Bài 1: Tìm x

a) Ta có: \(\dfrac{4}{3}:0.8=\dfrac{2}{3}:\left(0.1\cdot x\right)\)

\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}:\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}\cdot\dfrac{5}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow x\cdot\dfrac{1}{10}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{3}\cdot\dfrac{3}{5}=\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{2}{5}:\dfrac{1}{10}=\dfrac{2}{5}\cdot10=\dfrac{20}{5}=4\)

Vậy: x=4

b) Ta có: \(\left|x\right|=-1.2\)

mà \(\left|x\right|\ge0\forall x\)

nên \(x\in\varnothing\)

Vậy: \(x\in\varnothing\)

Bài 2: Tính

a) Ta có: \(\left(-2.5\right)\cdot\left(-4\right)\cdot\left(-7.9\right)\)

\(=\left(2.5\cdot4\right)\cdot\left(-7.9\right)\)

\(=-7.9\cdot10=-79\)

b) Ta có: \(\left(-0.375\right)\cdot\dfrac{13}{3}\cdot\left(-2\right)^3\)

\(=\dfrac{3}{8}\cdot8\cdot\dfrac{13}{3}\)

\(=3\cdot\dfrac{13}{3}=13\)

\(a)\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)

\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{6}{12}\right)\)

\(=\left(\dfrac{153}{54}+\dfrac{78}{54}\right):\left(1\dfrac{-5}{12}\right)\)

\(=\dfrac{231}{54}:\dfrac{7}{12}\)

\(=\dfrac{198}{27}\)

\(b)\dfrac{0,8\left(\dfrac{4}{5}:1,25\right)}{0,64-\dfrac{1}{25}}\)

\(=\dfrac{0,8\left(0,8:1,25\right)}{0,64-0,04}\)

\(=\dfrac{0,8.0,64}{0,6}\)

\(=\dfrac{0,512}{0,6}\)\(=\dfrac{64}{75}\)

tích trung tỉ bằng tihs ngoại tỉ là ra ý mà

\(\dfrac{x+1}{x+2}=\dfrac{0,8}{1,2}\) \(\Leftrightarrow0,8\left(x+2\right)=1,2\left(x+1\right)\)

\(\Leftrightarrow0,8x+1,6=1,2x+1,2\) \(\Leftrightarrow0,4x=0,4\Leftrightarrow x=1\)

vậy \(x=1\)

\(\dfrac{x+1}{x+2}=\dfrac{0,8}{1,2}\)

\(\Rightarrow\left(x+1\right)1,2=\left(x+2\right)0,8\)

\(1,2x+1,2=0,8x+1,6\)

\(1,2x-0,8x=1,6-1,2\)

\(0,4x=0,4\)

\(x=1\)

Cái này bn lầy máy tính ra tính tí là xong thôi

a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)

= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\dfrac{11}{4}\)

= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\left(2,75\right)\)

= \(2,75.\left\{\left(-0,4\right)-\left(1,6\right)+\left(-1,2\right)\right\}\)

= \(2,75.\left(-3,2\right)\)

= \(-8,8\)

b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)

= \(\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)

= \(\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}\)

= \(\dfrac{3}{7}-\dfrac{2}{3}\)

= \(-\dfrac{5}{21}\)

c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)

= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(\dfrac{4}{5}-2\dfrac{4}{15}\right):\dfrac{11}{3}\)

= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\)

= \(\left(-\dfrac{3}{4}\right)+\left(-\dfrac{2}{5}\right)\)

= \(-\dfrac{23}{20}\)

d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\)

= \(\dfrac{1}{50}.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-\dfrac{49}{20}\right).\dfrac{2}{7}\)

=\(\left(-\dfrac{1}{4}\right)+\dfrac{3}{8}+\left(-\dfrac{7}{10}\right)\)

= \(\dfrac{1}{8}+\left(-\dfrac{7}{10}=\right)\)

= \(-\dfrac{23}{40}\)

e) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)

= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)

= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)

= \(-\dfrac{5501}{225}\)

Xin mấy CTV giúp e với tặng 3 sp huhu

dài vậy?Ghi đáp án thôi nhé!

1: Ta có: \(23\dfrac{1}{4}\cdot\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)

\(=\dfrac{93}{4}\cdot\dfrac{7}{5}-\dfrac{53}{4}\cdot\dfrac{7}{5}\)

\(=\dfrac{7}{5}\cdot10=14\)

2: Ta có: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{12+8-3}{12}\cdot\dfrac{1}{400}\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

a) 15/11 - (5/7 - 18/11) + 27/7

= 15/11 - 5/7 + 18/11 + 27/7

= (15/11 + 18/11) + (-5/7 + 27/7)

= 3 + 22/7

= 43/7

b) 39/5 + (9/4 - 9/5) - (5/4 + 1,2)

= 39/5 + 9/4 - 9/5 - 5/4 - 6/5

= (39/5 - 9/5 - 6/5) + (9/4 - 5/4)

= 24/5 + 1

= 29/5

c) -1,2 - 0,8 + 0,25 + 5,75 - 2022

= (-1,2 - 0,8) + (0,25 + 5,76) - 2022

= -2 + 6 - 2022

= 4 - 2022

= -2018

d) 0,1 + 16/9 + 5,1 + (-20/9)

= (0,1 + 5,1) + (16/9 - 20/9)

= 5,2 - 4/9

= 419/90

a) \(\dfrac{15}{11}-\left(\dfrac{5}{7}-\dfrac{18}{11}\right)+\dfrac{27}{7}=\dfrac{22}{7}+3=\dfrac{43}{77}\)

b) \(\dfrac{39}{5}+\left(\dfrac{9}{4}-\dfrac{9}{5}\right)-\left(\dfrac{5}{4}+\dfrac{6}{5}\right)=\dfrac{24}{5}+1=\dfrac{29}{5}\)

c) \(-1,2-0,8+0,25+5,75-2022=-2+6-2022=-2018\)

d) \(0,1+\dfrac{16}{9}+5,1+\dfrac{-20}{9}=\dfrac{26}{5}-\dfrac{4}{9}=\dfrac{214}{45}\)

1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)

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

2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)

\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)

3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)

\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)

\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)

\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)

\(=12\cdot\dfrac{5}{3}=20\)

5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)

6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)

\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)

\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)