Tính giá trị các biểu thức sau theo cách hợp lí nhất.
a) $\mathrm{A}=\left(\dfrac{2}{7} \cdot \dfrac{1}{4}-\dfrac{1}{3} \cdot \dfrac{2}{7}\right):\left(\dfrac{2}{7} \cdot \dfrac{3}{9}-\dfrac{2}{7} \cdot \dfrac{2}{5}\right)$;
b) $\mathrm{B}=\dfrac{\left(\dfrac{1}{5}-\dfrac{2}{7}\right) \cdot \dfrac{3}{4}-\dfrac{3}{4} \cdot\left(\dfrac{1}{3}-\dfrac{2}{7}\right)}{\dfrac{1}{5} \cdot \dfrac{2}{7}-\dfrac{1}{3} \cdot\left(\dfrac{2}{7}+\dfrac{3}{9}\right)+\dfrac{3}{9} \cdot \dfrac{1}{5}} .$
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Giải:
a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)
\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)
\(\Leftrightarrow x=-\dfrac{79}{210}\)
Vậy \(x=-\dfrac{79}{210}\).
b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)
\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)
\(\Leftrightarrow3x=\dfrac{67}{75}\)
\(\Leftrightarrow x=\dfrac{67}{75}:3\)
\(\Leftrightarrow x=\dfrac{67}{225}\)
Vậy \(x=\dfrac{67}{225}\).
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NHỚ TRẢ LỜI NHANH GIÙM MK NHA
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
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a) \(\Rightarrow\dfrac{\dfrac{7}{2}x+\dfrac{59}{24}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Rightarrow\left(\dfrac{7}{2}x+\dfrac{59}{24}\right).52=\dfrac{13}{30}.137\)
\(\Rightarrow182x+\dfrac{767}{6}=\dfrac{1781}{30}\)
\(\Rightarrow x=\dfrac{-79}{210}\)
b) Tương tự câu a)
a) A=[27(14−13)]:[27(13−25)]=(14−13):(13−25)=114A=[27(14−13)]:[27(13−25)]=(14−13):(13−25)=114.
b) B=34(15−27−13+27)15(27+13)−13(27+13)=34(15−13)(15−13)(27+13)=11152B=34(15−27−13+27)15(27+13)−13(27+13)=34(15−13)(15−13)(27+13)=11152.
a) \mathrm{A}=\left[\dfrac{2}{7}\left(\dfrac{1}{4}-\dfrac{1}{3}\right)\right]:\left[\dfrac{2}{7}\left(\dfrac{1}{3}-\dfrac{2}{5}\right)\right]=\left(\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{1}{3}-\dfrac{2}{5}\right)=1 \dfrac{1}{4}A=[72(41−31)]:[72(31−52)]=(41−31):(31−52)=141.
b) \mathrm{B}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{3}+\dfrac{2}{7}\right)}{\dfrac{1}{5}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)-\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{1}{3}\right)}{\left(\dfrac{1}{5}-\dfrac{1}{3}\right)\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=1 \dfrac{11}{52}B=51(72+31)−31(72+31)43(51−72−31+72)=(51−31)(72+31)43(51−31)=15211