đề sai rồi bạn

\(H=\left[0,\left(32\right).1,\left(5\right)-0,\left(25\right)\right].\dfrac{11}{83}\)

\(\Leftrightarrow H=\left(\dfrac{32}{99}.\dfrac{14}{9}-\dfrac{25}{99}\right).\dfrac{11}{83}\)

\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{25}{99}\right).\dfrac{11}{83}\)

\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{225}{891}\right).\dfrac{11}{83}\)

\(\Leftrightarrow H=\dfrac{448-225}{891}.\dfrac{11}{83}\)

\(\Leftrightarrow H=\dfrac{223}{891}.\dfrac{11}{83}\)

\(\Leftrightarrow H=\dfrac{2453}{73953}\)

\(\Leftrightarrow H=\dfrac{223}{6723}\)

2) \(A=\dfrac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)

\(\Leftrightarrow A=\dfrac{\dfrac{3}{6}+\dfrac{2}{6}-\dfrac{1}{6}}{\dfrac{15}{6}+\dfrac{10}{6}-\dfrac{5}{6}}\)

\(\Leftrightarrow A=\dfrac{\dfrac{3+2-1}{6}}{\dfrac{15+10-5}{6}}\)

\(\Leftrightarrow A=\dfrac{\dfrac{4}{6}}{\dfrac{20}{6}}\)

\(\Leftrightarrow A=\dfrac{4}{6}.\dfrac{6}{20}\)

\(\Leftrightarrow A=\dfrac{24}{120}\)

\(\Leftrightarrow A=\dfrac{1}{5}\)

\(G=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{6}}=\dfrac{1}{5}\)

\(H=\left[\dfrac{32}{99}\cdot\dfrac{14}{9}-\dfrac{25}{99}\right]\cdot\dfrac{11}{83}\)

\(=\dfrac{223}{891}\cdot\dfrac{11}{83}=\dfrac{223}{6723}\)

H = [0,(32) . 1,(5) - 0,(25)] . \(\frac{11}{83}\)

H = \(\left(\frac{32}{99}.\frac{15}{9}-\frac{25}{99}\right).\frac{11}{83}\)

H = \(\left(\frac{160}{297}-\frac{75}{297}\right).\frac{11}{83}\)

H = \(\frac{85}{297}.\frac{11}{83}\)

H = \(\frac{85}{2241}\)

H = ( 0 . 5 . 0).11 phần 83 (mk ko bt ghi phân số)

H = 0 . 11

H = 0 

Study good

a: \(\left(\dfrac{5}{9}-\dfrac{\sqrt{9}}{12}\right):\dfrac{3}{4}+\dfrac{11}{3}:\dfrac{3}{4}\)

\(=\left(\dfrac{5}{9}-\dfrac{3}{12}\right)\cdot\dfrac{4}{3}+\dfrac{11}{3}\cdot\dfrac{4}{3}\)

\(=\left(\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{11}{3}\right)\cdot\dfrac{4}{3}\)

\(=\dfrac{20-9+132}{36}\cdot\dfrac{4}{3}\)

\(=\dfrac{143}{3}\cdot\dfrac{1}{9}=\dfrac{143}{27}\)

b: \(\left(0.\left(3\right)+\dfrac{\left|-2\right|}{3}\right):\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\)

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

\(=\dfrac{4}{5}-1=-\dfrac{1}{5}\)

\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)

=>Trong 2 số phải có 1 số âm và 1 số dương

Mà \(2-x>\dfrac{4}{5}-x\)

=>\(\dfrac{4}{5}< x< 2\)

Vậy...

a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)

Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)

b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)

c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x+3-\dfrac{1}{2}\right)\left(x+3+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)

a) \(\left|x-\dfrac{4}{11}\right|+\left|5+y\right|=0\)

<=>\(\left[{}\begin{matrix}x-\dfrac{4}{11}=0\\5+y=0\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=\dfrac{4}{11}\\y=-5\end{matrix}\right.\)

phần b, c tương tự

a) Vì \(0,\left(3\right)=\dfrac{3-0}{9}=\dfrac{3}{9}=\dfrac{1}{3}\) và \(-0,4\left(2\right)=-\dfrac{42-4}{90}=-\dfrac{38}{90}=-\dfrac{19}{45}\) nên:

\(0,\left(3\right)+3\dfrac{1}{3}-0,4\left(2\right)=\dfrac{1}{3}+\dfrac{10}{3}-\dfrac{19}{45}=\dfrac{11}{3}-\dfrac{49}{45}\)

\(=\dfrac{165-19}{45}=\dfrac{146}{45}\)

b) Vì \(0,\left(5\right)=\dfrac{5-0}{9}=\dfrac{5}{9}\) và \(0,\left(2\right)=\dfrac{2-0}{9}=\dfrac{2}{9}\) nên:

\(\left[0,\left(5\right).0,\left(2\right)\right]:\left(3\dfrac{1}{3}:\dfrac{33}{25}\right)=\left(\dfrac{5}{9}.\dfrac{2}{9}\right):\left(\dfrac{10}{3}.\dfrac{25}{33}\right)=\dfrac{10}{81}:\left(\dfrac{110.25}{33}\right)\)

\(=\dfrac{10}{81}.\dfrac{33}{110.25}=\dfrac{3}{81.25}=\dfrac{1}{27.25}=\dfrac{1}{675}\)