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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rêtrt

A= 5/4

B= 63/52

Hoàng Ngọc Anh bài này nè bn giúp mk nha!!! ngày mai mk phải nộp bài rùi =.=

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)

\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=2\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)

\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=-\dfrac{49}{8}\)

\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

a= 1/10

b= 18/5

c= -49/8

d= 413

.

A= 3/5.6/7+3/5.3/7-3/5.2/7

A= 3/5. (6/7+3/7-2/7) = 3/5

B= (-13.2/5+-2/9.2/5+2/5.11/9).5/2

B= (-13-2/9+11/9).2/5.5/2

B= -13+(11/9-2/9)= -12

C= (-4/5+5/7).3/2+(-1/5+2/7).3/2

C= (-4/5+5/7+-1/5+2/7).3/2

C= ((-4/5+-1/5)+(5/7+2/7)).3/2

C= 0

D= 4/9.-5/3+4/9.-22/3

D= 4/9.110/9

D= 440/81

.

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=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)

hôi lì sít

\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)

\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)

\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)

\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)

\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)

4.

\(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+5}{95}+1\right)=\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)+\left(\dfrac{x+11}{89}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{97}++\dfrac{x+100}{95}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89}\right)=0\\ \Leftrightarrow x+100=0\Leftrightarrow x=-100\)

\(\text{1) }\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\\ \Leftrightarrow\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\cdot24=\left[\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\right]24\\ \Leftrightarrow3\left(4x^2-9\right)=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\\ \Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\\ \Leftrightarrow12x^2-27=12x^2-64x+96\\ \Leftrightarrow12x^2-12x^2+64x=96+27\\ \Leftrightarrow64x=123\\ \Leftrightarrow x=\dfrac{123}{64}\\ \text{Vậy }S=\left\{\dfrac{123}{64}\right\}\\ \)

\(\text{2) }x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\\ \Leftrightarrow\left(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}\right)15=\dfrac{7x-\dfrac{x-3}{2}}{5}\cdot15\\ \Leftrightarrow15x+30-2x-\dfrac{2x-5}{6}=21x-\dfrac{3x-9}{2}\\ \Leftrightarrow15x-2x-\dfrac{2x-5}{6}-21x+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow\left(-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}\right)6=-30\cdot6\\ \Leftrightarrow-48x-2x+5+9x-27=-180\\ \Leftrightarrow-41x==-158\\ \Leftrightarrow x=\dfrac{158}{41}\\ \text{Vậy }S=\left\{\dfrac{158}{41}\right\}\)

\(\text{3) }1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\\ \Leftrightarrow\left(1-\dfrac{x-1-x}{3}\right)6=\left(\dfrac{x}{2}-\dfrac{2x-1}{2}\right)6\\ \Leftrightarrow6+2=-3x+3\\ \Leftrightarrow-3x=8-3\\ \Leftrightarrow-3x=5\\ \Leftrightarrow x=-\dfrac{5}{3}\\ \\ \text{Vậy }S=\left\{-\dfrac{5}{3}\right\}\)

1.\(\dfrac{4}{5}\)

2.\(\dfrac{2}{3}\)

3.\(\dfrac{39}{55}\)

a) \(\dfrac{3}{5}\times\dfrac{2}{3}:\dfrac{1}{2}=\dfrac{3}{5}\times\dfrac{2}{3}\times2=\dfrac{6}{15}\times2=\dfrac{12}{15}=\dfrac{4}{5}\)

b) \(\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}\)

c) \(\left(\dfrac{5}{7}+\dfrac{2}{5}\right):\dfrac{11}{7}=\left(\dfrac{25}{35}+\dfrac{14}{35}\right)\cdot\dfrac{7}{11}=\dfrac{39}{35}\cdot\dfrac{7}{11}=\dfrac{273}{385}=\dfrac{39}{55}\)

a) = \(\dfrac{6}{15}\) \(\div\dfrac{1}{2}=\) \(\dfrac{12}{15}\) rút gọn = \(\dfrac{4}{5}\)

b) = \(\dfrac{2}{6}+\dfrac{1}{3}\) = \(\dfrac{6}{18}+\dfrac{6}{18}\) = \(\dfrac{12}{18}\) = \(\dfrac{2}{3}\)

c) = \(\dfrac{5}{7}+\dfrac{2}{5}=\dfrac{25}{35}+\dfrac{14}{35}=\dfrac{39}{35}\) \(\div\dfrac{11}{7}=\dfrac{39}{35}\)