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\(\Rightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{3}{10}x-\dfrac{1}{4}x\)
\(\Rightarrow\dfrac{33}{20}=\dfrac{11}{20}x\)
\(\Rightarrow x=\dfrac{33}{20}\div\dfrac{11}{20}\)
\(\Rightarrow x=3\)
\(1\dfrac{1}{4}-x\dfrac{1}{4}=x\cdot30\%\cdot\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-x\dfrac{1}{4}=x\cdot\dfrac{3}{10}-\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Leftrightarrow25-5x=6x-8\)
\(\Leftrightarrow-5x-6x=-8-25\)
\(\Leftrightarrow-11x=-33\)
\(\Leftrightarrow x=3\)
Vậy x = 3
A = 10,11 + 11,12 + 12,13 + . . .+ 98,99 + 99,10
Ta có :
10,11 = 10 + 0,11
11,12 = 11 + 0,12
12,13 = 12 + 0,13
. . . . . . . . . . . . . .
97,98 = 97 + 0,98
98,99 = 98 + 0,99
99,10 = 99 + 0,10
Đặt B = 10 + 11 + 12 + 13 + . .. +98 + 99
và C = 0,11 + 0,12 + 0,13 + . . . .+ 0,98 + 0,99 + 0,10
- - > 100C = 11 + 12 + 13 + . . .+ 98 + 99 + 10
Ta chỉ việc tính B là suy ra C !
B = 10 + 11 + 12 + 13 + . .. +98 + 99
B = (10+99)+(11+98)+(12+97)+. . . +(44+65) + (45 + 64)
Vì từ 10 đến 99 có tất cả 90 số . Ta sẽ có 90/2 = 45 cặp
Mỗi cặp có tổng là 10 + 99 = 11 + 98 = . .= 45 +64 = 109
Vậy ta có B = 45.109 = 4905
Với A = 4905 . Ta thấy 100C = 10 + 11 + 12 +. . + 98 + 99 =B
- - > 100C = 4905 . Hay C = 4905/100 = 49,05
Vậy A = B + C = 4905 + 49,05 = 4954,05
Từ đề bài ta có:
\(T=\dfrac{1+2}{2}.\dfrac{1+3}{3}.\dfrac{1+4}{4}...\dfrac{1+98}{98}.\dfrac{1+99}{99}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}\)
\(=50\).
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5......99}{3.4.5......99}.\dfrac{100}{2}\)
\(T=50\)
Theo mk được biết thì Shinichi và Kid là hai anh em nên mk thích cả hai
ko đc viết mỗi kq




àm vòng tròn




giải giùm tớ nha


a: \(\Leftrightarrow x^2=16\)
hay \(x\in\left\{4;-4\right\}\)
b: \(\Leftrightarrow\dfrac{20}{6-5x}=\dfrac{5}{14}\)
=>6-5x=56
=>5x=-50
hay x=-10
c: =>7/17=x/119=-35/y
=>x=49; y=-85
d: =>x/27=1/9+1/9=2/9
hay x=6
\(a,\dfrac{x}{-2}=\dfrac{-8}{x}\)
\(\Leftrightarrow x^2=16\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(b,\dfrac{5}{14}=\dfrac{20}{6-5x}\)
\(\Leftrightarrow30-25x=280\)
\(\Leftrightarrow25x=30-280=-250\)
\(\Leftrightarrow x=-10\)
\(c,\dfrac{7}{17}=\dfrac{x}{119}=\dfrac{-35}{y}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{119.7}{17}=49\\y=\dfrac{-35.17}{7}=-85\end{matrix}\right.\)
\(d,\dfrac{x}{27}+\dfrac{-3}{27}=\dfrac{1}{9}\)
\(\dfrac{x-3}{27}=\dfrac{1}{9}\)
\(9x-27=27\)
\(9x=54\Rightarrow x=6\)
a)\(\dfrac{x}{-2}=-\dfrac{8}{x}\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=-4\) hoặc x=4
b)\(-\dfrac{5}{-14}=\dfrac{20}{6-5x}\)
\(\Rightarrow-5\left(6-5x\right)=-280\)
\(-30+25x=-280\)
25x=-250
x=-10
c)\(\dfrac{7}{17}=\dfrac{x}{119}=-\dfrac{35}{y}\)
\(\Rightarrow\dfrac{x}{119}=\dfrac{7}{17}\Rightarrow x=49\)
\(-\dfrac{35}{y}=\dfrac{7}{17}\Rightarrow y=-85\)
d)\(\dfrac{x}{27}+\dfrac{-3}{27}=\dfrac{1}{9}\)
\(\dfrac{x-3}{27}=\dfrac{1}{9}\)
\(x-3=3\)
x=6
\(a,\dfrac{x}{-2}=\dfrac{-8}{x}\)
\(\Rightarrow2x=-2.\left(-8\right)=16\)
\(\Rightarrow x=\dfrac{16}{2}=8\)
\(b,\dfrac{-5}{-14}=\dfrac{20}{6-5x}\)
\(\Rightarrow-5.\left(6-5x\right)=20.\left(-14\right)\)
\(\Rightarrow-30+25x=280\)
\(\Leftrightarrow25x=280-\left(-30\right)\)
\(\Rightarrow25x=310\)
\(\Rightarrow x=\dfrac{310}{25}=12,4\)
\(c,\dfrac{7}{17}=\dfrac{x}{119}=\dfrac{-35}{y}\)
\(\Rightarrow x=\dfrac{119.7}{17}=49\)
\(\Rightarrow y=\dfrac{-35.17}{7}=-85\)
\(d,\dfrac{x}{27}+\dfrac{3}{27}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{x}{27}=\dfrac{1}{9}-\dfrac{-3}{27}\)
\(\Rightarrow\dfrac{x}{27}=\dfrac{6}{27}\)
\(\Rightarrow x=6\)
\(a:\Rightarrow x^2=\left(-2\right)\cdot\left(-8\right)=16\\ x^2=8\\ b:\Rightarrow\left(-5\right)\cdot6-5x=\left(-14\right)\cdot20=-280\\ \Rightarrow6-5x=56\\ \Rightarrow5x=6-56\\ \Rightarrow5x=-50\\ \Rightarrow x=-10\\ c:\dfrac{7}{17}=\dfrac{x}{119}\Rightarrow7\cdot119=17\cdot x=833\\ \Rightarrow x=49\\ \dfrac{49}{119}=\dfrac{-35}{y}\Rightarrow49\cdot y=119\cdot\left(-35\right)=-4165\\ \Rightarrow y=-85\\ d:\dfrac{x}{27}=\dfrac{1}{9}-\dfrac{-3}{27}\\ \dfrac{x}{27}=\dfrac{0}{27}\)