1) Có 3 = (2² - 1)

=> BT = (2² - 1)(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ +1)

           = (2⁴ - 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ +1)

           = (2⁸ - 1)(2⁸ + 1)(2¹⁶ +1)

           = (2¹⁶ - 1)(2¹⁶ +1)

           = 2³² - 1

1, x³+ 6x²+11x+6

= x³ + 2x² + 4x² + 8x + 3x + 6 

= x²(x + 2) + 4x(x + 2) + 3(x + 2)

= (x + 2)(x² + 4x + 3)

2, x⁴+3x³-7x²-27x-18

= x⁴ + 3x³ - 9x² + 2x² - 27x -18

= (x⁴ - 9x²) + (3x³ - 27x) + (2x² - 18)

= x²(x² - 9) + 3x(x² - 9) + 2(x² - 9)

= (x² - 9)(x² + 3x + 2)

= (x + 3)(x - 3)(x² + 3x + 2)

3, x³-8x²+x+42

= x³ - 7x² - x² + 7x - 6x + 42

= (x³ - 7x²) - (x² - 7x) - (6x - 42)

= x²(x - 7) - x(x - 7) - 6(x - 7)

= (x - 7)(x² - x - 6) 

4, x⁴+5x³-7x²-41x-30 

= x⁴ + x³ + 4x³ - 4x² - 11x² - 11x - 30x - 30

= (x⁴ + x³) + (4x³ - 4x²) - (11x² + 11x) - (30x + 30)

= x³(x + 1) + 4x²(x + 1) - 11x(x + 1) - 30(x + 1)

= (x³ + 4x² - 11x - 30)(x + 1)

5, x⁵+x-1

= x⁵ - x⁴ + x³ + x⁴ - x³ + x² - x²+ x -1 

= x³(x² - x + 1)+ x²(x² - x + 1)- (x² - x + 1) 

= (x² - x + 1)(x³ + x² - 1)

6, x⁵-x⁴-1

= x⁵ - x³ - x² - x⁴ + x² + x + x³ - x - 1 

= x²(x³ - x - 1) - x(x³ - x - 1) + (x³ - x - 1)

= (x² - x + 1)(x³ - x - 1)

1, x 3+ 6x 2+11x+6

= x 3 + 2x 2 + 4x 2 + 8x + 3x + 6

= x 2 ﴾x + 2﴿ + 4x﴾x + 2﴿ + 3﴾x + 2﴿

= ﴾x + 2﴿﴾x 2 + 4x + 3﴿

2, x 4+3x 3‐7x 2‐27x‐18

= x 4 + 3x 3 ‐ 9x 2 + 2x 2 ‐ 27x ‐18

= ﴾x 4 ‐ 9x 2 ﴿ + ﴾3x 3 ‐ 27x﴿ + ﴾2x 2 ‐ 18﴿

= x 2 ﴾x 2 ‐ 9﴿ + 3x﴾x 2 ‐ 9﴿ + 2﴾x 2 ‐ 9﴿

= ﴾x 2 ‐ 9﴿﴾x 2 + 3x + 2﴿

=﴾x + 3﴿﴾x ‐ 3﴿﴾x 2 + 3x + 2﴿

3, x 3‐8x 2+x+42

= x 3 ‐ 7x 2 ‐ x 2 + 7x ‐ 6x + 42

= ﴾x 3 ‐ 7x 2 ﴿ ‐ ﴾x 2 ‐ 7x﴿ ‐ ﴾6x ‐ 42﴿

= x 2 ﴾x ‐ 7﴿ ‐ x﴾x ‐ 7﴿ ‐ 6﴾x ‐ 7﴿

= ﴾x ‐ 7﴿﴾x 2 ‐ x ‐ 6﴿

4, x 4+5x 3‐7x 2‐41x‐30

= x 4 + x 3 + 4x 3 ‐ 4x 2 ‐ 11x 2 ‐ 11x ‐ 30x ‐ 30

= ﴾x 4 + x 3 ﴿ + ﴾4x 3 ‐ 4x 2 ﴿ ‐ ﴾11x 2 + 11x﴿ ‐ ﴾30x + 30﴿

= x 3 ﴾x + 1﴿ + 4x 2 ﴾x + 1﴿ ‐ 11x﴾x + 1﴿ ‐ 30﴾x + 1﴿

= ﴾x 3 + 4x 2 ‐ 11x ‐ 30﴿﴾x + 1﴿

5, x 5+x‐1

= x 5 ‐ x 4 + x 3 + x 4 ‐ x 3 + x 2 ‐ x 2+ x ‐1

= x 3 ﴾x 2 ‐ x + 1﴿+ x 2 ﴾x 2 ‐ x + 1﴿‐ ﴾x 2 ‐ x + 1﴿

= ﴾x 2 ‐ x + 1﴿﴾x 3 + x 2 ‐ 1﴿ 6, x 5‐x 4‐1

= x 5 ‐ x 3 ‐ x 2 ‐ x 4 + x 2 + x + x 3 ‐ x ‐ 1

= x 2 ﴾x 3 ‐ x ‐ 1﴿ ‐ x﴾x 3 ‐ x ‐ 1﴿ + ﴾x 3 ‐ x ‐ 1﴿

= ﴾x 2 ‐ x + 1﴿﴾x 3 ‐ x ‐ 1﴿ 

\(a,=\left(x-3\right)\left(x+3\right)\\ b,=\left(2x-5\right)\left(2x+5\right)\\ c,=\left(x^3-y^3\right)\left(x^3+y^3\right)\\ =\left(x-y\right)\left(x+y\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\\ d,=\left(3x+y\right)^2\\ e,=-\left(x-3\right)^2\\ f,=\left(x+2y\right)^2\)

\(a.\\ x^2-9=\left(x-3\right)\left(x+3\right)\\ b.\\ 4x^2-25=\left(2x\right)^2-5^2\\ =\left(2x-5\right)\left(2x+5\right)\\ c.\\ x^6-y^6=\left(x^2\right)^3-\left(y^2\right)^3\\ =\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)\\ =\left(x-y\right)\left(x+y\right)\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)\\ \)

\(d.\\ 9x^2+6xy+y^2=\left(3x\right)^2+2\cdot3\cdot x\cdot y+y^2\\ =\left(3x+y\right)^2\\ e.\\ 6x-9-x^2=-\left(x^2-6x+9\right)\\=-\left(x-3\right)^2\\ f.\\ x^2+4y^2+4xy=x^2+4xy+\left(2y\right)^2=\left(x+2y\right)^2\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)

Đặt \(x^2+7x=t\)

\(\left(t+10\right)\left(t+12\right)-8=t^2+22t+120-8\)

\(=t^2+22t+112=\left(t+8\right)\left(t+14\right)\)

Theo cách đặt \(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

CHAO BAN

1+1bang may

\(-3x^2+4x-2020\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{2020}{3}\right)\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{4}{9}+\frac{6056}{9}\right)\)

\(=-3\left[\left(x-\frac{2}{3}\right)^2+\frac{6056}{9}\right]\)

\(=-3\left(x-\frac{2}{3}\right)^2-\frac{6056}{3}\ge-\frac{6056}{3}\)

(Dấu "=" \(\Leftrightarrow x-\frac{2}{3}=0\Leftrightarrow x=\frac{2}{3}\))

trả lời :

bài đâu bn

^HT^

@Lê Mạnh Hùng đây nhé

sorry mk ngủ đây

a, Những công nhân đang làm việc

=> Những công nhân_CN// đang làm việc_VN 

b, Rất nhiều sen nở trong đầm 

=> Rất nhiều sen_CN// nở trong đầm_VN  

C1

a) -7x(3x-2)=-21x^2+14x

b) 87^2+26.87+13^2=87^2+2.13.87+13^2=(87+13)^2=100^2

C2

a) (x-5)(x+5)

b)3x(x+5)-2(x+5)=(3x-2)(x+5)=0

\(\Rightarrow\left[\begin{array}{nghiempt}3x-2=0\\x+5=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-5\end{array}\right.\)

Vậy S={-5;2/3}

C3:

a)3x^3-2x^2+2=(x+1)(3x^2-5x-5)-3

b) Để A chia hết cho B=> x+1\(\inƯ\left(-3\right)\)

\(\Rightarrow\begin{cases}x+1=3\\x+1=-3\\x+1=1\\x+1=-1\end{cases}\)\(\Rightarrow\begin{cases}x=2\\x=-4\\x=0\\x=-2\end{cases}\)

a) 3x3-2x2+2 chia x+1= 3x2-5x+5 dư -3 b) -3 chia hết x+1 vậy chon x =2

1)

a) \(-7x\left(3x-2\right)\)

\(=-21x^2+14x\)

b) \(87^2+26.87+13^2\)

\(=87^2+2.87.13+13^2\)

\(=\left(87+13\right)^2\)

\(=100^2\)

\(=10000\)

2)

a) \(x^2-25\)

\(=x^2-5^2\)

\(=\left(x-5\right)\left(x+5\right)\)

b) \(3x\left(x+5\right)-2x-10=0\)

\(\Leftrightarrow3x\left(x+5\right)-\left(2x-10\right)=0\)

\(\Leftrightarrow3x\left(x+5\right)-2\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy..........

3)

a) \(A:B=\left(3x^3-2x^2+2\right):\left(x+1\right)\)

Vậy \(\left(3x^3-2x^2+2\right):\left(x+1\right)=\left(3x^2-5x-5\right)+7\)

b)

Để \(A⋮B\Rightarrow7⋮\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)\in U\left(7\right)=\left\{-1;1-7;7\right\}\)

Vì x là số nguyên nên x=0 ; x=6 thì \(A⋮B\)