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27 tháng 3 2023

`3/5=(-x)/20`

`=>12/20=(-x)/20`

`=>-x=12`

`=>x=-12`

27 tháng 3 2023

⇒3. 20 = 5. (-x)

⇒60 = 5. (-x)

⇒-x = 60 : 5

⇒-x = 12

vậy x = 12

 

b: \(\Leftrightarrow x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)

\(\Leftrightarrow x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)

\(\Leftrightarrow x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)

=>x=3/11+20/55=3/11+4/11=7/11

c: \(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-2}{98}-1\right)+\left(\dfrac{x-5}{95}-1\right)=\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{95}\)

\(\Leftrightarrow x-100=1\)

hay x=101

20 tháng 4 2017

các bạn phại đổi kiểu chữ để làm bài này (VNI) , (TELEX)

20 tháng 4 2017

vni

28 tháng 4 2017

a) \(\dfrac{x}{2}+\dfrac{y}{3}=\dfrac{x+y}{2+3}\)

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

\(\dfrac{x}{2}=\dfrac{x+y}{5}-\dfrac{y}{3}\)

\(\dfrac{x}{2}=\dfrac{3\left(x+y\right)}{15}-\dfrac{5y}{15}\)

\(\dfrac{x}{2}=\dfrac{3x-2y}{15}\)

\(\Rightarrow15x=2\left(3x-2y\right)\)

\(15x=6x-4y\)

\(15x-6x=4y\)

\(9x=4y\)

(CÒN LẠI MÌNH KHÔNG BIẾT LÀM)

b) \(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)

\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{4}{y}\\ \)

\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{20}{5y}\)

\(\dfrac{x}{3}=\dfrac{1+4}{y+1}\)

\(\Rightarrow x\left(y+1\right)=15\)

(CÒN NHIÊU TỰ LÀM NHÉ)

15 tháng 4 2018

c)

\(\left|\dfrac{2}{3}x-\dfrac{1}{2}\right|-1=\dfrac{5}{6}\)

\(\Leftrightarrow\left|\dfrac{4x-3}{6}\right|=1+\dfrac{5}{6}=\dfrac{11}{6}\)

\(\Leftrightarrow\left|4x-3\right|=11\Leftrightarrow\left[{}\begin{matrix}4x-3=11=>x=\dfrac{11+3}{4}=\dfrac{7}{2}\\4x-3=-11=>x=\dfrac{-8}{2}=-4\end{matrix}\right.\)

26 tháng 6 2017

Bài 2: a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)

\(\Leftrightarrow7x-21=5x+25\)

\(\Leftrightarrow7x-5x=21+25\)

\(\Leftrightarrow2x=46\)

\(\Rightarrow x=46:2=23\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=63\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

\(\Rightarrow x^2=\left(\pm8\right)^2\)

\(\Rightarrow x=8\) hoặc \(x=-8\)

26 tháng 6 2017

2)a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

\(\Leftrightarrow7\left(x-3\right)=5\left(x+5\right)\)

\(7x-21=5x+25\)

\(7x-5x+25=21\)

\(2x+25=21\)

\(2x=-4\Rightarrow x=-2\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(7.9=\left(x+1\right)\left(x-1\right)\)

\(63=x\left(x-1\right)+1\left(x-1\right)\)

\(63=x^2-x+x-1\)

\(x^2=63+1=64\)

\(x=\left\{\pm8\right\}\)

c) \(\dfrac{x+4}{20}=\dfrac{2}{x+4}\)

\(\Leftrightarrow\left(x+4\right)\left(x+4\right)=2.20=40\)

\(x\left(x+4\right)+4\left(x+4\right)=40\)

\(x^2+4x+4x+16=40\)

\(x^2+8x=40-16=24\)

\(x\left(x+8\right)=24\)

\(x\in\left\{\varnothing\right\}\)

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

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=\left(x-1\right)\left(x+3\right)\)

\(x\left(x-2\right)+2\left(x-2\right)=x\left(x+3\right)-1\left(x+3\right)\)

\(x^2-2x+2x-4=x^2+3x-x-3\)

\(\)\(x^2-4=x^2+2x-3\)

\(\Leftrightarrow x^2-x^2-2x+3=4\)

\(-2x+3=4\)

\(-2x=1\)

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

6 tháng 3 2017

b, \(\dfrac{x-3}{4}=\dfrac{15}{20}\)

<=> \(\dfrac{x-3}{4}=\dfrac{3}{4}\)

=> x-3=3

<=> x=6

Vậy x=6

9 tháng 6 2017

\(a,\dfrac{x}{15}=\dfrac{4}{y}=\dfrac{-2}{5}\)

* \(\dfrac{x}{15}=\dfrac{-2}{5}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{-6}{15}\)

\(\Rightarrow x=-6\)

*\(\dfrac{4}{y}=\dfrac{-2}{5}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{4}{-10}\)

\(\Rightarrow y=-10\)

Vậy x = - 6 ; y = - 10

\(b,\dfrac{x-3}{4}=\dfrac{15}{20}\)

=> ( x - 3 ) . 20 = 4. 15

=> 20x - 60 = 60

=> 20x = 60 + 60

=> 20x = 120

=> x = 120 : 20

=> x = 6

Vậy x = 6

\(c,\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{22}{-9}+\dfrac{-7}{15}< x\le\dfrac{-1}{3}+\dfrac{-1}{4}+\dfrac{-5}{12}\)

\(\Rightarrow\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{-22}{9}+\dfrac{-7}{15}< x\le\dfrac{-4}{12}+\dfrac{-3}{12}+\dfrac{-5}{12}\)

\(\Rightarrow\left(\dfrac{-5}{9}+\dfrac{-22}{9}\right)+\left(\dfrac{-8}{15}+\dfrac{-7}{15}\right)< x\le-1\)

\(\Rightarrow-3+\left(-1\right)< x\le-1\)

\(\Rightarrow-4< x\le-1\)

\(\Rightarrow x=-3;-2;-1\)

20 tháng 5 2017

a)

\(\dfrac{-2}{3}\cdot\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\cdot\left(2x-1\right)\\ \dfrac{-2}{3}x-\left(\dfrac{-1}{6}\right)=\dfrac{2}{3}x-\dfrac{1}{3}\\ \dfrac{-2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\\ \dfrac{-2}{3}x-\dfrac{2}{3}x=\dfrac{-1}{3}-\dfrac{1}{6}\\ -x\cdot\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{-2}{6}-\dfrac{1}{6}\\ \dfrac{-4}{3}x=\dfrac{-1}{2}\\ x=\dfrac{-1}{2}:\dfrac{-4}{3}\\ x=\dfrac{-1}{2}\cdot\dfrac{-3}{4}\\ x=\dfrac{3}{8}\)

19 tháng 7 2018

a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)

\(\Rightarrow-4< x< \dfrac{-3}{10}\)

\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)

\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)

b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)

\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)

\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)

\(\Rightarrow x=\varnothing\)

c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)

\(\Rightarrow x\in\left\{1;2\right\}\)

+) Với \(x=1\)

\(\Rightarrow y\in\left\{1;2\right\}\)

+) Với \(x=2\)

\(\Rightarrow y=2\)

Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).

28 tháng 4 2017

Bài 1:

a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{1}{2}\)

\(=\dfrac{1}{4}+\dfrac{2}{4}\)

\(=\dfrac{3}{4}\)

b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)

\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)

\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)

\(=\dfrac{1}{2}+\dfrac{4}{5}\)

\(=\dfrac{5}{10}+\dfrac{8}{10}\)

\(=\dfrac{9}{5}\)

c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)

\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)

\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)

\(=\dfrac{7}{3}+\dfrac{28}{3}\)

\(=\dfrac{35}{3}\)

d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)

\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)

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

\(=\dfrac{1}{6}-\dfrac{21}{6}\)

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

e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)

\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)

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

f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{3}{2}\)

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

g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)

\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)

\(=\dfrac{1}{2}-\dfrac{3}{4}\)

\(=\dfrac{2}{4}-\dfrac{3}{4}\)

\(=\dfrac{-1}{4}\)

h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)

\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{9}{28}\)

\(=\dfrac{196}{140}-\dfrac{45}{140}\)

\(=\dfrac{151}{140}\)

i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)

\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)

\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)

\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)

k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)

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

29 tháng 4 2017

\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)

\(A=\dfrac{1}{8}.1.20\)

\(A=\dfrac{20}{8}=\dfrac{5}{2}\)

\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)

\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)

\(B=\left(16+1\right)+4,03\)

\(B=17+4,03\)

\(B=21,03\)

\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)

\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)

\(C=390.\dfrac{15}{78}\)

\(C=75\)

17 tháng 4 2017

Giải bài 90 trang 43 SGK Toán 6 Tập 2 | Giải toán lớp 6

1 tháng 5 2018

Giải bà i 90 trang 43 SGK Toán 6 Tập 2 | Giải toán lá»p 6