a: Hiệu vận tốc hai xe là:

45:3=15(km/h)

Hiệu số phần bằng nhau là 2-1=1(phần)

Vận tốc của ô tô đi từ A là:

15:1x2=30(km/h)

Vận tốc của ô tô đi từ B là:

30-15=15(km/h)

b: Độ dài quãng đường BC là:

15x3=45(km)

a: Xét ΔAHB và ΔAHC có

AH chung

HB=HC

AB=AC

Do đó: ΔAHB=ΔAHC

b: ΔAHB=ΔAHC

=>\(\widehat{AHB}=\widehat{AHC}\)

mà \(\widehat{AHB}+\widehat{AHC}=180^0\)(hai góc kề bù)

nên \(\widehat{AHB}=\widehat{AHC}=\dfrac{180^0}{2}=90^0\)

=>AH\(\perp\)BC

c: H là trung điểm của BC

=>\(HB=HC=\dfrac{BC}{2}=3\left(cm\right)\)

ΔAHB vuông tại H

=>\(HA^2+HB^2=AB^2\)

=>\(HA=\sqrt{5^2-3^2}=4\left(cm\right)\)

d: ΔAHB=ΔAHC

=>\(\widehat{HAB}=\widehat{HAC}\)

Xét ΔAEH vuông tại E và ΔAKH vuông tại K có

AH chung

\(\widehat{EAH}=\widehat{KAH}\)

Do đó: ΔAEH=ΔAKH

=>HE=HK

e: ΔAEH=ΔAKH

=>AE=AK

Xét ΔABC có \(\dfrac{AE}{AB}=\dfrac{AK}{AC}\)

nên EK//BC

1. They are speaking English now.

2.The sun is shining

3. It is raining right at the moment.

4. The wind is blowing right now.

5. Is she decorating the room now?

6. He is looking at the Christmas tree.

7. Is Mr. picke doing his homework?

8. She and her friend are swimming in the river.

9. We are watching television right now?

10. Are they playing in the garden now?

1. They/ speak/be/English/now.

They are speaking English now

2. Shine/ the sun/be.

The sun is shining

3. Be/it/rain/ right at the moment.

It is raining right at the moment

4. Blow/ the wind/ right now/be.

The wind is blowing right now

5. Decorate/she/be/now/ the/room?

Is she decorating the room now?

Bài 3: Gọi H là giao điểm của CD với AB

\(\widehat{HCB}+\widehat{DCB}=180^0\)(hai góc kề bù)

=>\(\widehat{HCB}+143^0=180^0\)

=>\(\widehat{HCB}=180^0-143^0=37^0\)

Xét ΔHCB có \(\widehat{HCB}+\widehat{HBC}=37^0+53^0=90^0\)

nên ΔHCB vuông tại H

=>CD\(\perp\)AB tại H

Bài 2:

a: Ta có: \(\widehat{DAB}=\widehat{xAM}\)(hai góc đối đỉnh)

mà \(\widehat{xAm}=124^0\)

nên \(\widehat{DAB}=124^0\)

Ta có: \(\widehat{DAB}+\widehat{ABC}=124^0+56^0=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên AD//BC

=>xy//zt

b: xy//zt

=>\(\widehat{BCD}+\widehat{ADC}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{BCD}+90^0=180^0\)

=>\(\widehat{BCD}=90^0\)

Ak là phân giác của góc DAB

=>\(\widehat{DAC}=\dfrac{124^0}{2}=62^0\)

ΔDAC vuông tại D

=>\(\widehat{DAC}+\widehat{DCA}=90^0\)

=>\(\widehat{DCA}+62^0=90^0\)

=>\(\widehat{DCA}=28^0\)

\(1,a)\dfrac{15}{12}-\dfrac{-1}{4}\\ =\dfrac{15}{12}+\dfrac{1}{2}\\ =\dfrac{15}{12}+\dfrac{6}{12}\\ =\dfrac{21}{12}=\dfrac{7}{4}\\ b)-\dfrac{5}{12}+0,75\\ =-\dfrac{5}{12}+\dfrac{3}{4}\\ =\dfrac{-5}{12}+\dfrac{9}{12}\\ =\dfrac{4}{12}=\dfrac{1}{3}\\ c)\dfrac{15}{12}+\dfrac{5}{13}-\left(\dfrac{3}{12}+\dfrac{18}{13}\right)\\ =\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}\\ =\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\\ =\dfrac{12}{12}-\dfrac{13}{13}\\ =1-1=0\)

2: a: \(-\dfrac{16}{42}-\dfrac{5}{8}=\dfrac{-64}{168}-\dfrac{105}{168}=\dfrac{-169}{168}\)

b: \(3,5-\left(-\dfrac{2}{7}\right)=3,5+\dfrac{2}{7}=\dfrac{7}{2}+\dfrac{2}{7}=\dfrac{7^2+2^2}{14}=\dfrac{53}{14}\)

c: \(\left(-\dfrac{1}{2}+\dfrac{3}{4}\right)-\left(-\dfrac{4}{5}+\dfrac{5}{6}\right)\)

\(=\dfrac{-1}{2}+\dfrac{3}{4}+\dfrac{4}{5}-\dfrac{5}{6}\)

\(=\dfrac{-30}{60}+\dfrac{45}{60}+\dfrac{48}{60}-\dfrac{50}{60}\)

\(=\dfrac{15}{60}-\dfrac{2}{60}=\dfrac{13}{60}\)

3:

a: \(\dfrac{2}{21}-\dfrac{-1}{28}=\dfrac{2}{21}+\dfrac{1}{28}=\dfrac{8}{84}+\dfrac{3}{84}=\dfrac{11}{84}\)

b: \(-4.75-1\dfrac{7}{12}=-\dfrac{57}{12}-\dfrac{19}{12}=-\dfrac{76}{12}=-\dfrac{19}{3}\)

c: \(-\left(\dfrac{3}{5}+\dfrac{5}{4}\right)-\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\)

\(=-\dfrac{3}{5}-\dfrac{5}{4}+\dfrac{3}{4}-\dfrac{2}{5}\)

\(=-1-\dfrac{2}{4}=-\dfrac{3}{2}\)

4:

a: \(-\dfrac{2}{33}+\dfrac{5}{55}=\dfrac{-10}{165}+\dfrac{15}{165}=\dfrac{5}{165}=\dfrac{1}{33}\)

b: \(0,4+\left(-2\dfrac{4}{5}\right)=0,4-2,8=-2,4\)

c: \(-\left(\dfrac{3}{7}+\dfrac{3}{8}\right)-\left(-\dfrac{3}{8}+\dfrac{4}{7}\right)\)

\(=\dfrac{-3}{7}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{7}\)

\(=-\dfrac{3}{7}-\dfrac{4}{7}=-\dfrac{7}{7}=-1\)

\(10x^2+y^2+4z^2+6x-4y-4xz=-5\\ =>10x^2+y^2+4z^2+6x-4y-4xz+5=0\\ =>\left(9x^2+6x+1\right)+\left(x^2-4xz+4z^2\right)+\left(y^2-4y+4\right)=0\\ =>\left(3x+1\right)^2+\left(x-2z\right)^2+\left(y-2\right)^2=0\)

Mà: \(\left\{{}\begin{matrix}\left(3x+1\right)^2\ge0\forall x\\\left(x-2z\right)^2\ge0\forall x,z\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.=>\left(3x+1\right)^2+\left(x-2z\right)^2+\left(y-2\right)^2\ge0\forall x,y,z\) 

\(=>\left\{{}\begin{matrix}3x+1=0\\x-2z=0\\y-2=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\z=-\dfrac{1}{6}\\y=2\end{matrix}\right.\)

\(10x^2+y^2+4z^2+6x-4y-4xz=-5\\ \Leftrightarrow\left(x^2-4xz+4z^2\right)+\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)=0\\ \Leftrightarrow\left(x-2z\right)^2+\left(3x+1\right)^2+\left(y-2\right)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2z\right)^2\ge0\forall x,z\\\left(3x+1\right)^2\ge0\forall x\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2z\right)^2+\left(3x+1\right)^2+\left(y-2\right)^2\ge0\forall x,y,z\)

Mà: \(\left(x-2z\right)^2+\left(3x+1\right)^2+\left(y-2\right)^2=0\)

Do đó: \(\left\{{}\begin{matrix}x-2z=0\\3x+1=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=2\\z=-\dfrac{1}{6}\end{matrix}\right.\)

#$\mathtt{Toru}$

Bài 3: Gọi H là giao điểm của CD với AB

\(\widehat{HCB}+\widehat{DCB}=180^0\)(hai góc kề bù)

=>\(\widehat{HCB}+143^0=180^0\)

=>\(\widehat{HCB}=180^0-143^0=37^0\)

Xét ΔHCB có \(\widehat{HCB}+\widehat{HBC}=37^0+53^0=90^0\)

nên ΔHCB vuông tại H

=>CD\(\perp\)AB tại H

Bài 2:

a: Ta có: \(\widehat{DAB}=\widehat{xAM}\)(hai góc đối đỉnh)

mà \(\widehat{xAm}=124^0\)

nên \(\widehat{DAB}=124^0\)

Ta có: \(\widehat{DAB}+\widehat{ABC}=124^0+56^0=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên AD//BC

=>xy//zt

b: xy//zt

=>\(\widehat{BCD}+\widehat{ADC}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{BCD}+90^0=180^0\)

=>\(\widehat{BCD}=90^0\)

Ak là phân giác của góc DAB

=>\(\widehat{DAC}=\dfrac{124^0}{2}=62^0\)

ΔDAC vuông tại D

=>\(\widehat{DAC}+\widehat{DCA}=90^0\)

=>\(\widehat{DCA}+62^0=90^0\)

=>\(\widehat{DCA}=28^0\)

c;     C = \(\dfrac{28^{28}+28^{24}+...+28^4+1}{28^{30}+28^{28}+...+28^2+1}\)

        A =         1 + 28⁴ + 28⁸ + 28¹² + ...28²⁸

  28⁴A = 28⁴ + 28⁸ + 28¹² + ... + 28²⁸ + 28³²

28⁴A - A = 28⁴+ 28⁸+...+28²⁸+ 28³²- (1 + 28⁴ + 28⁸+...+28²⁸)

(28⁴ - 1)A = 28⁴ + 28⁸+ ...+ 28²⁸ + 28³² - 1 - 28⁴- ...- 28²⁸

(28⁴ - 1)A = (28³² - 1) + (28⁴ - 28⁴) + (28⁸ - 28⁸) + ... + (28²⁸ - 28²⁸)

(28⁴ - 1)A = 28³² - 1 + 0 + 0... + 0

            A = (28³² - 1): (28⁴ - 1)

  Đặt B = 28³⁰ + 28²⁸ + ... + 28² + 1

  28²B = 28³² + 28³⁰ + ... + 28⁴ + 28²

28²B - B = 28³² + 28³⁰ + ... + 28⁴ + 28² - (28³⁰ + 28²⁸ +...+1)

(28² - 1)B = 28³² + 28³⁰+...+28⁴ + 28² - 28³⁰ - 28²⁸ - ... 28²- 1

(28² - 1)B = (28³² - 1) + (28³⁰ - 28³⁰) +...+(28² - 28²)

(28² - 1)B = (28³² - 1) + 0 + 0 +...+ 0

(28² - 1)B = 28³² - 1 

             B = (28³² - 1): (28² - 1)

C = \(\dfrac{A}{B}\) = \(\dfrac{28^{32}-1}{28^4-1}\) : \(\dfrac{28^{32}-1}{28^2-1}\)

C = \(\dfrac{28^{32}-1}{28^4-1}\) \(\times\) \(\dfrac{28^2-1}{28^{32}-1}\)

C = \(\dfrac{28^2-1}{28^4-1}\)

C = \(\dfrac{1}{785}\) 

                Câu d:

 \(\dfrac{x-1}{99}\) + \(\dfrac{x-2}{98}\) + \(\dfrac{x-3}{97}\) = \(\dfrac{x-4}{96}\) + \(\dfrac{x-5}{95}\) + \(\dfrac{x-6}{94}\)

(\(\dfrac{x-1}{99}\)-1)+(\(\dfrac{x-2}{98}\)-1)+(\(\dfrac{x-3}{97}\)-1) = (\(\dfrac{x-4}{96}\)-1) + (\(\dfrac{x-5}{95}\)-1)+(\(\dfrac{x-6}{94}\)-1)

\(\dfrac{x-100}{99}\)+\(\dfrac{x-100}{98}\)+\(\dfrac{x-100}{97}\) = \(\dfrac{x-100}{96}\)+\(\dfrac{x-100}{95}\)+\(\dfrac{x-100}{94}\)

\(\dfrac{x-100}{99}\)+\(\dfrac{x-100}{98}\)+\(\dfrac{x-100}{97}\)- \(\dfrac{x-100}{96}\)-\(\dfrac{x-100}{95}\)-\(\dfrac{x-100}{94}\) = 0

(\(x-100\)).(\(\dfrac{1}{99}\)+\(\dfrac{1}{98}\)+\(\dfrac{1}{97}\) - \(\dfrac{1}{96}\)-\(\dfrac{1}{95}\)-\(\dfrac{1}{94}\)) = 0

Vì\(\dfrac{1}{98}< \dfrac{1}{98}< \dfrac{1}{97}< \dfrac{1}{96}< \dfrac{1}{95}< \dfrac{1}{94}\)

Nên (\(\dfrac{1}{99}\) + \(\dfrac{1}{98}\) + \(\dfrac{1}{97}\) )- (\(\dfrac{1}{96}\) + \(\dfrac{1}{95}\) +\(\dfrac{1}{94}\) )< 0 

⇒\(x-100\) = 0

Vậy \(x\) = 100

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

=>\(2x^2-3x-6+2x^2+1=0\)

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

\(\text{Δ}=\left(-3\right)^2-4\cdot4\cdot\left(-5\right)=9+80=89>0\)

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

\(\left[{}\begin{matrix}x=\dfrac{3-\sqrt{89}}{2\cdot4}=\dfrac{3-\sqrt{89}}{8}\\x=\dfrac{3+\sqrt{89}}{2\cdot4}=\dfrac{3+\sqrt{89}}{8}\end{matrix}\right.\)

a) \(\lim\limits_{ }\left(\sqrt{n^2-n+1}-n\right)\)

\(=\lim\limits_{ }\left[\dfrac{\left(\sqrt{n^2-n+1}-n\right)\left(\sqrt{n^2-n+1}+n\right)}{\sqrt{n^2-n+1}+n}\right]\)

\(=\lim\limits_{ }\left(\dfrac{1-n}{\sqrt{n^2-n+1}+n}\right)\)

\(=\lim\limits_{ }\left(\dfrac{\dfrac{1}{n}-1}{\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\right)\)

\(=-\dfrac{1}{2}\)

b) \(\lim\limits_{ }\left(\dfrac{-3}{4n^2-2n+1}\right)=0\)

c) \(\lim\limits_{ }\dfrac{n^2+n+5}{2n+1}=+\infty\)

d) \(\lim\limits_{ }\left(\sqrt{n^2-1}-\sqrt{3n^2+2}\right)\)

\(=\lim\limits_{ }\left(\dfrac{-2n^2-3}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)\)

\(\lim\limits_{ }\left(\dfrac{-2n-\dfrac{3}{n}}{\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}}\right)\)

\(=-\infty\)

a: \(lim\left(\sqrt{n^2-n+1}-n\right)\)

\(=\lim\limits\dfrac{n^2-n+1-n^2}{\sqrt{n^2-n+1}+n}=\lim\limits\dfrac{-n+1}{\sqrt{n^2-n+1}+n}\)

\(=\lim\limits\dfrac{-1+\dfrac{1}{n}}{\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}+1}=\dfrac{-1+0}{\sqrt{1-0+0}+1}=\dfrac{-1}{2}\)

b: \(\lim\limits\dfrac{-3}{4n^2-2n+1}\)

\(=\lim\limits\dfrac{-\dfrac{3}{n^2}}{4-\dfrac{2}{n}+\dfrac{1}{n^2}}=\dfrac{0}{4-0+0}=0\)

c: \(\lim\limits\dfrac{n^2+n+5}{2n+1}=\lim\limits\dfrac{n^2\left(1+\dfrac{1}{n}+\dfrac{5}{n^2}\right)}{n\left(2+\dfrac{1}{n}\right)}\)

\(=\lim\limits\dfrac{n\left(1+\dfrac{1}{n}+\dfrac{5}{n^2}\right)}{2+\dfrac{1}{n}}=+\infty\)

d: \(\lim\limits\left(\sqrt{n^2-1}-\sqrt{3n^2+2}\right)\)

\(=\lim\limits\left(\dfrac{n^2-1-3n^2-2}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)=\lim\limits\left(\dfrac{-2n^2-3}{\sqrt{n^2-1}+\sqrt{3n^2+2}}\right)\)

\(=\lim\limits\left(\dfrac{n^2\left(-2-\dfrac{3}{n^2}\right)}{n\cdot\left(\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}\right)}\right)\)

\(=\lim\limits\left(\dfrac{n\left(-2-\dfrac{3}{n^2}\right)}{\sqrt{1-\dfrac{1}{n^2}}+\sqrt{3+\dfrac{2}{n^2}}}\right)=+\infty\)