Câu 1 : 

a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)

\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)

\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)

\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)

\(< =>45x-90+50x-150+12x-60=0\)

\(< =>107x-300=0< =>x=\frac{300}{107}\)

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\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)

\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)

\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)

\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)

\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)

\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)

\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)

\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)

\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)

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a: x*3/4=1/5

=>x=1/5:3/4=1/5*4/3=4/15

b: x*3/7=2/5

=>x=2/5:3/7=2/5*7/3=14/15

c: 1/3+2/9=2/12x

=>1/6x=3/9+2/9=5/9

=>x=5/9*6=30/9=10/3

d: 4/15*x-2/3=1/5

=>4/15*x=2/3+1/5=10/15+3/15=13/15

=>4x=13

=>x=13/4

e: x:1/7=2/3

=>x=2/3*1/7=2/21

f: 1/9:x=7/3

=>x=1/9:7/3=1/9*3/7=3/63=1/21

j: 1/4+5/12=8/3:x

=>8/3:x=3/12+5/12=8/12=2/3

=>x=4

h: =>7/4:x=1/5+1/2=7/10

=>x=7/4:7/10=10/4=5/2

`@` `\text {Ans}`

`\downarrow`

`1)`

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

`\Rightarrow` \(x=\dfrac{5}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(x=\dfrac{7}{6}\)

Vậy, `x =`\(\dfrac{7}{6}\)

`2)`

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

`\Rightarrow`\(x=\dfrac{3}{5}-\dfrac{1}{3}\)

`\Rightarrow`\(x=\dfrac{4}{15}\)

Vậy, `x =`\(\dfrac{4}{15}\)

`3)`

\(\dfrac{3}{4}+x=\dfrac{7}{2}\)

`\Rightarrow`\(x=\dfrac{7}{2}-\dfrac{3}{4}\)

`\Rightarrow`\(x=\dfrac{11}{4}\)

Vậy, \(x=\dfrac{11}{4}\)

`4)`

\(x-\dfrac{4}{3}=\dfrac{7}{9}\)

`\Rightarrow`\(x=\dfrac{7}{9}+\dfrac{4}{3}\)

`\Rightarrow`\(x=\dfrac{19}{9}\)

Vậy, `x=`\(\dfrac{19}{9}\)

`5)`

\(x-\dfrac{5}{6}=\dfrac{7}{3}\)

`\Rightarrow`\(x=\dfrac{7}{3}+\dfrac{5}{6}\)

`\Rightarrow x =`\(\dfrac{19}{6}\)

Vậy, `x=`\(\dfrac{19}{6}\)

`6)`

\(x-\dfrac{1}{5}=\dfrac{9}{10}\)

`\Rightarrow x=`\(\dfrac{9}{10}+\dfrac{1}{5}\)

`\Rightarrow x=`\(\dfrac{11}{10}\)

Vậy, `x=`\(\dfrac{11}{10}\)

Nhập biểu thức ở dấu Σ cho dễ nhìn em ạ

\(P=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-2x}{x-9}:\dfrac{\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-x-3\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\sqrt{x}+5}\)

\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)\cdot\sqrt{x}\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(-\sqrt{x}+5\right)}=\dfrac{-x}{-\sqrt{x}+5}\)

\(A=\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}+\dfrac{x+9}{25-x}\right):\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)

\(=\dfrac{x-7\sqrt{x}+10+x+5\sqrt{x}-x-9}{\left(x-25\right)}\cdot\dfrac{\sqrt{x}+5}{1-\sqrt{x}}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}=\dfrac{1-\sqrt{x}}{\sqrt{x}-5}\)

\(B=\left(\dfrac{1}{x-4}-\dfrac{1}{x-4\sqrt{x}+4}\right):\dfrac{\sqrt[2]{x}}{2\sqrt{x}-x}\)

\(=\dfrac{\sqrt{x}-2-\sqrt{x}-2}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)}\cdot\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}}\)

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

rảnh à

100

a) x - 3/2 = 4/3 

    x = 4/3 + 3/2

    x = 8/6 + 9/6 = 17/6

b) 2/5 * x = 1/3

    x = 1/3 : 2/5

    x = 1/3 x 5/2 = 5/6

c) x - 4/9 = 3/7 : 9/14

    x - 4/9 = 2/3

    x = 2/3 + 4/9

    x = 6/9 + 4/9 = 10/9

d) 3/5 * x - 1/2 = 1/5

    3/5 * x = 1/5 + 1/2 = 7/10

    x = 7/10 : 3/5

    x = 7/10 * 5/3 = 7/6

a/\(x-\dfrac{3}{2}=\dfrac{4}{3}\)
\(x=\dfrac{4}{3}+\dfrac{3}{2}\)
\(x=\dfrac{17}{6}\)
b/\(\dfrac{2}{5}\times x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:\dfrac{2}{5}\)
\(x=\dfrac{5}{6}\)
c/\(x-\dfrac{4}{9}=\dfrac{3}{7}:\dfrac{9}{14}\)
\(x-\dfrac{4}{9}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+\dfrac{4}{9}\)
\(x=\dfrac{10}{9}\)
d/\(\dfrac{3}{5}\times x-\dfrac{1}{2}=\dfrac{1}{5}\)
\(\dfrac{3}{5}\times x=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{3}{5}\times x=\dfrac{7}{10}\)
\(x=\dfrac{7}{10}:\dfrac{3}{5}\)
\(x=\dfrac{7}{6}\)

a,x=\(\dfrac{17}{6}\)b,x=\(\dfrac{5}{6}\)c,x=\(\dfrac{10}{9}\)d,x=\(\dfrac{7}{6}\)

\(a,\frac{1}{2}+\frac{3}{4}x=\frac{1}{4}\)

\(\Rightarrow\frac{3}{4}x=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\)

\(\Rightarrow x=-\frac{1}{3}\)

Bạn Hồng có một tờ bìa hình chữ nhật có chiều rộng 1/5 m và chiều dài 1/4 m . bạn Hồng muốn cắt từ tấm bìa đó thành một hình chữ nhật có diện tích bằng một nửa diện tích tấm bìa và độ dài một cạnh là 1/4 m để làm hộp đồ chơi.

a) tính diện tích hình chữ nhật bạn Hồng đã cắt

b) bạn Hồng có thể cắt như thế nào ? Vẽ hình minh họa

Một vòi nước chảy vào bể giờ thứ nhất chảy được 1/2 bể giờ thứ hai chảy được 2/5 bể. Hỏi sau 2 giờ thì phần bể còn trống chiếm bao nhiêu phần bé tích và biết lúc đầu để không có nước ?