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\(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)
\(=x+\dfrac{\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}}{-\left(\dfrac{3}{10}-\dfrac{9}{16}+\dfrac{15}{22}\right)}\)
\(=x+\dfrac{\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}}{-\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}\right)}\)
\(=x+\dfrac{1}{-\dfrac{3}{2}}\)
\(=x+\dfrac{-2}{3}\)
Với \(x=-\dfrac{1}{3}\), ta được:
\(B=-\dfrac{1}{3}+\dfrac{-2}{3}=-\dfrac{3}{3}=-1\)
\(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)
\(=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)
\(=x+\dfrac{-\dfrac{7}{40}+\dfrac{5}{11}}{-\dfrac{369}{880}}\)
\(=x+\dfrac{\dfrac{123}{440}}{\dfrac{369}{880}}\)
\(=x-\dfrac{2}{3}\)
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 \(x\) ∈ \(\left\{3;1\right\}\)
\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)
=1/64
7) 5x=4y ⇒\(\dfrac{x}{4}=\dfrac{y}{5}\)
Nhân cả hai vế với \(\dfrac{x}{4}\), ta có: \(\left(\dfrac{x}{4}\right)^2=\dfrac{x}{4}.\dfrac{y}{5}=\dfrac{xy}{20}=\dfrac{20}{20}=1\)
\(\left(\dfrac{x}{4}\right)^2=1\Rightarrow\left[{}\begin{matrix}\dfrac{x}{4}=1\\\dfrac{x}{4}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)
4) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{0,5}=\dfrac{y}{0,3}=\dfrac{z}{0,2}=\dfrac{z-y+x}{0,2-0,3+0,5}=\dfrac{1}{\dfrac{2}{5}}=\dfrac{5}{2}\)
\(\dfrac{x}{0,5}=\dfrac{5}{2}\Rightarrow x=\dfrac{5}{4}\)
\(\dfrac{y}{0,3}=\dfrac{5}{2}\Rightarrow y=\dfrac{3}{4}\)
\(\dfrac{z}{0,2}=\dfrac{5}{2}\Rightarrow z=\dfrac{1}{2}\)
6) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x+11}{13}=\dfrac{y+12}{14}=\dfrac{z+13}{15}=\dfrac{x+11+y+12+z+13}{13+14+15}=\dfrac{42}{42}=1\)
\(\dfrac{x+11}{13}=1\Rightarrow x=2\)
\(\dfrac{y+12}{13}=1\Rightarrow y=1\)
\(\dfrac{z+13}{15}=1\Rightarrow z=2\)
7) \(5x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}=k\)
\(\Rightarrow x=4k,y=5k\)
\(x.y=20\\ \Rightarrow4k.5k=20\\ \Rightarrow20k^2=20\\ \Rightarrow k^2=1\\ \Rightarrow\left[{}\begin{matrix}k=-1\\k=1\end{matrix}\right.\)
\(x=4k\Rightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
\(y=5k\Rightarrow\left[{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)
Vậy \(\left(x,y\right)=\left\{\left(-4;-5\right);\left(4;5\right)\right\}\)
\(\dfrac{5}{6}x-\dfrac{3}{4}=\dfrac{-1}{4}+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{5}{6}x=\dfrac{7}{6}\)
\(\Rightarrow x=\dfrac{7}{5}\)
b) \(-1\dfrac{1}{2}-\dfrac{2}{3}x=\dfrac{5}{6}-\left(\dfrac{-2}{5}\right)\)
\(\Leftrightarrow\dfrac{2}{3}x=-\dfrac{41}{15}\)
\(\Rightarrow x=-\dfrac{41}{10}\)
c) \(\left(\dfrac{4}{5}:x+1,5\right):\dfrac{2}{3}=-1,5\)
\(\Leftrightarrow\dfrac{8+15x}{10x}.\dfrac{3}{2}=\dfrac{-3}{2}\)
\(\Leftrightarrow\dfrac{24+45x}{20x}=\dfrac{-3}{2}\)
\(\Leftrightarrow-60x=48+90x\)
\(\Rightarrow x=-0,32\)
d) \(\dfrac{4}{3}x-\dfrac{2}{3}=\dfrac{1}{4}-x\)
\(\Leftrightarrow\dfrac{4x-2}{3}=\dfrac{1-4x}{4}\)
\(\Rightarrow16x-8=3-12x\)
\(\Rightarrow x=\dfrac{11}{28}\)
\(A=x+152=\dfrac{1}{3}+152=\dfrac{457}{3}\)