Ôn tập chương III

a)

\(\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\\ x\left(\dfrac{1}{2}+\dfrac{1}{8}\right)=\dfrac{3}{4}\\ x\cdot\dfrac{5}{8}=\dfrac{3}{4}\\ x=\dfrac{3}{4}:\dfrac{5}{8}=\dfrac{6}{5}\)

b)

\(\left(2x-4,5\right):\dfrac{3}{4}-\dfrac{1}{3}=1\\\left(2x-\dfrac{9}{2}\right):\dfrac{3}{4}-\dfrac{1}{3}=1\\ \left(2x-\dfrac{9}{2}\right):\dfrac{3}{4}=1+\dfrac{1}{3}=\dfrac{4}{3}\\ \left(2x-\dfrac{9}{2}\right)=\dfrac{4}{3}\cdot\dfrac{3}{4}=1\\ 2x=1+\dfrac{9}{2}=\dfrac{11}{2}\\ x=\dfrac{11}{2}:2=\dfrac{11}{4}\)

a) 1/2 . x + 1/8 . x = 3/4

<=> ( 1/2 + 1/8 ) . x = 3/4

<=> ( 4/8 + 1/8 ) . x = 3/4

<=> 5/8 . x = 3/4

<=> x = 3/4 . 8/5

<=> x = 6/5

Vậy x = 6/5

b) ( 2x - 4,5 ) : 3/4 - 1/3 = 1

<=> ( 2x - 45/10 ) : 3/4 - 1/3 = 1

<=> ( 2x - 9/2 ) : 3/4 - 1/3 = 1

<=> ( 2x - 9/2 ) : 3/4 = 1 + 1/3

<=> ( 2x - 9/2 ) : 3/4 = 4/3

<=> 2x - 9/2 = 4/3 . 3/4

<=> 2x - 9/2 = 1

<=> 2x = 1 + 9/2

<=> 2x = 11/2

<=> x = 11/2 . 1/2

<=> x = 11/4

Vậy x = 11/4

a: \(=\dfrac{-7}{25}\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}=\dfrac{-7}{25}-\dfrac{18}{25}=-1\)

c: \(=\dfrac{5}{7}\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)=0\)

c: \(=\dfrac{27}{5}\cdot\dfrac{30}{7}+\dfrac{40}{7}\cdot\dfrac{37}{5}\)

\(=\dfrac{27\cdot30+40\cdot37}{35}=\dfrac{458}{7}\)

d: \(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}\cdot\dfrac{12}{5}-\dfrac{1}{4}\)

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

\(\Leftrightarrow x\left(-\dfrac{7}{2}-\dfrac{5}{4}\right)=\dfrac{1}{4}\cdot\dfrac{-4}{3}+\dfrac{1}{6}+\dfrac{3}{4}\)

\(\Leftrightarrow x\cdot\dfrac{-19}{4}=\dfrac{7}{12}\)

hay x=-7/57

Gọi \(d=ƯCLN\left(2n+1;3n+2\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2n+1⋮d\\3n+2⋮d\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6n+3⋮d\\6n+4⋮d\end{matrix}\right.\)

\(\Leftrightarrow1⋮d\)

\(\Leftrightarrow d=1\)

\(\LeftrightarrowƯCLN\left(2n+1;3n+2\right)=1\)

\(\Leftrightarrow\dfrac{2n+1}{3n+2}\) toois giản

Gọi d là ƯCLN(2n + 1;3n + 2)

=> 2n + 1 chia hết cho d => 3(2n + 1) = 6n + 3 chia hết cho d

3n + 2 chia hết cho d => 2(3n + 1) = 6n + 2 chia hết cho d

=> [(6n + 3) - (6n + 2)] chia hết cho d

=> 1 chia hết cho d

=> \(d=\left[{}\begin{matrix}1\\-1\end{matrix}\right.\)

Vậy phân số \(\dfrac{2n+1}{3n+1}\) là phân số tối giản với mọi n thuộc Z

Gọi ƯCLN(2n + 1; 3n + 2) là d.

\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\3n+2⋮d\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}6n+3⋮d\\6n+4⋮d\end{matrix}\right.\)

\(\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\)

\(\Rightarrow6n+4-6n-3⋮d\)

\(\Rightarrow1⋮d\)

\(\Rightarrow d\inƯ\left(1\right)=\left\{1\right\}\)

Vì d = 1 hay ƯCLN(2n + 1; 3n + 2) = 1

\(\Rightarrow\dfrac{2n+1}{3n+2}\) là phân số tối giản.

x - [ 1 - x + (-x) -( 3 - x ) ] = 2[ x - 2( x - 1 )]

=> x - (1 - 2x - 3 + x)= 2.(x - 2x + 2)

=> x - (-x - 2) = 2.(2 - x)

=> 2x + 2= 2.(2 - x)

=> 2. (x + 1) = 2.(2-x)

=> x+1 = 2-x

=> 2 - 2x= 1

=> 2x = 1

=> x = 1/2

Vậy...

\(\Leftrightarrow-3\cdot\left(x+x-1\right)-\left[-x+3-x\right]=5-\left[x\right]\)

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

=>-6x+3+2x-3=5-x

=>-4x+x=5

=>-3x=5

hay x=-5/3

b) sai

phải là b) \(\dfrac{1-n}{4-n}\)

Người đó đi từ A đến B mất:

7 giờ 30 phút - 6 giờ = 1 giờ 30 phút

Đổi 1 giờ 30 phút = 1,5 giờ

Quãng đường AB dài số ki-lô-mét là:

12 . 1,5 = 18 ( km )

Vận tốc của người ấy lúc về là:

18 : \(\dfrac{2}{3}\)= 27 ( km/h )

Đ/S: 27 km/h