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

b: \(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

c: \(x^8+x^4+1\)

\(=\left(x^8+2x^4+1\right)-x^4\)

\(=\left(x^4-x^2+1\right)\cdot\left(x^4+x^2+1\right)\)

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

a)\(x^4+4\\ =\left(x^2\right)^2+4x^2+4-4x^2\\ =\left[\left(x^2\right)^2+4x^2+4\right]-\left(2x\right)^2\\ =\left(x^2+2\right)^2-\left(2x\right)^2\\ =\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)

\(a)\; x^4+4 \\= x^4+4x^2+4-4x^2\\=(x^2+2)^2-4x^2\\=(x^2+2-2x)(x^2+2+2x)\)

Đáp án đúng: c

b. Ta có:

A(x) + B(x) = x2 + 2x + 1 + x2 + 1 = 2x2 + 2x + 2 (0.5 điểm)

A(x) - B(x) = x2 + 2x + 1 - (x2 + 1) = 2x (0.5 điểm)

Thu gọn, sắp xếp đa thức theo lũy thừa giảm của biến:

* Ta có: f(x) = x5 – 3x2 + x3 – x2 – 2x + 5

= x5 – (3x2 + x2 ) + x3 - 2x + 5

= x5 – 4x2 + x3 – 2x + 5

= x5 + x3 – 4x2 – 2x + 5

Và g(x) = x2 – 3x + 1 + x2 – x4 + x5

= (x2 + x2 ) – 3x + 1 – x4 + x5

= 2x2 – 3x + 1 – x4 + x5

= x5 – x4 + 2x2 – 3x + 1

* f(x) + g(x):

`@` `\text {Ans}`

`\downarrow`

`1,`

`a)`

`5 \times 72 \times 10 \times 2`

`= 5 \times 2 \times 10 \times 72`

`= 10 \times 10 \times 72`

`= 100 \times 72`

`= 7200`

`b)`

`40 \times 125`

`= 4 \times 10 \times 25 \times 5`

`= (5 \times 10) \times (4 \times 25)`

`= 50 \times 100`

`= 5000`

`c)`

`4 \times 2021 \times 25`

`= (4 \times 25) \times 2021`

`= 100 \times 2021`

`= 202100`

`d)`

`16 \times 6 \times 25`

`= 4 \times 4 \times 6 \times 25`

`= (4 \times 25) \times 4 \times 6`

`= 100 \times 24`

`= 2400`

`2,`

`a)`

`24 \times 57 + 43 \times 24`

`= 24 \times (57+43)`

`= 24 \times 100`

`= 2400`

`b)`

`12 \times 19 + 80 \times 12 +12`

`= 12 \times (19 + 80 + 1)`

`= 12 \times 100`

`= 1200`

`c)`

`(36 \times 15 \times 169) \div (5 \times 18 \times 13)`

`= 36 \times 15 \times 169 \div 5 \div 18 \div 13`

`= 6 \times 6 \times 3 \times 5 \times 13 \times 13 \div 5 \div 3 \times 6 \div 13`

`= (6 \div 6) \times (3 \div 3) \times (5 \div 5) \times (13 \div 13) \times 6 \times 13`

`= 6 \times 13`

`= 78`

`d)`

`(44 \times 52 \times 60) \div ( 11 \times 13 \times 15)`

`= 44 \times 52 \times 60 \div 11 \div 13 \div 15`

`= 4 \times 11 \times 13 \times 4 \times 15 \times 4 \div 11 \div 13 \div 15`

`= (11 \div 11) \times (13 \div 13) \times (15 \div 15) \times 4 \times 4 \times`

`= 4 \times 4 \times 4`

`= 64`

`3,`

`a)`

`x - 280 \div 35 = 5 \times 54`

`x - 8 = 270`

`x = 270 + 8`

`x = 278`

`b)`

`(x - 280) \div 35 = 54 \div 4`

`(x - 280) \div 35 = 13,5`

`x - 280 = 13,5 \times 35`

`x - 280 = 472,5`

`x = 472,5 + 280`

`x = 752,5`

`c)`

`(x - 128 + 20) \div 192 = 0`

`x - 128 + 20 = 0 \times 192`

`x - 128 + 20 = 0`

`x - 108 = 0`

`x = 0 + 108`

`x = 108`

`d)`

`4 \times (x + 200) = 460 + 85 \times 4`

`4 \times (x+200) = 460 + 340`

`4 \times (x+200) = 800`

`x + 200 = 800 \div 4`

`x + 200 = 200`

`x = 200 - 200`

`x = 0`

`4,`

`a)`

`7/12 - 5/12`

`= (7 - 5)/12`

`= 2/12`

`= 1/6`

`b)`

`8/11 + 19/11`

`= (8+19)/11`

`= 27/11`

`c)`

`3/8 + 5/12`

`= 9/24 + 10/24`

`= 19/24`

`d)`

`3/4 + 7/12`

`= 9/12 + 7/12`

`= 16/12`

`= 4/3`

`5,`

`a)`

`x - 6/7 = 5/2`

`x = 5/2 + 6/7`

`x = 47/14`

`b)`

`12/7 \div x + 2/3 = 7/5`

`12/7 \div x = 7/5 - 2/3`

`12/7 \div x = 11/15`

`x = 12/7 \div 11/15`

`x = 180/77`

`@` `\text {Kaizuu lv uuu}`

a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)

=> x=-1  

với \(3x^2+x-2=0\)

ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)

Vậy  ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)

b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)

\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)

\(\Leftrightarrow3x^2=3\)

hay \(x\in\left\{1;-1\right\}\)

c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)

hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)

Ta có: f(x) + g(x) – h(x)

      = (x5 – 4x3 + x2 – 2x + 1) + (x5 – 2x4 + x2 – 5x + 3) – (x4 – 3x2 + 2x – 5)

      = x5 – 4x3 + x2 – 2x + 1 + x5 – 2x4 + x2 – 5x + 3 – x4 + 3x2 - 2x + 5

      = (x5 +x5) – (2x4 + x4) – 4x3 + (x2 + x2 + 3x2)- (2x + 5x + 2x) + (1 + 3 + 5)

      = (1 + 1)x5 – (2 + 1)x4 – 4x3 + (1 + 1 + 3)x2 - (2 + 5 + 2)x + (1 + 3 + 5)

      = 2x5 – 3x4 – 4x3 + 5x2 – 9x + 9