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

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

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

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

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

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.

\(\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{5}{11}}{-\dfrac{3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}-\dfrac{5}{22}\right)}=\dfrac{-2}{3}\)

\(\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{5}{11}}{-\dfrac{3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}-\dfrac{5}{22}\right)=\dfrac{-2}{3}}\)

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

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

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

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

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 \(x\) ∈ \(\left\{3;1\right\}\)

\(\dfrac{x+3}{15}=\dfrac{1}{3}\left(1\right)\\ \Leftrightarrow x+3=\dfrac{1}{3}\cdot15\\ \Leftrightarrow x+3=5\\ \Rightarrow x=5-3\\ \Rightarrow x=2\)

Vậy tập nghiệm phương trình (1) là \(\left\{2\right\}\)

(X+3)3=15x1

3X+9=15

3X=15-9

3X=6

X=6:3

X=2

\(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{x}{15}+\dfrac{3}{15}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{1}{3}-\dfrac{3}{15}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{2}{15}\)
\(\Rightarrow x=2\)

\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)

\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)

=1/64

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

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)