Yêu cầu đề là gì vậy bạn?

phân tích đa thức thành phân tử nha                                                        quên mất không ghi đề =))

\(C=16x^2-8x+2024\)

\(\Rightarrow C=16x^2-8x+1+2023\)

\(\Rightarrow C=\left(4x-1\right)^2+2023\ge2023\left(\left(4x-1\right)^2\ge0\right)\)

\(\Rightarrow Min\left(C\right)=2023\)

\(D=-25x^2+50x-2023\)

\(\Rightarrow D=-\left(25x^2-50x+25\right)-1998\)

\(\Rightarrow D=-\left(5x-5\right)^2-1998\le1998\left(-\left(5x-5\right)^2\le0\right)\)

\(\Rightarrow Max\left(D\right)=1998\)

\(B=-x^2+20x+100=-\left(x^2-20x+100\right)+200=-\left(x-10\right)^2+200\le200\left(-\left(x-10\right)^2\le0\right)\)

\(\Rightarrow Max\left(B\right)=200\)

\(E=\left(2x-1\right)^2-\left(3x+2\right)\left(x-5\right)\)

\(\Rightarrow E=4x^2-4x+1-\left(3x^2-13x-10\right)\)

\(\Rightarrow E=4x^2-4x+1-3x^2+13x+10\)

\(\Rightarrow E=x^2+9x+11=x^2+9x+\dfrac{81}{4}-\dfrac{81}{4}+11\)

\(\Rightarrow E=\left(x+\dfrac{9}{2}\right)^2-\dfrac{37}{4}\ge-\dfrac{37}{4}\left(\left(x+\dfrac{9}{2}\right)^2\ge0\right)\)

\(\Rightarrow Min\left(E\right)=-\dfrac{37}{4}\)

\(F=\left(3x-5\right)^2-\left(3x+2\right)\left(4x-1\right)\)

\(\Rightarrow F=9x^2-30x+25-\left(12x^2+3x-2\right)\)

\(\Rightarrow F=-3x^2-33x+27=-3\left(x^2-10x+9\right)\)

\(\Rightarrow F=-3\left(x^2-10x+25\right)+48=-3\left(x-5\right)^2+48\le48\left(-3\left(x-5\right)^2\le0\right)\)

\(\Rightarrow Max\left(F\right)=48\)

Phân tích đa thức:

x^4 + 2x^3 - x^2 - 2x + 1

= (x^4 + 2x^3) - (x^2 + 2x) + 1

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

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

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

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

Vậy phương trình đã cho có các nghiệm là x = -2, x = -1, x = 0 và x = 1.

(x + 2y)² - (x - y)²

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

= 3y(2x + y)

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

\(=\left[\left(x+2y\right)-\left(x-y\right)\right]\left[\left(x+2y\right)+\left(x-y\right)\right]\)

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

\(=3y\left(2x+y\right)\)

Bài 2

a) 3x(x - 1) - 3(x - 1) = 0

(x - 1)(3x - 3) = 0

3(x - 1)(x - 1) = 0

3(x - 1)² = 0

x - 1 = 0

x = 1

b) x² - x = 0

x(x - 1) = 0

x = 0 hoặc x - 1 = 0

*) x - 1 = 0

x = 1

Vậy x = 0; x = 1

c) 25x² - 100x = 0

25x(x - 4) = 0

25x = 0 hoặc x - 4 = 0

*) 25x = 0

x = 0

*) x - 4 = 0

x = 4

Vậy x = 0; x = 4

d) (2x - 1)² - 64 = 0

(2x - 1 - 8)(2x - 1 + 8) = 0

(2x - 9)(2x + 7) = 0

*) 2x - 9 = 0

2x = 9

x = 9/2

*) 2x + 7 = 0

2x = -7

x = -7/2

Vậy x = -7/2; x = 9/2

giúp tui với mng ơi tui đang cần gấp ạaaa

Bài 4

1) x⁴ - 9x³ + x² - 9x

= x(x³ - 9x² + x - 9)

= x[(x³ - 9x²) + (x - 9)]

= x[x²(x - 9) + (x - 9)]

= x(x - 9)(x² + 1)

2) 3x² + 5y - 3xy - 5x

= (3x² - 3xy) - (5x - 5y)

= 3x(x - y) - 5(x - y)

= (x - y)(3x - 5)

3) xy + z + y + xz

= (xy + xz) + (y + z)

= x(y + z) + (y + z)

= (y + z)(x + 1)

4) x² + y - xy - x

= (x² - xy) - (x - y)

= x(x - y) - (x - y)

= (x - y)(x - 1) 

Bài 5

1) x² + 2x - 8

= x² - 2x + 4x - 8

= (x² - 2x) + (4x - 8)

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

= (x - 2)(x + 4)

2) x² + 5x + 6

= x² + 2x + 3x + 6

= (x² + 2x) + (3x + 6)

= x(x + 2) + 3(x + 2)

= (x + 2)(x + 3)

3) 4x² - 12x + 8

= 4(x² - 3x + 2)

= 4(x² - x - 2x + 2)

= 4[(x² - x) - (2x - 2)]

= 4[x(x - 1) - 2(x - 1)]

= 4(x - 1)(x - 2)

4) x² + 6x + 8

= x² + 2x + 4x + 8

= (x² + 2x) + (4x + 8)

= x(x + 2) + 4(x + 2)

= (x + 2)(x + 4)

5) x² + 10x + 16

= x² + 2x + 8x + 16

= (x² + 2x) + (8x + 16)

= x(x + 2) + 8(x + 2)

= (x + 2)(x + 8)

6) (x + 1)² - 4

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

= (x - 1)(x + 3)

7) 9 - 6x + x² - y²

= (9 - 6x + x²) - y²

= (3 - x)² - y²

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

8) x² + 3x + 2

= x² + x + 2x + 2

= (x² + x) + (2x + 2)

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

= (x + 1)(x + 2)

9) x² - x - 2

= x² + x - 2x - 2

= (x² + x) - (2x + 2)

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

= (x + 1)(x - 2)

10) x² - x - 6

= x² + 2x - 3x - 6

= (x² + 2x) - (3x + 6)

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

= (x + 2)(x - 3)

cô làm rồi em ơi https://olm.vn/cau-hoi/bai-3-tu-giac-abcd-co-goc-c-goc-d-90-do-chung-minh-rang-ac2-bd-ab2cd2.8140260328277

\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)

\(=\dfrac{bxz-cxy}{ax}=\dfrac{cyx-ayz}{by}=\dfrac{azy-bxz}{cz}\)

\(=\dfrac{bxz-cxy+cyx-ayz+azy-bxz}{ax+by+cz}=0\)

\(\Rightarrow bz-cy=0\Rightarrow bz=cy\Rightarrow\dfrac{y}{b}=\dfrac{z}{c}\)

Tương tự...

\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\left(dpcm\right)\)