi − 2 x + 1 x+1 −2 hay x + 1 x+1 là ước của − 2 −2. Do đó x + 1 x+1 1 1 − 1 −1 2 2 − 2 −2 x x 0 0 − 2 −2 1 1 − 3 −3 Đối chiếu điều kiện ta thấy x x có giá trị − 2 ; − 3 ; 0 −2;−3;0 thì biểu thức A A nguyên.

a) 7 x + 2 = 0 7x+2=0 7 x = − 2 7x=−2 x = − 2 7 x=− 7 2 . b) 18 − 5 x = 7 + 3 x 18−5x=7+3x − 5 x − 3 x = 7 − 18 −5x−3x=7−18 − 8 x = − 11 −8x=−11 x = 11 8 x= 8 11 .

a) \(x^{2} + 25 - 10 x = x^{2} - 2.5. x + 5^{2} = \left(\left(\right. x - 5 \left.\right)\right)^{2}\)

b) \(- 8 y^{3} + x^{3} = x^{3} - \left(\left(\right. 2 y \left.\right)\right)^{3} = \left(\right. x - 2 y \left.\right) \left(\right. x^{2} + 2 x y + 4 y^{2} \left.\right)\).

a) \(\left(\left(\right. 2 x + 1 \left.\right)\right)^{2} = 4 x^{2} + 4 x + 1\).

b) \(\left(\left(\right. a - \frac{b}{2} \left.\right)\right)^{3} = a^{3} - 3 a^{2} . \frac{b}{2} + 3 a \left(\left(\right. \frac{b}{2} \left.\right)\right)^{2} - \left(\left(\right. \frac{b}{2} \left.\right)\right)^{3}\)

\(= a^{3} - \frac{3}{2} a^{2} b + \frac{3}{4} a b^{2} - \frac{1}{8} b^{3}\).

?

a) \(x^{2} + 2 x y + y^{2} - x - y = \left(\right. x + y \left.\right) \left(\right. x + y - 1 \left.\right)\);

b) \(2 x^{3} + 6 x^{2} + 12 x + 8 = \&\text{nbsp}; \left(\right. 2 x + 2 \left.\right) \left(\right. x^{2} + 2 x + 4 \left.\right)\).