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Bài 1: tìm lỗi sai

1. She goes to visit her grandmother and grandfather next week

=> will go

2. Which place is Phuong and Mai going to visit firts?

=> are

3. They are going in vacation this summer

=> on

4. He wants visit the temple and take some photos

=> to visit

5. They always are going to have a picnic in the park on the weekend

=>  \(\varnothing\)

6. It is on of the higher mountains in the world

=> highest

7. This city has a population for 6.3 milion

=> of

8. She always go to Hong Kong on summer vacation

=> goes

9. The River Nile is 6,437 kilometers length

=> long

10. Tomorrow, I am going to visiting the Tower of LonDon

=> am going to visit

a) \(\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)

 0,5 x + 0,6 ( x - 2 ) = 3

0,5 x + 0.6 x - 1,2 = 3

1,1 x = 4,2

x = \(\frac{42}{11}\)

Kết luận: 

b) \(\frac{1}{3}x-0,5x=0,75\)

\(\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)

\(-\frac{1}{6}x=\frac{3}{4}\)

\(x=-\frac{9}{2}\)

Kết luận: 

c) \(\frac{3}{-2}x-0,5x=75\%\)

-1,5x - 0,5x = 0,75 

-2x = 0,75 

x = -0,375

Kết luận: 

d) \(-\frac{2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)

-0,4 x + 0,25 = 0,75 - 0,75 x 

-0,4 x + 0,75 x = 0,75 - 0,25

0,35 x = 0,5 

x = \(\frac{10}{7}\)

Kết luận: 

\(a,\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)

\(\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)

\(\left(\frac{1}{2}+\frac{3}{5}\right)x-\frac{6}{5}=3\)

\(\frac{11}{10}x-\frac{6}{5}=3\)

\(\frac{11}{10}x=\frac{21}{5}\)

\(x=\frac{42}{11}\)

\(b,\frac{1}{3}x-\frac{1}{2}x=\frac{3}{4}\)

\(\left(\frac{1}{3}-\frac{1}{2}\right)x=\frac{3}{4}\)

\(\frac{-1}{6}x=\frac{3}{4}\)

\(x=\frac{-9}{2}\)

\(c,\frac{3}{-2}x-\frac{1}{2}x=75\%\)

\(\left(\frac{3}{-2}-\frac{1}{2}\right)x=\frac{3}{4}\)

\(-2x=\frac{3}{4}\)

\(x=\frac{-3}{8}\)

\(\frac{-2}{5}x+\frac{1}{4}=75\%-\frac{3}{4}x\)

\(\frac{-2}{5}x+\frac{3}{4}x=\frac{3}{4}-\frac{1}{4}\)

\(\left(\frac{-2}{5}+\frac{3}{4}\right)x=\frac{1}{2}\)

\(\frac{7}{20}x=\frac{1}{2}\)

\(x=\frac{10}{7}\)

\(A=\frac{23n+1}{n-2}=\frac{23n-46+46+1}{n-2}=\frac{23\left(n-2\right)+47}{n-2}=23+\frac{47}{n-2}\)

A là số nguyên <=> \(\frac{47}{n-2}\) là số nguyên <=> \(47⋮n-2\) hay \(n-2\inƯ\left(47\right)=\left\{-47;-1;1;47\right\}\)

<=> \(n\in\left\{-45;1;3;49\right\}\)

Kết luận:...

\(A=\frac{23n+1}{n-2}=\frac{23\left(n-2\right)+47}{n-2}=23+\frac{47}{n-2}\)

A nguyên <=> \(\frac{47}{n-2}\)nguyên

=> \(47⋮n-2\)=> \(n-2\inƯ\left(47\right)=\left\{\pm1;\pm47\right\}\)