Bài 6

a) (3x² + 5) + [(2x² - 5x) - (5x² + 4)]

= 3x² + 5 + (2x² - 5x - 5x² - 4)

= 3x² + 5 + 2x² - 5x - 5x² - 4

= (3x² + 2x² - 5x²) - 5x + (5 - 4)

= -5x + 1

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b) (x + 2)(x² - 2x + 4)

= x.x² - x.2x + x.4 + 2.x² - 2.2x + 2.4

= x³ - 2x² + 4x + 2x² - 4x + 8

= x³ + (-2x² + 2x²) + (4x - 4x) + 8

= x³ + 8

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c) (4x³ - 8x² + 13x - 5) : (2x - 1)

= (4x³ - 2x² - 6x² + 3x + 10x - 5) : (2x - 1)

= [(4x³ - 2x²) - (6x² - 3x) + (10x - 5)] : (2x - 1)

= [2x²(2x - 1) - 3x(2x - 1) + 5(2x - 1)] : (2x - 1)

= (2x - 1)(2x² - 3x + 5) : (2x - 1)

= 2x² - 3x + 5

C

BẠN ƠI, LỖI RỒI KÌA

\(a,A\left(-1;3\right)\in\left(1\right)\Leftrightarrow-2+m=3\Leftrightarrow m=5\)

Trong 1 giờ đi được: 8:2/3=12(km)

12 Km

HT

8 : 2/3 = 12 (km)

\(A=\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{9}}{\dfrac{4}{3}+\dfrac{4}{5}-\dfrac{4}{9}}\\ =\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{9}\right):\left(\dfrac{4}{3}+\dfrac{4}{5}-\dfrac{4}{9}\right)\\ =\left[2\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{9}\right)\right]:\left[4\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{9}\right)\right]\\ =2\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{9}\right):4\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{9}\right)\\ =\dfrac{2}{4}\\ =\dfrac{1}{2}\)

\(A=\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{9}}{\dfrac{4}{3}+\dfrac{4}{5}-\dfrac{4}{9}}\\ A=\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{9}\right):\left(\dfrac{4}{3}+\dfrac{4}{5}-\dfrac{4}{9}\right)\\ A=\left(\dfrac{30}{45}+\dfrac{18}{45}-\dfrac{10}{45}\right):\left(\dfrac{60}{45}+\dfrac{36}{45}-\dfrac{20}{45}\right)\\ A=\dfrac{38}{45}:\dfrac{76}{45}\\ A=\dfrac{38}{76}\)

\(\Leftrightarrow x\left(x-1\right)+3\left(x+3\right)=6x\)

\(\Leftrightarrow x^2-x+3x+9-6x=0\)

\(\Leftrightarrow x^2-4x+9=0\)(vô lý)

\(ĐK:x\ne0\)

\(\Leftrightarrow\dfrac{x\left(x-1\right)+3\left(x+3\right)}{3x}=\dfrac{6x}{3x}\)

\(\Leftrightarrow x\left(x-1\right)+3\left(x+3\right)=6x\)

\(\Leftrightarrow x^2-x+3x+9=6x\)

\(\Leftrightarrow x^2-4x+9=0\)

11B 12B 13C 14d 15b 16c 17a 18B 19A 20a