a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

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

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

a) x - 1/2 = 3/5

x = 3/5 + 1/2

x = 11/10

b) x - 1/2 = -2/3

x = -2/3 + 1/2

x = -1/6

c) 2/5 - x = 0,25

x = 2/5 - 0,25

x = 2/5 - 1/4

x = 3/20

a, \(\sqrt{25}-3\sqrt{\dfrac{4}{9}}=5-3.\dfrac{2}{3}=3\)

b, \(\left(2-\dfrac{5}{3}\right):\left(\dfrac{2}{7}+\dfrac{5}{21}-1\right)\)

\(=\dfrac{1}{3}:\dfrac{6+5-21}{21}\)

\(=-\dfrac{1}{3}.\dfrac{21}{10}\)

\(=-\dfrac{7}{10}\)

Theo t/c dãy tỉ số bằng nhau, ta có:

\(\frac{x}{2}-\frac{y}{3}=\frac{y^2-x^2}{3^2-2^2}=\frac{20}{5}=4\)

\(=>\hept{\begin{cases}\frac{x}{2}=4\\\frac{y}{3}=4\end{cases}}=>\hept{\begin{cases}x=8\\y=12\end{cases}}\)

lộn r lộn r

Ta có: \(\frac{x}{2}=\frac{y}{3}\)

\(=>\frac{x^2}{4}=\frac{y^2}{9}\)

Theo t/c dãy tỉ số bằng nhau: \(\frac{x^2}{4}=\frac{y^2}{9}=>\frac{y^2-x^2}{9-4}=\frac{20}{5}=4\)

Khi đó: \(\hept{\begin{cases}\frac{x^2}{4}=4\\\frac{y^2}{9}=4\end{cases}}=>\hept{\begin{cases}x=4\\y=6\end{cases}}\)

Vậy ...

\(\dfrac{21}{36}-\left(-\dfrac{11}{30}\right)=\dfrac{7}{12}+\dfrac{11}{30}=\dfrac{7.5+11.2}{60}=\dfrac{57}{60}=\dfrac{19}{20}\\ ----\\\dfrac{-4}{8}+\left(-\dfrac{3}{10}\right)=\dfrac{-1}{2}-\dfrac{3}{10}=\dfrac{-1.5-3}{10}=\dfrac{-8}{10}=-\dfrac{4}{5}\\ ----\\ \dfrac{7}{12}-\left(-\dfrac{9}{20}\right)=\dfrac{7}{12}+\dfrac{9}{20}=\dfrac{7.5+9.3}{60}=\dfrac{62}{60}=\dfrac{31}{30}\\ ---\\ \dfrac{-2}{5}+\left(-\dfrac{11}{30}\right)=-\dfrac{2}{5}-\dfrac{11}{30}=\dfrac{-2.6-11}{30}=-\dfrac{29}{30}\)

Gọi x,y,z là số học sinh khối 6, 7, 8

(x,y,z>0, đvị là học sinh)

Đã biết khối học sinh lớp 8 ít hơn số hs khối 6 là 120 hs

x-z=120

x, y, z tỉ lệ với 8, 7, 5

x/8=y/7=z/5

Áp dụng tính chất dãy tỉ số bằng nhau có:

x/8=y/7=z/5= x-z/8-5=120/3=40

=> x/8= 40       => x=40.8=320        => số hs khối 6 là 320 hs

y/7= 40                 y=40.7= 280            số hs khối 7 là 280 hs

z/5= 40                 z=40.5=200              số hs khối 8 là 200 hs

`x^2+[-18]/[x^2+x]=3-x`         `ĐK: x \ne -1,x \ne 0`

`<=>[x^2(x^2+x)-18]/[x^2+x]=[(3-x)(x^2+x)]/[x^2+x]`

   `=>x^4+x^3-18=3x^2+3x-x^3-x^2`

`<=>x^4+2x^3-2x^2-3x-18=0`

`<=>x^4-2x^3+4x^3-8x^2+6x^2-12x+9x-18=0`

`<=>x^3(x-2)+4x^2(x-2)+6x(x-2)+9(x-2)=0`

`<=>(x-2)(x^3+4x^2+6x+9)=0`

`<=>(x-2)(x^3+3x^2+x^2+3x+3x+9)=0`

`<=>(x-2)[x^2(x+3)+x(x+3)+3(x+3)]=0`

`<=>(x-2)(x+3)(x^2+x+3)=0`

`<=>` $\left[\begin{matrix} x=2 (t/m)\\ x=-3 (t/m)\\x^2+x+3=0\text{ (Vô nghiệm)}\end{matrix}\right.$

Vậy `S={-3;2}`

\(x^2+\dfrac{-18}{x^2+x}=3-x\)

\(\Leftrightarrow x^2-\dfrac{18}{x\left(x+1\right)}=3-x\);\(ĐK:x\ne0;-1\)

\(\Leftrightarrow-\dfrac{18}{x\left(x+1\right)}=3-x-x^2\)

\(\Leftrightarrow\dfrac{18}{x\left(x+1\right)}=x^2+x-3\)

\(\Leftrightarrow\dfrac{18}{x\left(x+1\right)}=x\left(x+1\right)-3\)

Đặt \(x\left(x+1\right)=a\)

\(\Leftrightarrow\dfrac{18}{a}=a-3\)

\(\Leftrightarrow a^2-3a-18=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-3\end{matrix}\right.\)

Với `x=6`

`=>`\(x^2+x=6\)

`<=>x^2+x-6=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\) \((tm)\)

Với `x=-3`

`=>`\(x^2+x=-3\)

`<=>x^2+x+3=0` ( vô lý )

Vậy \(S=\left\{2;-3\right\}\)

ai đó cứu với TvT