đề bài đúng đây:

cho hình vẽ biết: x//y, biết góc M3= 50 độ. Tính góc M2, M4, N1, N2, N3 ?

Bài 1.

Ta có : B = ( x + 2 )² + ( x - 2 )² - 2( x + 2 )( x - 2 )

= [ ( x + 2 ) - ( x - 2 ) ]²

= ( x + 2 - x + 2 )²

= 4² = 16

=> B không phụ thuộc vào x

Vậy với x = -4 thì B vẫn bằng 16

Bài 2.

4x² - 4x + 1 = ( 2x )² - 2.2x.1 + 1² = ( 2x - 1 )²

Bài 3.

Ta có : \(A=\frac{3}{2}x^2+2x+3\)

\(=\frac{3}{2}\left(x^2+\frac{4}{3}x+\frac{4}{9}\right)+\frac{7}{3}\)

\(=\frac{3}{2}\left(x+\frac{2}{3}\right)^2+\frac{7}{3}\ge\frac{7}{3}\forall x\)

Dấu "=" xảy ra khi x = -2/3

=> MinA = 7/3 <=> x = -2/3

Bài 1.

3x² + 2 có bậc thấp hơn x³ + 1 nên không thể chia tiếp

Vậy x³ + 3x² + 3 = 1( x³ + 1 ) + 3x² + 2

Bài 2.

Ta có : x³ + 3x² + 3x + a có bậc là 3

x + 2 có bậc là 1

=> Thương bậc 2

lại có hệ số cao nhất của đa thức bị chia là 1

Đặt đa thức thương là x² + bx + c

khi đó : x³ + 3x² + 3x + a chia hết cho x + 2

<=> x³ + 3x² + 3x + a = ( x + 2 )( x² + bx + c )

<=> x³ + 3x² + 3x + a = x³ + bx² + cx + 2x² + 2bx + 2c

<=> x³ + 3x² + 3x + a = x³ + ( b + 2 )x² + ( c + 2b )x + 2c

Đồng nhất hệ số ta được :

\(\hept{\begin{cases}b+2=3\\c+2b=3\\2c=a\end{cases}}\Leftrightarrow\hept{\begin{cases}b=1\\c=1\\a=2\end{cases}}\Rightarrow a=2\)

Vậy a = 2

\(2.2^2.2^3.2^4...2^x=\left(2^3\right)^{12}\)

<=> \(2^{1+2+3+...+x}=2^{36}\)

<=> 1 + 2 + 3 + ...+ x = 36

<=> x.(x+1): 2 = 36 

<=> x .(x + 1) = 8.9 

<=> x = 8

cần gấp

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a) 2x³ + 8x² - 8x

= 2x(x² + 4x - 4)

= 2x(x² + 4x + 4 - 8)

= 2x[(x + 2)² - 8]

= \(2x\left(x+2-\sqrt{8}\right)\left(x+2+\sqrt{8}\right)\)

b) a² - b² + 4a + 4b

= (a - b)(a + b) + 4(a + b)

= (a + b)(a - b + 4)

c) x² - 2x - 3

= x² + x - 3x - 3

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

= (x + 1)(x - 3)

d) x² - 4x - 3

= x² - 4x + 4 - 7

= (x + 2)² - 7

= \(\left(x+2-\sqrt{7}\right)\left(x+2+\sqrt{7}\right)\)