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\(\left[0,\left(32\right).1,\left(5\right)-0,\left(25\right)\right].\dfrac{11}{83}\)
\(=\left[\dfrac{32}{99}.\left(1+\dfrac{5}{9}\right)-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\left[\dfrac{32}{99}.\dfrac{14}{9}-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\left[\dfrac{448}{891}-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\dfrac{223}{891}.\dfrac{11}{83}\)
\(=\dfrac{223}{6723}\)
\(H=\left[0,\left(32\right).1,\left(5\right)-0,\left(25\right)\right].\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{32}{99}.\dfrac{14}{9}-\dfrac{25}{99}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{25}{99}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{225}{891}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{448-225}{891}.\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{223}{891}.\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{2453}{73953}\)
\(\Leftrightarrow H=\dfrac{223}{6723}\)
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
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>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) \(\left|x-\dfrac{4}{11}\right|+\left|5+y\right|=0\)
<=>\(\left[{}\begin{matrix}x-\dfrac{4}{11}=0\\5+y=0\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=\dfrac{4}{11}\\y=-5\end{matrix}\right.\)
phần b, c tương tự
a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)
=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)
=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)
=> \(x=\dfrac{-1}{11}\)
Đây toán 8 mà? :v
a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)
\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)
\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)
\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)
\(\Leftrightarrow\left(11+1\right)x=0\)
\(\Leftrightarrow11x+1=0;x=0\)
\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)
Vậy....
1)(-1/2)^2:1/4-2.(-1/2)^3+căn 4
=1/4:1/4-2.-1/8+2
= 1-(-1/4)+2
=1+1/4+2=13/4
2) 3-(-6/7)^0+căn 9 :2
= 3-1+3:2
=3-1+3/2=7/2
3) (-2)^3+1/2:1/8-căn 25 + |-64|
= -8+4-5+64= 55
4) (-1/2)^4+|-2/3|-2007^0
= 1/16+2/3-1
= -13/48
5) = 178/495:623/495-17/60:119/120
= 2/7-2/7=0
6) [2^3.(-1/2)^3+1/2]+[25/22+6/25-3/22+19/25+1/2]
= [-1+1/2]+[(25/22-3/22)+(6/25+19/25)+1/2]
= -1/2+[1+1+1/2]
= -1/2+5/2=2
Mấy cái dấu chấm đó là nhân nha bn!
a: =>|3x-5|=|x+2|
=>3x-5=x+2 hoặc 3x-5=-x-2
=>2x=7 hoặc 4x=3
=>x=7/2 hoặc x=3/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
c: \(\Leftrightarrow\left|3x-5\right|=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(3x-5-x+2\right)\left(3x-5+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(2x-3\right)\left(4x-7\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
d: \(\dfrac{11}{2}\le\left|x\right|< \dfrac{17}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{11}{2}< =x< \dfrac{17}{2}\\-\dfrac{17}{2}< x< =-\dfrac{11}{2}\end{matrix}\right.\)
đề sai rồi bạn