Ta có: 

\(C=1-x-\frac{x^2}{3}-\frac{x^3}{27}\)

\(C=1-\left(-27\right)-\frac{\left(-27\right)^2}{3}-\frac{\left(-27\right)^3}{27}\) tại x = -27

\(C=1+27-\frac{3^6}{3}+27^2\)

\(C=1+27-243+729\)

\(C=514\)

1) \(x^3+2x-3\)

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

\(=x^2\left(x-1\right)+x\left(x-1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+3\right)\)

2) \(x^3-6x+4\)

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

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

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

3) \(x^3-2x^2+1\)

\(=\left(x^3-x^2\right)-\left(x^2-x\right)-\left(x-1\right)\)

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

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

4) \(x^3+5x^2-12\)

\(=\left(x^3+2x^2\right)+\left(3x^2+6x\right)-\left(6x+12\right)\)

\(=x^2\left(x+2\right)+3x\left(x+2\right)-6\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+3x-6\right)\)

Ta có:

\(-25x^6-y^8+10x^3y^4\)

\(=-\left[\left(5x^3\right)^2-10x^3y^4+\left(y^4\right)^2\right]\)

\(=-\left(5x^3-y^4\right)^2\)

a. Gọi thời gian đoạn xuống dốc là t (h), thời gian đoạn lên dốc là \(\frac{4}{3}.t\left(h\right)\)

Quãng đường các đoạn xuống dốc và lên dốc lần lượt là: 

      \(s_1=50t\) (km)

      \(s_2=30.\frac{4t}{3}=40t\) (km) 

Ta có: 

 \(\frac{s_1}{s_2}=\frac{5}{4}\)

\(\Rightarrow s_1=\frac{5s_2}{4}\)

Vậy đoạn xuống dốc dài hơn đoạn lên dốc và bằng 5454 đoạn lên dốc. .

lỡ bấm Gửi trả lời , câu b này :

b,Vận tốc trung bình trên cả quãng đường là : 

\(v_{tb}=\frac{s}{t}=\frac{s_1+s_2}{t+\frac{5t}{4}}=\frac{50t+40t}{\frac{9t}{4}}=\frac{90t.4}{9t}=40\left(km/h\right)\)

học tốt

a) (2x - 3)² = (x + 5)²

=> 4x² - 12x + 9 = x² + 10x + 25

=> 4x² - 12x + 9 - (x² + 10x + 25) = 0

=> 3x² - 22x - 16 = 0

=> 3x² - 24x + 2x - 16 = 0

=> 3x(x - 8) + 2(x - 8) = 0

=> (3x + 2)(x - 8) = 0

=> \(\orbr{\begin{cases}3x+2=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=8\end{cases}}\)

b) x²(x - 1) - 4x² + 8x - 4 = 0

=> x²(x - 1) - (2x  - 2)² = 0

=> x²(x - 1) - [2(x- 1)]² = 0

=> x²(x - 1) - 4(x - 1)² = 0

=> (x - 1)(x² - 4(x - 1) = 0

=> (x - 1)(x² - 4x + 4) = 0

=> (x - 1)(x - 2)² = 0

=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

c) x² + 7x + 12 = 0

=> x² + 3x + 4x + 12 = 0

=> x(x + 3) + 4(x + 3) = 0

=> (x + 4)(x + 3) = 0

=> \(\orbr{\begin{cases}x+4=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-4\\x=-3\end{cases}}\)

d) x² + 3x - 18 = 0

=> x² + 6x - 3x - 18 = 0

=> x(x + 6) - 3(x + 6) = 0

=> (x - 3)(x + 6) = 0

=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

e) x(x + 6) - 7x - 42 = 0

=> x(x + 6) - 7(x + 6) = 0

=> (x - 7)(x + 6) = 0

=> \(\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)

1. ( 2x - 3 )² = ( x + 5 )²

<=> ( 2x - 3 )² - ( x + 5 )² = 0

<=> [ ( 2x - 3 ) - ( x + 5 ) ][ ( 2x - 3 ) + ( x + 5 ) ] = 0

<=> ( 2x - 3 - x - 5 )( 2x - 3 + x + 5 ) = 0

<=> ( x - 8 )( 3x + 2 ) = 0

<=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

2. x²( x - 1 ) - 4x² + 8x - 4 = 0

<=> x²( x - 1 ) - ( 4x² - 8x + 4 ) = 0

<=> x²( x - 1 ) - 4( x² - 2x + 1 ) = 0

<=> x²( x - 1 ) - 4( x - 1 )² = 0

<=> ( x - 1 )[ x² - 4( x - 1 ) ] = 0

<=> ( x - 1 )( x² - 4x + 4 ) = 0

<=> ( x - 1 )( x - 2 )² = 0

<=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

3. x² + 7x + 12 = 0

<=> x² + 3x + 4x + 12 = 0

<=> x( x + 3 ) + 4( x + 3 ) = 0

<=> ( x + 3 )( x + 4 ) = 0

<=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)

4. x² + 3x - 18 = 0

<=> x² - 3x + 6x - 18 = 0

<=> x( x - 3 ) + 6( x - 3 ) = 0

<=> ( x - 3 )( x + 6 ) = 0

<=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

5. x( x + 6 ) - 7x - 42 = 0

<=> x( x + 6 ) - 7( x + 6 ) = 0

<=> ( x + 6 )( x - 7 ) = 0

<=> \(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}\)

1. x² - 16 - 4xy + 4y²

= ( x² - 4xy + 4y² ) - 16

= ( x - 2y )² - 4²

= ( x - 2y - 4 )( x - 2y + 4 )

2. 4x² + 4x - 3

= ( 4x² + 4x + 1 ) - 4

= ( 2x + 1 )² - 2

= ( 2x + 1 - 2 )( 2x + 1 + 2 )

= ( 2x - 1 )( 2x + 3 )

3. x² - x - 12

= x² + 3x - 4x - 12

= x( x + 3 ) - 4( x + 3 )

= ( x + 3 )( x - 4 )

4. 3x + 3y - x² - 2xy - y²

= ( 3x + 3y ) - ( x² + 2xy + y² )

= 3( x + y ) - ( x + y )²

= ( x + y )( 3 - x - y )

5. 4y⁴ + 16 

= 4( y⁴ + 4 )

= 4( y⁴ + 4y² + 4 - 4y² )

= 4[ ( y⁴ + 4y² + 4 ) - 4y² ]

= 4[ ( y² + 2 )² - ( 2y )² ]

= 4( y² - 2y + 2 )( y² + 2y + 2 )

a,\(x^2-16-4xy+4y^2\)

\(=\left(x^2-4xy+4y^2\right)-16\)

\(=\left(x-2y\right)^2-4^2\)

\(=\left(x-2y-4\right)\left(x-2y+4\right)\)

b,\(4x^2+4x-3\)

\(=4x^2-2x+6x-3\)

\(=\left(4x^2-2x\right)+\left(6x-3\right)\)

\(=2x\left(2x-1\right)+3\left(2x-1\right)\)

\(=\left(2x+3\right)\left(2x-1\right)\)

c,\(x^2-x-12\)

\(=x^2-4x+3x-12\)

\(=\left(x^2+3x\right)-\left(4x-12\right)\)

\(=x\left(x+3\right)-4\left(x+3\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

2x - 2x² - 5

= -2( x² - x + 1/4 ) - 9/2

= -2( x - 1/2 )² - 9/2 ≤ -9/2 ∀ x

Dấu "=" xảy ra <=> x = 1/2

Vậy GTLN của biểu thức = -9/2 <=> x = 1/2

x-x\(^2\)=0

Ta có:

\(x-x^2\)

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

\(=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(-\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)

Vậy Max = 1/4 khi x = 1/2

Ta có: \(4x-x^2+3\)

\(=-\left(x^2-4x+4\right)+7\)

\(=-\left(x-2\right)^2+7\le7\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(-\left(x-2\right)^2=0\Rightarrow x=2\)

Vậy Max = 7 khi x = 2

4x - x² + 3

= -( x² - 4x + 4 ) + 7

= -( x - 2 )² + 7 ≤ 7 ∀ x

Dấu = xảy ra <=> x = 2

Vậy GTLN của đa thức = 7 <=> x = 2

Ta có: 

\(M=x^2+y^2-x+6y+10\)

\(M=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)

\(M=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)

M = x² + y² - x + 6y + 10

= ( x² - x + 1/4 ) + ( y² + 6y + 9 ) + 3/4

= ( x - 1/2 )² + ( y + 3 )² + 3/4 ≥ 3/4 ∀ x

Dấu "=" xảy ra <=> x = 1/2 ; y = -3

=> MinM = 3/4 <=> x = 1/2 ; y = -3