Ta có:

\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)

\(P=\frac{5^{32}-1}{2}\)

Ta có: \(x\left(x-1\right)\)

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

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

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

Vậy \(Min=-\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)

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

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

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

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

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

Dấu " = " xảy ra khi và chỉ khi (x - 1/2)² = 0 => x = 1/2

Vậy giá trị nhỏ nhất là -1/4 khi x = 1/2

A = 5x( x - 1 )( 2x + 3 ) - 10x( x - 4 )

= 5x( 2x² + x - 3 ) - 10x² + 40x

= 10x³ + 5x² - 15x - 10x² + 40x

= 10x³ - 5x² + 25x

Thế x = -1/3 ta được

A = \(10\times\left(-\frac{1}{3}\right)^3-5\times\left(-\frac{1}{3}\right)^2+25\times\left(-\frac{1}{3}\right)\)

= \(10\times\left(-\frac{1}{27}\right)-5\times\frac{1}{9}-\frac{25}{3}\)

= \(-\frac{10}{27}-\frac{5}{9}-\frac{25}{3}\)

= \(-\frac{250}{27}\)

b) Đề sai . Tính khó

c) x = 14

=> 13 = x - 1

15 = x + 1

16 = x + 2

29 = 2x + 1 

Thế vào C ta được :

C = x⁵ - ( x + 1 )x⁴ + ( x + 2 )x³ - ( 2x + 1 )x² + ( x - 1 )x

= x⁵ - x⁵ - x⁴ + x⁴ + 2x³ - 2x³ - x² + x² - x

= -x = -14

           Bài làm :

a) M là trung điểm BC ; N là trung điểm DC

=> MN là đường trung bình tam giác CBD =>MN // BD

=> Điều phải chứng minh

b) Sai đề

Vì BD là đường trung tuyến ứng với AC ; AM là đường trung tuyến ứng với BC

=> Theo tính chất đường trung tuyến thì IA = 2/3 IM

a) ( x - 1 )² - ( x - 1 )( x + 1 ) = 0

<=> x² - 2x + 1 - ( x² - 1 ) = 0

<=> x² - 2x + 1 - x² + 1 = 0

<=> 2 - 2x = 0

<=> 2x = 2 

<=> x = 1

b) ( 2x - 1 )² - ( 2x + 1 )² = 0

<=> [ ( 2x - 1 ) - ( 2x + 1 ) ][ ( 2x - 1 ) + ( 2x + 1 ) ] = 0

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

<=> -2.4x = 0

<=> -8x = 0

<=> x = 0

c) 25( x + 3 )² + ( 1 - 5x )( 1 + 5x ) = 8

<=> 5²( x + 3 )² + 1² - 25x² = 8

<=> [ 5( x + 3 ) ]² + 1 - 25x² = 8

<=> ( 5x + 15 )² + 1 - 25x² = 8

<=> 25x² + 150x + 225 + 1 - 25x² = 8

<=> 150x + 226 = 8

<=> 150x = -218

<=> x = -218/150 = -109/75

d) 9( x + 1 )² - ( 3x - 2 )( 3x + 2 ) = 10

<=> 3²( x + 1 )² - ( 9x² - 4 ) = 10

<=> [ 3( x + 1 ) ]² - 9x² + 4 = 10

<=> ( 3x + 3 )² - 9x² + 4 = 10

<=> 9x² + 18x + 9 - 9x² + 4 = 10

<=> 18x + 13 = 10

<=> 18x = -3

<=> x = -3/18 = -1/6

a) (x - 1)² - (x - 1)(x + 1) = 0

=> (x - 1)² - (x² - 1²) = 0

=> x² - 2.x.1 + 1² - x² + 1 = 0

=> x² - 2x + 1 - x² + 1 = 0

=> -2x + 1 + 1 = 0

=> -2x + 2 = 0

=> -2x = -2 => x = 1

b) (2x - 1)² - (2x + 1)² = 0

=> (2x - 1 - 2x + 1)(2x - 1 + 2x + 1) = 0

=> 0 = 0(đúng)

c) 25(x + 3)² + (1 - 5x)(1 + 5x) = 8

=> 25(x² + 2.x.3 + 3²) + (1² - (5x)²) = 8

=> 25x² + 150x + 225 + 1 - 25x² = 8

=> 150x +225 + 1 = 8

=> 150x = -218

=> x = -109/75

d) 9(x + 1)² - (3x - 2)(3x + 2) = 10

=> 9(x² + 2x + 1) - [(3x)² - 2²  ] = 10

=> 9x² + 18x + 9 - (9x² - 4) = 10

=> 9x² + 18x + 9 - 9x² + 4 = 10

=> 18x + 9 + 4 = 10

=> 18x = -3

=> x = -1/6

a) \(\left(x^2-1\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-1=0\\x^2-25=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=1\\x^2=25\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=\pm5\end{cases}}\)

b) \(x^2-8x+16=0\)

\(\Leftrightarrow\left(x-4\right)^2=0\)

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)

c) \(x^3+3x^2+3x+1=0\)

\(\Leftrightarrow\left(x+1\right)^3=0\)

\(\Leftrightarrow x+1=0\)

\(\Rightarrow x=-1\)

d) \(x^3+10x^2+25x=0\)

\(\Leftrightarrow x\left(x+5\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

a) ( x² - 1 )( x² - 25 ) = 0

<=> \(\orbr{\begin{cases}x^2-1=0\\x^2-25=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=\pm5\end{cases}}\)

b) x² - 8x + 16 = 0

<=> ( x - 4 )² = 0

<=> x - 4 = 0 

<=> x = 4

c) x³ + 3x² + 3x + 1 = 0

<=> ( x + 1 )³ = 0

<=> x + 1 = 0

<=> x = -1

d) x³ + 10x² + 25x = 0

<=> x( x² + 10x + 25 ) = 0

<=> x( x + 5 )² = 0

<=> \(\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

\(x^3+17x\)

\(x.x.x+x.17\)

Vì \(17+1\)mới chia hết cho 3.

Nếu x > 3  1 đơn vị thì có x³ = 68 là nhỏ nhất

Ta có x nhỏ nhất là 4,thỏa mãn điều kiện x . 17 : 3 dư 2.

Vậy ta có \(x\in N\left|x⋮4\right|x^3+17x⋮3|x\ge4\)

a) \(\left(x+2\right)\left(x+3\right)-\left(x+1\right)\left(x+7\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-8x-7=6\)

\(\Leftrightarrow-3x=7\)

\(\Leftrightarrow x=-\frac{7}{3}\)

b) \(\left(8x-3\right)\left(3x+2\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)

\(\Leftrightarrow\left(8x-3\right)\left(9x^2+12x+4\right)-4x^2-23x-28=10x^2+3x-1-33\)

\(\Leftrightarrow72x^3+69x^2-4x-12-14x^2-26x+6=0\)

\(\Leftrightarrow72x^3+55x^2-30x-6=0\)

Nghiệm vô tỉ: \(x_1=-1,078...\) ; \(x_2=0,476...\) ; \(x_3=-0,162...\)

a) (x + 2)(x + 3) - (x + 1)(x + 7) = 6

=> x(x + 3) + 2(x + 3) - x(x + 7) - 1(x + 7) = 6

=> x² + 3x + 2x + 6 - x² - 7x - x - 7 = 6

=> x² + 5x + 6 - x² - 7x - x - 7 = 6

=> (x² - x²) + (5x - 7x - x) + (6 - 7) = 6

=> -3x - 1 = 6

=> -3x = 7

=> x = -7/3

b) (8x - 3)(3x + 2)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33

=> (8x - 3)(9x² + 12x + 4) - [4x(x + 4) + 7(x + 4)] = 2x(5x - 1) + 1(5x - 1) - 33

=> 8x(9x² + 12x + 4) - 3(9x² + 12x + 4) - (4x² + 16x + 7x + 28) = 10x² - 2x + 5x - 1 - 33

=> 72x³ + 96x² + 32x - 27x² - 36x - 12  - 4x² - 16x - 7x - 28 - 10x² + 2x - 5x + 1 + 33 = 0

=> 72x³ + (96x² - 27x² - 10x² - 4x²) + (32x - 36x  - 16x -  7x + 2x - 5x)  + (-12  - 28 + 1 +  33) = 0

=> 72x³ + 55x² - 30x - 6 = 0

=> x vô nghiệm

Ta có: \(M-N=77^2+75^2+....+1^2-\left(76^2+74^2+...+2^2\right)\)

\(=77^2+75^2+....+1^2-76^2-74^2-...-2^2\)

\(=\left(77^2-76^2\right)+\left(75^2-74^2\right)+...+\left(3^2-2^2\right)+1^2\)

\(=\left(77-76\right)\left(77+76\right)+\left(75-74\right)\left(75+74\right)+...+\left(3-2\right)\left(3+2\right)+1\)

\(=77+76+75+74+...+3+2+1\)

\(=\frac{\left[\left(77-1\right):1+1\right].\left(1+77\right)}{2}=\frac{77.78}{2}=3003\)

Thay vào S, ta có: \(S=\frac{M-N-3}{3000}=\frac{3003-3}{3000}=\frac{3000}{3000}=1\)

A = x² - 4xy + 5y² + 10x - 22y + 2044

= ( x² - 4xy + 4y² + 10x - 20y + 25 ) + ( y² - 2y + 1 ) + 2018

= [ ( x² - 4xy + 4y² ) + ( 10x - 20y ) + 25 ] + ( y - 1 )² + 2018

= [ ( x - 2y )² + 2( x - 2y ).5 + 5² ] + ( y - 1 )² + 2018

= ( x - 2y + 5 )² + ( y - 1 )² + 2018 ≥ 2018 ∀ x, y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

=> MinA = 2018 <=> x = -3 ; y = 1