a: Xét ΔMDA và ΔMCB có

MD=MC

\(\hat{MDA}=\hat{MCB}\) (ABCD là hình thang cân)

DA=CB

Do đó: ΔMDA=ΔMCB

=>MA=MB

b: Sửa đề: Chứng minh MN⊥DC

Ta có: ΔMAB cân tại M

mà MN là đường trung tuyến

nên MN⊥AB

Ta có; MN⊥AB

AB//DC

Do đó: MN⊥DC

1, \(x^2\) - 9 = 0

 (\(x\) - 3)(\(x\) + 3) = 0

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

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

 vậy \(x\) \(\in\) {-3; 3}

5, 4\(x^2\) - 36 = 0

    4.(\(x^2\) - 9) = 0

       \(x^2\) - 9 = 0

       (\(x\) - 3)(\(x\) + 3) = 0

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

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

Vậy \(x\) \(\in\) {-3; 3}

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

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

=-x+1

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

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

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

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

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

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

\(=2x^2-1\)

7: \(\left(1-2x\right)\left(5-3x\right)+\left(4-x\right)^2\)

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

\(=6x^2-10x-3x+5+x^2-8x+16\)

\(=7x^2-21x+21\)

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

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

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

11: \(\left(x+4\right)^2+\left(x+5\right)\left(x-5\right)-2x\left(x+1\right)\)

\(=x_{}^2+8x+16+x^2-25-2x^2-2x\)

=6x-9

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

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

\(=5x^2-18x+1-2x^2+18=3x^2-18x+19\)

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

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

=4x+2

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

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

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

6: (x-3)(x+3)-x(x+23)

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

=-23x-9

8: (x-2)(x+2)-(x-3)(x+1)

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

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

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

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

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

\(=x^2+4x+14-x^2+9=4x+23\)

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

\(=x^2-2x+1-\left(x^2-16\right)+x^2+6x+9\)

\(=2x^2+4x+10-x^2+16=x^2+4x+26\)

2.My apartment is cheaper than hers

A.cheap than B.The more hard C.hardert D.The hardest

3.Her school is....from home than mine

A.father   B.More far    C. Farther   D.farer

4.Tom is.....than any students is his class

A.handsome  B.the more handsome C. more handsome D.the most handsome

5.He did the exam worse than I did

A.as bad as  B.badder  C.more bably than  D.wores than

6.travelling by car is.... than travelling by plane

A.slower  B .slowest  C.more slow  D. more slower

7.My new bed is....than the old one

A.more comfortable  B.comfotbably C.more comfortable  D.comfortable

8.My sister plays the piano....than me

A.gooder  B.weller C.better  D.more good

9.this road is.... than that road,so we can't drive here

A.narrower   B.narrow   C.the most mot narrow  D.more narrower

10.Antony drives...his brother

A.more careful  B.more carefully   C.more carefully than   D. as careful as

11.It is...in january than in november

A.the colder  B.the coldest  C.colaest  D.colder

12.she is....than other students in my class

A.most hard-working  B.more hard-working  C.the most hard-working

D.as hard-working

13.No other planet in the solar system is...than jupiter

A.the biggest   B.the bigger  C.bigger   D.giger

#\(yGLinh\)

1.A

A

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