\(\text{Bài 1:a)}25\dfrac{3}{19}.\left(-\dfrac{4}{5}\right)-35\dfrac{3}{19}.\left(-\dfrac{4}{5}\right)\)

         \(=\dfrac{478}{19}.\left(-\dfrac{4}{5}\right)-\dfrac{668}{19}.\left(-\dfrac{4}{5}\right)\)

         \(=\left(-\dfrac{4}{5}\right).\left(\dfrac{478}{19}-\dfrac{668}{19}\right)\)

         \(=\left(-\dfrac{4}{5}\right).\left(\dfrac{-190}{19}\right)\)

         \(=\left(-\dfrac{4}{5}\right).\left(-10\right)=8\)

\(\text{b)}5:\left(-\dfrac{5}{2}\right)^2+\dfrac{2}{15}.\sqrt{\dfrac{9}{4}}-\left(-2021\right)^0+0,25\)

\(=5:\dfrac{25}{4}+\dfrac{2}{15}.\dfrac{3}{2}-1+\dfrac{1}{4}\)

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

\(=1-1+\dfrac{1}{4}\)

\(=0+\dfrac{1}{4}=\dfrac{1}{4}\)

\(\text{Bài 2:a)}\dfrac{8}{5}-\dfrac{3}{5}:x=0,4\)

                    \(\dfrac{3}{5}:x=\dfrac{8}{5}-0,4=\dfrac{6}{5}\)

                          \(x=\dfrac{3}{5}.\dfrac{5}{6}=\dfrac{1}{2}\)

\(\text{b)}\left(3x-\dfrac{1}{2}\right)^2+\dfrac{21}{25}=1\)

   \(\left(3x-\dfrac{1}{2}\right)^2\)           \(=1-\dfrac{21}{25}=\dfrac{4}{25}=\pm\left(\dfrac{2}{5}\right)^2\)

\(\text{Vậy }3x-\dfrac{1}{2}=\dfrac{2}{5}\)

       \(3x\)         \(=\dfrac{2}{5}+\dfrac{1}{2}=\dfrac{9}{10}\)

        \(x\)          \(=\dfrac{9}{10}.\dfrac{1}{3}=\dfrac{3}{10}\)

\(\text{hoặc }3x-\dfrac{1}{2}=\dfrac{-2}{5}\)

        \(3x\)         \(=\left(\dfrac{-2}{5}\right)+\dfrac{1}{2}=\dfrac{1}{10}\)

          \(x\)         \(=\dfrac{1}{10}.\dfrac{1}{3}=\dfrac{1}{30}\)

\(\Rightarrow x\in\left\{\dfrac{3}{10};\dfrac{1}{30}\right\}\)

Bài 2: 

a: =>3/5:x=6/5

hay x=3/5:6/5=1/2

b: \(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^2=\dfrac{4}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{1}{2}=\dfrac{2}{5}\\3x-\dfrac{1}{2}=-\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=\dfrac{1}{30}\end{matrix}\right.\)

Bài 4: 

a: Xét ΔABI và ΔACI có 

AB=AC

AI chung

BI=CI

Do đó: ΔABI=ΔACI

b: Xét tứ giác ABDC có 

I là trung điểm của BC

I là trung điểm của AD

Do đó: ABDC là hình bình hành

Suy ra: AB=CD

Độ dãn của lò xo:

\(F_{đh}=k\cdot\Delta l\Rightarrow\Delta l=\dfrac{F_{đh}}{k}=\dfrac{2}{100}=0,02\)m=2cm

B ... Thì lo xo dài 35cm tính f2

b)Lò xo dài 35cm.

   \(\Rightarrow\) Độ dãn của lò xo: \(\Delta l'=l'-l_0=35-30=5cm=0,05m\)

   \(F_2=k\cdot\Delta l'=100\cdot0,05=5N\)

hai bài câu a mik lm đc r nhe mn lm giúp mik câu b thôi ạ mik ko bt lm;-;

Bài 3:

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

Dấu \("="\Leftrightarrow x=1\)

\(b,=-2\left(x^2+2\cdot\dfrac{1}{4}x+\dfrac{1}{16}\right)+\dfrac{9}{8}=-2\left(x+\dfrac{1}{4}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\)

Dấu \("="\Leftrightarrow x=-\dfrac{1}{4}\)

Bài 4:

\(a,=\left(x^2+2\cdot\dfrac{5}{2}x+\dfrac{25}{4}\right)-\dfrac{21}{4}=\left(x+\dfrac{5}{2}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{5}{2}\)

\(b,=\left(x^2-8x+16\right)+1=\left(x-4\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow x=4\)

bài 1

1 an

2 any

3 any

4 some

5 any

6 any

7 some

8 some

9 any

10 some

11 any

12 sone

13 some

14 any

bài 2

1 how many

2 how much

3 how much

4 how many

5 how many

6 how much

7 how much

8 how much

9 how many

10 how many

11 how many

12 how many

bài 3

how many eggs are there ?

how much tomato juice is there ?

how many packet of pasta is there ?

how many red peper are there ?

how many beans are there ?

how many pizza is there ?

how much salt is there ?

bài 4

1 are

2 aren't

3 a

4 an

5 any

6 isn't

7 many

8 is

9 some

10 much

11 any

12 some

bài 4

mình làm những cái đếm dcd còn mấy cái còn lại là ko đếm dcd 

2 C

5 C

6 C

7 c

8 c

10 c

11 c

14c

15c

16c

18c

20c

22c

23c

24c

25c

26c 

Câu 4: 

a: Xét ΔABD và ΔAED có 

AB=AE

\(\widehat{BAD}=\widehat{EAD}\)

AD chung

Do đó: ΔABD=ΔAED

Câu 1:

\(a,=\dfrac{1}{2}+9\cdot\dfrac{1}{9}-18=\dfrac{1}{2}+1-18=-\dfrac{33}{2}\\ b,=2-1+4\cdot\dfrac{1}{4}+9\cdot\dfrac{1}{9}\cdot9=1+1+9=11\\ c,=-21,3\left(54,6+45,4\right)=-21,3\cdot100=-2130\\ d,B=\left(\dfrac{1}{16}+\dfrac{1}{2}-\dfrac{1}{16}\right):\left(\dfrac{1}{8}-\dfrac{1}{8}+1\right)=\dfrac{1}{2}:1=\dfrac{1}{2}\)

đề khó nhìn quá

I.

1A

2C

3A

4C

5B

6B

7C

8B

9B

10C

II. 

1T

2F

3F

4T

5F

6T

7F

8T

9F

10F