Câu 1: 10km

Câu 2: 1,8 giờ

Câu 3: vận tốc đơn vị j chứ nếu hỏi thế thì 1km/ 2p 5s vẫn là vận tốc mà

Câu 4: 4,5 N

A = x² - 2x + 9 = ( x² - 2x + 1 ) + 8 = ( x - 1 )² + 8 ≥ 8 ∀ x

Dấu "=" xảy ra khi x = 1

=> MinA = 8 <=> x = 1

B = x² + 6x - 3 = ( x² + 6x + 9 ) - 12 = ( x + 3 )² - 12 ≥ -12 ∀ x

Dấu "=" xảy ra khi x = -3

=> MinB = -12 <=> x = -3

C = ( x - 1 )( x - 3 ) + 9 = x² - 4x + 3 + 9 = ( x² - 4x + 4 ) + 8 = ( x - 2 )² + 8 ≥ 8 ∀ x

Dấu "=" xảy ra khi x = 2

=> MinC = 8 <=> x = 2

D = -x² - 4x + 7 = -( x² + 4x + 4 ) + 11 = -( x + 2 )² + 11 ≤ 11 ∀ x

Dấu "=" xảy ra khi x = -2

=> MaxD = 11 <=> x = -2

hello, cần lm j z?

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( 2x² - y )³ - 64y³

= ( 2x² - y )³ - ( 4y )³

= ( 2x² - y - 4y )[ ( 2x² - y )² + ( 2x² - y ).4y + ( 4y )² ]

= ( 2x² - 5y )( 4x⁴ - 4x²y + y² + 8x²y - 4y² + 16y² )

= ( 2x² - 5y )( 4x⁴ + 4x²y + 13y² )

Ta có: \(4x^2+y^2=8+3xy\Leftrightarrow4x^2-4xy+y^2=8-xy\)

\(\Leftrightarrow\left(2x-y\right)^2=8-xy\ge0\forall x,y\inℝ\Rightarrow xy\le8\)

\(\Rightarrow P=xy+2020\le8+2020=2028\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}2x=y\\xy=8\end{cases}}\Rightarrow\left(x,y\right)\in\left\{\left(2;4\right);\left(-2;-4\right)\right\}\)

a)2x(x²-3y)

=2x³-6xy

-18x³y⁴:3xy⁴

=(-18:3) (x³:x) (y⁴:y⁴)

=-6x²

\(C=\left(1-\frac{x}{x+1}\right)\div\left(\frac{x+3}{x-2}+\frac{2+x}{3-x}+\frac{x+2}{x^2-5x+6}\right)\)

ĐKXĐ : x ≠ -1 ; x ≠ 2 ; x ≠ 3 ; x ≠ 11/5

\(=\left(\frac{x+1}{x+1}-\frac{x}{x+1}\right)\div\left(\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right)\)

\(=\frac{1}{x+1}\div\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right)\)

\(=\frac{1}{x+1}\div\left(\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-4}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right)\)

\(=\frac{1}{x+1}\div\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right)\)

\(=\frac{1}{x+1}\div\frac{x-3}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{1}{x+1}\times\frac{x-2}{1}\)

\(=\frac{x-2}{x+1}\)

ĐKXĐ bạn bỏ cái x ≠ 11/5 hộ mình nhé ;-;