Gọi N là trung điểm của HD

Xét ΔHDC có

M,N lần lượt là trung điểm của HC,HD

=>MN là đường trung bình của ΔHDC

=>MN//DC và \(MN=\frac{DC}{2}\)

MN//DC

DC⊥ AD

Do đó: MN⊥AD

Xét ΔAMD có

MN,DH là các đường cao

MN cắt DH tại N

Do đó: N là trực tâm của ΔAMD

=>AN⊥DM

MN//DC

DC//AB

Do đó: MN//AB

\(MN=\frac{DC}{2}\)

\(AB=\frac{DC}{2}\)

Do đó: MN=AB

Xét tứ giác ABMN có

AB//MN

AB=MN

Do đó: ABMN là hình bình hành

=>AN//BM

mà AN⊥DM

nên BM⊥MD

=>\(\hat{BMD}=90^0\)

Bài 1;

1; Đặt \(A=x\left(x+1\right)-\left(x+2\right)^2\)

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

=-3x-4

Khi x=1 thì \(A=-3\cdot1-4=-3-4=-7\)

2: Đặt B=(5x+2)(2x-7)-(2x+5)(2x-5)

\(=10x^2-35x+4x-14-\left(4x^2-25\right)\)

\(=10x^2-31x-14-4x^2+25=6x^2-31x+11\)

Khi x=-3 thì \(B=6\cdot\left(-3\right)^2-31\cdot\left(-3\right)+11\)

=54+93+11

=65+93

=158

3: Đặt \(C=\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)\)

\(=x^2-4x+4-\left(x^2-9\right)\)

\(=x^2-4x+4-x^2+9=-4x+13\)

Khi x=2 thì \(C=-4\cdot2+13=-8+13=5\)

BÀi 2:

1: \(\left(x+7\right)^2-x\left(x-3\right)=12\)

=>\(x^2+14x+49-x^2+3x=12\)

=>17x=12-49=-37

=>\(x=-\frac{37}{17}\)

2: \(\left(2x+3\right)^2-4x^2=10\)

=>\(4x^2+12x+9-4x^2=10\)

=>12x=10-9=1

=>\(x=\frac{1}{12}\)

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

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

=>\(x^2+4x+4-x^2+x+2=3\)

=>5x+5=3

=>5x=-2

=>\(x=-\frac25\)

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

=>\(4x^2+12x+9-4\left(x^2-2x+1\right)=2\)

=>\(4x^2+12x+9-4x^2+8x-4=2\)

=>20x+5=2

=>20x=-3

=>\(x=-\frac{3}{20}\)

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

=>\(x^2-4x+4-\left(x^2+4x+3\right)=-7\)

=>\(x^2-4x+4-x^2-4x-3=-7\)

=>-8x+1=-7

=>-8x=-8

=>x=1

6: \(\left(x+1\right)^2-\left(x-2\right)\left(x+2\right)=0\)

=>\(x^2+2x+1-\left(x^2-4\right)=0\)

=>\(x^2+2x+1-x^2+4=0\)

=>2x=-5

=>x=-5/2

7: \(\left(2x+3\right)^2-4\left(x-1\right)^2=16\)

=>\(4x^2+12x+9-4\left(x^2-2x+1\right)=16\)

=>\(4x^2+12x+9-4x^2+8x-4=16\)

=>20x+5=16

=>20x=11

=>\(x=\frac{11}{20}\)

8: \(\left(x-2\right)^2-x\left(x-5\right)-3=0\)

=>\(x^2-4x+4-x^2+5x-3=0\)

=>x+1=0

=>x=-1

9: \(\left(3x-2\right)^2-\left(x+9\right)\left(3x+2\right)=0\)

=>\(9x^2-12x+4-3x^2-2x-27x-18=0\)

=>\(6x^2-41x-14=0\)

\(\Delta=\left(-41\right)^2-4\cdot6\cdot\left(-14\right)=2017>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{41-\sqrt{2017}}{2\cdot6}=\frac{41-\sqrt{2017}}{12}\\ x=\frac{41+\sqrt{2017}}{2\cdot6}=\frac{41+\sqrt{2017}}{12}\end{array}\right.\)

X=0 y=3

x=0 y=3

BÀi 4:

1: \(x\left(1-x\right)+\left(x-1\right)^2\)

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

=-x+1

2: \(\left(x-3\right)^2-x^2+10x-7\)

\(=x^2-6x+9-x^2+10x-7\)

=4x+2

3: \(\left(x+2\right)^2-\left(x-3\right)\left(x+1\right)\)

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

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

4: (x+4)(x-2)-\(\left(x-3\right)^2\)

\(=x^2-2x+4x-8-\left(x^2-6x+9\right)\)

\(=x^2+2x-8-x^2+6x-9=8x-17\)

5: \(\left(x-2\right)^2+\left(x-1\right)\left(x+5\right)\)

\(=x^2-4x+4+x^2+4x-5=2x^2-1\)

6: \(\left(x+3\right)\left(x-3\right)-x\left(x+23\right)\)

\(=x^2-9-x^2-23x\)

=-23x-9

Bài 3:

1: \(\left(2x+1\right)^2+\left(2x-1\right)^2\)

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

2: \(-\left(x+1\right)^2-\left(x-1\right)^2\)

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

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

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

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

\(=4y\cdot2x=8xy\)

4: \(\left(3x+y\right)^2+\left(x-y\right)^2\)

\(=9x^2+6xy+y^2+x^2-2xy+y^2=10x^2+4xy+2y^2\)

5: \(-\left(x+5\right)^2-\left(x-3\right)^2\)

\(=-\left(x^2+10x+25\right)-\left(x^2-6x+9\right)\)

\(=-x^2-10x-25-x^2+6x-9=-2x^2-4x-34\)

6: \(\left(3x-2\right)^2-\left(3x-1\right)^2\)

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

7: \(\left(x-4y\right)^2+\left(x+4y\right)^2\)

\(=x^2-8xy+16y^2+x^2+8xy+16y^2=2x^2+32y^2\)

8: \(-\left(-2x+3\right)^2-\left(5x-3\right)^2\)

\(=-\left(4x^2-12x+9\right)-\left(25x^2-30x+9\right)\)

\(=-4x^2+12x-9-25x^2+30x-9=-29x^2+42x-18\)

9: \(\left(-2x+3\right)^2-\left(5x-3\right)^2\)

\(=4x^2-12x+9-\left(25x^2-30x+9\right)\)

\(=4x_{}^2-12x+9-25x^2+30x-9=-21x^2+18x\)

10: \(\left(2x+1\right)^2+\left(-3x-1\right)^2\)

\(=4x^2+4x+1+9x^2+6x+1=13x^2+10x+2\)

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

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

\(=-x^2+2xy-y^2-4x^2-4xy-4y^2=-5x^2-2xy-5y^2\)

12: \(-\left(x+1\right)^2+\left(x-1\right)^2\)

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

13: \(\left(2x+7\right)^2+\left(-2x-3\right)^2\)

\(=4x^2+28x+49+4x_{}^2-12x+9=8x^2+16x+58\)

14: \(-\left(2x-y\right)^2-\left(x+3y\right)^2\)

\(=-\left(4x^2-4xy+y^2\right)-\left(x^2+6xy+9y^2\right)\)

\(=-4x^2+4xy-y^2-x^2-6xy-9y^2=-5x^2-2xy-10y^2\)

15: \(-\left(2x+7\right)^2+\left(-2x-3\right)^2\)

\(=-4x^2-28x-49+4x^2+12x+9=-16x-40\)

Bài 1:

1: \(\left(x+1\right)^2=x^2+2x+1\)

2: \(\left(4+x\right)^2=\left(x+4\right)^2=x^2+8x+16\)

3: \(\left(5x+1\right)^2=\left(5x\right)^2+2\cdot5x\cdot1+1^2=25x^2+10x+1\)

4: \(\left(2x+3\right)^2=\left(2x\right)^2+2\cdot2x\cdot3+3^2=4x^2+12x+9\)

5: \(\left(x+2y\right)^2=x^2+2\cdot x\cdot2y+\left(2y\right)^2=x^2+4xy+4y^2\)

6: \(\left(x+5y\right)^2=x^2+2\cdot x\cdot5y+\left(5y\right)^2=x^2+10xy+25y^2\)

7: \(\left(3x+5y\right)^2=\left(3x\right)^2+2\cdot3x\cdot5y+\left(5y\right)^2=9x^2+30xy+25y^2\)

8: \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)

9: \(\left(x^2+9\right)^2=\left(x^2\right)^2+2\cdot x^2\cdot9+9^2=x^4+18x^2+81\)

10: \(\left(2x^2+1\right)^2=\left(2x^2\right)^2+2\cdot2x^2\cdot1+1^2=4x^4+4x^2+1\)

11: \(\left(x+2y^2\right)^2=x^2+2\cdot x\cdot2y^2+\left(2y^2\right)^2=x^2+4xy^2+4y^4\)

12: \(\left(2x+3y^2\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y^2+\left(3y^2\right)^2\)

\(=4x^2+12xy^2+9y^4\)

Bài 2:

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

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

3: \(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

4: \(9-25x^2=3^2-\left(5x\right)^2=\left(3-5x\right)\left(3+5x\right)\)

5: \(4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)

6: \(9x^2-36=9\left(x^2-4\right)=9\left(x-2\right)\left(x+2\right)\)

7: \(\left(3x\right)^2-y^2=\left(3x-y\right)\left(3x+y\right)\)

8: \(x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)

9: \(\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)

10: \(\left(3x\right)^2-9y^4=9x^2-9y^4=9\left(x^2-y^4\right)=9\left(x-y^2\right)\left(x+y^2\right)\)

11: \(16x^2-\left(y^2\right)^2=\left(4x\right)^2-\left(y^2\right)^2=\left(4x-y^2\right)\left(4x+y^2\right)\)

12: \(x^4-\left(3y^2\right)^2=\left(x^2\right)^2-\left(3y^2\right)^2=\left(x^2-3y^2\right)\left(x^2+3y^2\right)\)

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

14: (x-5)(x+5)

\(=x^2-5^2=x^2-25\)

15: \(\left(x-6\right)\left(6+x\right)=x^2-6^2=x^2-36\)

16: (2x+1)(2x-1)=\(\left(2x\right)^2-1^2=4x^2-1\)

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

18: (5x-3y)(3y+5x)

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

\(=\left(5x\right)^2-\left(3y\right)^2=25x^2-9y^2\)

19: \(\left(\frac{1}{x}-5\right)\left(\frac{1}{x}+5\right)=\left(\frac{1}{x}\right)^2-5^2=\frac{1}{x^2}-25\)

20: \(\left(x-\frac32\right)\left(x+\frac32\right)=x^2-\left(\frac32\right)^2=x^2-\frac94\)

21: \(\left(\frac{x}{3}-\frac{y}{4}\right)\left(\frac{x}{3}+\frac{y}{4}\right)=\left(\frac{x}{3}\right)^2-\left(\frac{y}{4}\right)^2=\frac{x^2}{9}-\frac{y^2}{16}\)

22: \(\left(\frac{x}{y}-\frac23\right)\left(\frac{x}{y}+\frac23\right)=\left(\frac{x}{y}\right)^2-\left(\frac23\right)^2=\frac{x^2}{y^2}-\frac49\)

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

24: \(\left(2x-\frac23\right)\left(\frac23+2x\right)=\left(2x\right)^2-\left(\frac23\right)^2=4x^2-\frac49\)

25: \(\left(2x+\frac35\right)\left(\frac35-2x\right)=\left(\frac35+2x\right)\left(\frac35-2x\right)=\left(\frac35\right)^2-\left(2x\right)^2=\frac{9}{25}-4x^2\)

26: \(\left(\frac12x-\frac43\right)\left(\frac43+\frac12x\right)=\left(\frac12x\right)^2-\left(\frac43\right)^2=\frac14x^2-\frac{16}{9}\)

27: \(\left(\frac23x^2-\frac{y}{2}\right)\left(\frac23x^2+\frac{y}{2}\right)=\left(\frac23x^2\right)^2-\left(\frac{y}{2}\right)^2=\frac49x^4-\frac{y^2}{4}\)

28: \(\left(3x-y^2\right)\left(3x+y^2\right)=\left(3x\right)^2-\left(y^2\right)^2=9x^2-y^4\)

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

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

Bài 1:

1, \((x+1)^2 = x^2 + 2x + 1\)

2, \((4+x)^2 = 16 + 8x + x^2\)

3, \((5x+1)^2 = 25x^2 + 10x + 1\)

4, \((2x+3)^2 = 4x^2 + 12x + 9\)

5, \((x+2y)^2 = x^2 + 4xy + 4y^2\)

6, \((x+5y)^2 = x^2 + 10xy + 25y^2\)

7, \((3x+5y)^2 = 9x^2 + 30xy + 25y^2\)

8, \((2x+3y)^2 = 4x^2 + 12xy + 9y^2\)

9, \((x^2+9)^2 = x^4 + 18x^2 + 81\)

10, \((2x^2+1)^2 = 4x^4 + 4x^2 + 1\)

11, \((x+2y^2)^2 = x^2 + 4xy^2 + 4y^4\)

12, \(x^4 - (3y^2)^2 = (x^2-3y^2)(x^2+3y^2)\)

Bài 2:

Từ câu 1 đến câu 12 (Phân tích thành tích \((A-B)(A+B)\) )

1, \(x^2 - 4 = (x-2)(x+2)\)

2, \(1 - 4x^2 = (1-2x)(1+2x)\)

3, \(4x^2 - 9 = (2x-3)(2x+3)\)

4, \(9 - 25x^2 = (3-5x)(3+5x)\)

5, \(4x^2 - 25 = (2x-5)(2x+5)\)

6, \(9x^2 - 36 = (3x-6)(3x+6) = 9(x-2)(x+2)\)

7, \((3x)^2 - y^2 = (3x-y)(3x+y)\)

8, \(x^2 - (2y)^2 = (x-2y)(x+2y)\)

9, \((2x)^2 - y^2 = (2x-y)(2x+y)\)

10, \((3x)^2 - 9y^4 = (3x-3y^2)(3x+3y^2)\)

11, \(16x^2 - (y^2)^2 = (4x-y^2)(4x+y^2)\)

12, \(x^4 - (3y^2)^2 = (x^2-3y^2)(x^2+3y^2)\)

Từ câu 13 đến câu 30 (Triển khai thành hiệu \(A^2 - B^2\) ):

13, \((x-1)(x+1) = x^2 - 1\)

14, \((x-5)(x+5) = x^2 - 25\)

15, \((x-6)(6+x) = x^2 - 36\)

16, \((2x+1)(2x-1) = 4x^2 - 1\)

17, \((x-2y)(2y+x) = x^2 - 4y^2\)

18, \((5x-3y)(3y+5x) = 25x^2 - 9y^2\)

19, \(\left(\frac{1}{x}-5\right)\left(\frac{1}{x}+5\right) = \frac{1}{x^2} - 25\)

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

21, \(\left(\frac{x}{3}-\frac{y}{4}\right)\left(\frac{x}{3}+\frac{y}{4}\right) = \frac{x^2}{9} - \frac{y^2}{16}\)

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

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

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

25, \(\left(2x+\frac{3}{5}\right)\left(\frac{3}{5}-2x\right) = \frac{9}{25} - 4x^2\)

26, \(\left(\frac{1}{2}x-\frac{4}{3}\right)\left(\frac{4}{3}+\frac{1}{2}x\right) = \frac{1}{4}x^2 - \frac{16}{9}\)

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

28, \((3x-y^2)(3x+y^2) = 9x^2 - y^4\)

29, \((x^2-2y)(x^2+2y) = x^4 - 4y^2\)

30, \((x^2-y^2)(x^2+y^2) = x^4 - y^4\)

`x^2 -2x = 24`

` x^2 -2x - 24 = 0`

`x^2 + 4x - 6x - 24 = 0 `

`x(x+4) - 6(x+4)= 0 `

`(x+4) (x-6) = 0`

TH`1`
`x+4 = 0 => x=-4 `
TH`2`
`x-6 = 0 => x=6 `
Vậy....

x^2-2x=24

x^2-2x-24

△=b^2-4ac=(-2)^2-4*1*(-24)=100

⇒△>0 hay pt trên có hai nghiệm phân biệt x1,x2

x1=(-b+√△)/2a=(-(-2)+√100 )/2*1=6

x2=(-b-√△)/2a=(-(-2)-√100 )/2*1=-4

vậy x1=6 ; x2=-4

mình đánh máy bạn thông cảm =))

\(x^2-2x=24\)

\(\Rightarrow x^2-2x+1=24+1\)

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

\(\Rightarrow\left(x-1\right)^2=5^2\)

\(\Rightarrow\left[\begin{array}{l}x-1=5\\ x-1=-5\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{l}x=6\\ x=-4\end{array}\right.\)

Vậy \(x\in\left\lbrace6;-4\right\rbrace\)

\(199^2=\left(200-1\right)^2\)

\(=200^2-2\cdot200\cdot1+1^2\)

=40000-400+1

=39601

1\(199^2=\left(200-1\right)^2\) khai triển HĐT số 2

=\(200^2-2\times200\times1+1^2\)

=\(40000-400+1\)

=\(39601\) ok chưa bạn???

`a)` vì `ΔODC` cân tại `O`

`=> OD = OC`

Xét `ΔODC` có :

`OB/OD = OA/OC ( OD = OC , OB = OA)`

`=> AB // DC` ( định lý thales đảo)

`b)` Ta có :\(\begin{cases}OB+BD=OD\\ OA+AC=OC\end{cases}\Rightarrow\begin{cases}BD=OD-OB\\ AC=OC-OA\end{cases}\)

Mà `OD=OC ; OB=OA`

`=> DB=AC`

lại có : \(\begin{cases}\hat{OBA}+\hat{DBA}=180^0\\ \hat{OAB}+\hat{CAB}=180^0\end{cases}\) (kể bù)

`=>`\(\begin{cases}\hat{DBA}=180^0-\hat{OBA}\\ \hat{CAB}=180^0-\hat{OAB}\end{cases}\)

Mà \(\hat{OBA}=\hat{OAB}\) ( Vì `ΔOBA` cân tại `O`)

`=>`\(\hat{DBA}=\hat{CAB}\)

Xét `ΔDBA` và `ΔCAB` có :

\(\begin{cases}DB=AC\left(\operatorname{cm}t\right)\\ \hat{DBA}=\hat{CAB}\left(\operatorname{cm}t\right)\\ AB\ch ung\end{cases}\)

`=> ΔDBA=ΔCAB`(c-g-c)`

Vậy ...