\(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)

\(=x+\dfrac{\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}}{-\left(\dfrac{3}{10}-\dfrac{9}{16}+\dfrac{15}{22}\right)}\)

\(=x+\dfrac{\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}}{-\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}\right)}\)

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

\(=x+\dfrac{-2}{3}\)

Với \(x=-\dfrac{1}{3}\), ta được:

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

B = -1

\(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)

\(=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)

\(=x+\dfrac{-\dfrac{7}{40}+\dfrac{5}{11}}{-\dfrac{369}{880}}\)

\(=x+\dfrac{\dfrac{123}{440}}{\dfrac{369}{880}}\)

\(=x-\dfrac{2}{3}\)

Thay \(x=-\dfrac{1}{3}\) vào biểu thức B.

Ta có: \(-\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{3}{3}=-1\)

Vậy giá trị biểu thức B tại \(x=-\dfrac{1}{3}\) là -1.

\(B=\dfrac{\dfrac{2}{10}-\dfrac{3}{8}+\dfrac{5}{11}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)

\(B=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{10}{22}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)

\(B=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}\)\(-\dfrac{1}{3}\)

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

6)a) \(\left|\dfrac{5}{3}:x\right|=\left|\dfrac{-1}{6}\right|\)

⇒ \(\left|\dfrac{5}{3}:x\right|=\dfrac{1}{6}\)

⇒ \(\dfrac{5}{3}:x=\dfrac{1}{6}\) hoặc \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)

*TH1 : \(\dfrac{5}{3}:x=\dfrac{1}{6}\)

⇒ \(x=\dfrac{5}{3}:\dfrac{1}{6}=10\)

*TH2 : \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)

⇒ \(x=\dfrac{5}{3}:\dfrac{-1}{6}=-10\)

Vậy \(x\) ∈ \(\left\{10;-10\right\}\)

\(b,\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|\dfrac{-3}{4}\right|\)

⇒ \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)

⇒\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)

⇒ \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\) hoặc \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)

TH1 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\)

⇒ \(\dfrac{3}{4}x=\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{9}{4}\)

⇒\(x=\dfrac{9}{4}:\dfrac{3}{4}=3\)

TH2 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)

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

⇒ \(x=\dfrac{-3}{4}:\dfrac{3}{4}=-1\)

Vậy \(x\) ∈ \(\left\{3;1\right\}\)

\(A=x+152=\dfrac{1}{3}+152=\dfrac{457}{3}\)

a) (1/7.x-2/7).(-1/5.x-2/5)=0

=> 1/7.x-2/7=0hoặc-1/5.x-2/5=0

*1/7.x-2/7=0

1/7.x=0+2/7

1/7.x=2/7

x=2/7:1/7

x=2

b)1/6.x+1/10.x-4/5.x+1=0

(1/6+1/10-4/5).x+1=0

(1/6+1/10-4/5).x=0-1

(1/6+1/10-4/5).x=-1

(-8/15).x=-1

x=-1:(-8/15) =15/8

`# \text {DNamNgV}`

`16/21 + 6/7`

`= 16/21 + 18/21`

`= 34/21`

__

`15/22 \times 11/35`

`= (15 \times 11)/(22 \times 35)`

`= (5 \times 3 \times 11)/(2 \times 11 \times 5 \times 7)`

`= (3 \times 1)/(2 \times 7)`

`= 3/14`

___

`8/11 \div 5/22`

`= 8/11 \times 22/5`

`= (8 \times 22)/(11 \times 5)`

`= (8 \times 11 \times 2)/(11 \times 5)`

`= (8 \times 2)/5`

`= 16/5`

___

`9/13 \div 27/39`

`= 9/13 \times 39/27`

`= 9/13 \times 13/9`

`= 1`

a: =13/26*9/18*2/3=2/3*1/4=2/12=1/6

b: =44/22*5/15*3/2=2*1/3*3/2=2*1/2=1

\(\dfrac{0.375-0.3+\dfrac{3}{11}+\dfrac{3}{12}}{-0.625+0.5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1.5+1-0.75}{2.5+\dfrac{5}{3}-1.25}\)

=\(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)

=\(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)

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

=\(0\)