a: 3/5=9/15

7/3=35/15

b: 9/10=36/40

5/4=50/40

c: 2/7=8/28

5/4=35/28

d: 8/9=16/18

11/6=33/18

e: 6/5=48/40

13/8=65/40

g: 4/5=8/10

3/2=15/10

h: 5/12=10/24

7/8=21/24

i: 11/4=33/12

13/6=26/12

\(\dfrac{2}{3};\dfrac{4}{6}\)

Cíu tui cái coi 

4 phần 6

\(\dfrac{7}{13}\)

\(\dfrac{7}{13}\)

\(\dfrac{7}{13}\)

1.There is a small pond in front of Lam's house.

2.You can play games in the afternoon but you must do your homework in the evening.

3.Does Lan walk or ride a bike to school?

4.When it is hot,We often go swimming.

5.What is there in front of your house?

6.Where is your father sitting now?

7.My class starts at seven in the morning.

8.I do not often go swimming with my friends.

Bài 2:

\(a,\dfrac{2\times6\times13}{5\times13\times6}=\dfrac{2}{5}\\ b,\dfrac{8\times11\times13}{11\times13\times16}=\dfrac{8}{16}=\dfrac{8}{8\times2}=\dfrac{1}{2}\\ c,\dfrac{5\times18\times12}{15\times24\times16}=\dfrac{5\times3\times3\times6\times3\times4}{3\times5\times4\times6\times4\times4}=\dfrac{3\times3}{4\times4}=\dfrac{9}{16}\\ d,\dfrac{15\times8\times4}{32\times3\times5}=\dfrac{5\times3\times8\times4}{4\times8\times3\times5}=\dfrac{1}{1}=1\)

Bài 1:

\(a,\dfrac{8}{12}=\dfrac{8:4}{12:4}=\dfrac{2}{3}\\ \dfrac{20}{48}=\dfrac{20:4}{48:4}=\dfrac{5}{12}\\ \dfrac{74}{100}=\dfrac{74:2}{100:2}=\dfrac{37}{50}\\ \dfrac{18}{21}=\dfrac{18:3}{21:3}=\dfrac{6}{7}\\ \dfrac{40}{54}=\dfrac{40:2}{54:2}=\dfrac{20}{27}\\ \dfrac{30}{45}=\dfrac{30:15}{45:3}=\dfrac{2}{3}\\ b,\dfrac{18}{24}=\dfrac{18:6}{24:6}=\dfrac{3}{4}\\ \dfrac{49}{28}=\dfrac{49:7}{28:7}=\dfrac{7}{4}\\ \dfrac{15}{35}=\dfrac{15:5}{35:5}=\dfrac{3}{7}\\ \dfrac{28}{40}=\dfrac{28:4}{40:4}=\dfrac{7}{10}\\ \dfrac{36}{27}=\dfrac{36:9}{27:9}=\dfrac{4}{3}\\ \dfrac{56}{49}=\dfrac{56:7}{49:7}=\dfrac{8}{7}\)

\(c,\dfrac{85}{51}=\dfrac{85:17}{51:17}=\dfrac{5}{3}\\ \dfrac{1515}{2323}=\dfrac{1515:101}{2323:101}=\dfrac{15}{23}\\ \dfrac{39}{26}=\dfrac{39:13}{26:13}=\dfrac{3}{2}\\ \dfrac{323232}{515151}=\dfrac{323232:10101}{515151:10101}=\dfrac{32}{51}\)

Sau 1 giờ vòi chảy được:

1/3+1/5+1/6=10/30+6/30+5/30=7/10(bể)

Sau 1 giờ thì bể còn:

1-7/10=3/10(bể không chứa nước)

1: Xét ΔMNP vuông tại M có MH là đường cao

nên MH^2=HN*HP; MN^2=NH*NP; PM^2=PH*PN

=>MH=căn 3,6*6,4=4,8cm; MN=căn 3,6*10=6cm; PM=căn 6,4*10=8cm

2: MK=8/2=4cm

Xét ΔMNK vuông tại M có tan MNK=MK/MN=4/6=2/3

nên \(\widehat{MNK}\simeq33^041'\)

3: ΔMNK vuông tại M có MF là đường cao

nên NF*NK=NM^2

ΔMNP vuông tại M có MH là đường cao

nên NH*NP=NM^2

=>NF*NK=NH*NP

Chọn đáp án B

câu 2 chọn C

a) \(\sqrt[]{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)

\(\Leftrightarrow\sqrt[]{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+9}=-\left(x^2+2x+1\right)+5\)

\(\Leftrightarrow\sqrt[]{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=-\left(x+1\right)^2+5\left(1\right)\)

Ta có :

\(\left\{{}\begin{matrix}\sqrt[]{3\left(x+1\right)^2+4}\ge2,\forall x\in R\\\sqrt[]{5\left(x+1\right)^2+9}\ge3,\forall x\in R\end{matrix}\right.\)

\(\Rightarrow VT=\sqrt[]{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\ge5,\forall x\in R\)

\(VP=-\left(x+1\right)^2+5\le5,\forall x\in R\)

Dấu "=" xảy ra thì \(VT=VP=5\)

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

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy nghiệm của phương trình đã cho là \(x=-1\)