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\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ =>\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ =>\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{2}{3}-\dfrac{1}{25}=\dfrac{3}{5}\\ =>x=1\)

\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ \Rightarrow\left(\dfrac{1}{5}x+\dfrac{2}{5}x\right)+\left(\dfrac{1}{25}+\dfrac{2}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x+\dfrac{53}{75}=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{53}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{45}{75}=\dfrac{3}{5}\\ \Rightarrow x=\dfrac{3}{5}:\dfrac{3}{5}\\ \Rightarrow x=1\)

\(\dfrac{1}{5}\)x(X+\(\dfrac{1}{5}\))+\(\dfrac{2}{5}\)x(X+\(\dfrac{5}{3}\))           = \(\dfrac{98}{75}\)
=> \(\dfrac{1}{5}\)X+\(\dfrac{2}{5}\)+\(\dfrac{6}{15}\)X+\(\dfrac{31}{15}\)           = \(\dfrac{98}{75}\)
=> (\(\dfrac{1}{5}\)X+\(\dfrac{6}{15}\)X)+(\(\dfrac{2}{5}\)+\(\dfrac{31}{15}\))      =\(\dfrac{98}{75}\)
=> X x(\(\dfrac{1}{5}\)+\(\dfrac{6}{15}\))+(\(\dfrac{6}{15}\)+\(\dfrac{31}{15}\))    =\(\dfrac{98}{75}\)
=> X x(\(\dfrac{3}{15}\)+\(\dfrac{6}{15}\))+\(\dfrac{37}{15}\)            = \(\dfrac{98}{75}\)
=>X x\(\dfrac{9}{15}\)+\(\dfrac{37}{15}\)                      =\(\dfrac{98}{75}\)
=>X x\(\dfrac{9}{15}\)                             =\(\dfrac{98}{75}\)-\(\dfrac{37}{15}\)
=>X                                     =\(\dfrac{-29}{25}\):\(\dfrac{9}{15}\)
=>X                                     =\(\dfrac{-29}{15}\)

a, Ta có: \(\left(x-\dfrac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow A=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu " = " khi \(\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(MIN_A=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)

b, Để B lớn nhất thì \(\left(x-\dfrac{2}{3}\right)^2+9\) nhỏ nhất

Ta có: \(\left(x-\dfrac{2}{3}\right)^2+9\ge9\)

\(\Leftrightarrow B=\dfrac{4}{\left(x-\dfrac{2}{3}\right)^2+9}\le\dfrac{4}{9}\)

Dấu " = " khi \(\left(x-\dfrac{2}{3}\right)^2=0\Leftrightarrow x=\dfrac{2}{3}\)

Vậy \(MAX_B=\dfrac{4}{9}\) khi \(x=\dfrac{2}{3}\)

Ta có:

\(\dfrac{\left(x-1\right)^2}{9}=\dfrac{3}{x-1}\)

=> (x - 1)² . (x - 1) = 9 . 3

(x - 1)³ = 27

=> \(x=\sqrt[3]{27}+1=3+1=4\)

Vậy x = 4

X = 4

\(A=\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+9\right)}+\dfrac{1}{\left(x+9\right)\left(x+11\right)}\)\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}+\dfrac{1}{x+9}-\dfrac{1}{x+11}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+11}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{x+11}{\left(x+1\right)\left(x+11\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+11\right)}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{x+11-x-1}{\left(x+1\right)\left(x+11\right)}\right)=\dfrac{1}{2}.\dfrac{10}{\left(x+1\right)\left(x+11\right)}=\dfrac{10}{2\left(x+1\right)\left(x+11\right)}\)

1 nghịch biến(a<0) 

2 đồng biến

3,4 thay các g trị tm đk vào

hojk tốt

\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{3}{130}\)

ĐK: \(\left\{{}\begin{matrix}x\ne-1\\x\ne-2\\x\ne-3\\x\ne-4\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{130\left(x+3\right)\left(x+4\right)+130\left(x+1\right)\left(x+4\right)+130\left(x+1\right)\left(x+2\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{3\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}\)

\(\Leftrightarrow3x^2+15x-378=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-14\end{matrix}\right.\)

@ngonhuminh @Nguyễn Huy Thắng @Đức Minh@Hoang Hung Quan@Nguyễn Huy Tú@Hoàng Thị Ngọc Anh.... và mb khác giúp mik đi mà, cần gấp lắm T_T

Do mỗi số hạng ở vế trái nằm trong dấu giá trị tuyệt đối mà vế phải 100 là số dương nên x cũng là số dương

Do x dương nên ta có:

\(x+\dfrac{1}{1.2}+x+\dfrac{1}{2.3}+...+x+\dfrac{1}{99.100}=100x\)

Dãy trên có 99 số hạng nên

\(99x+\left(x-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(1-\dfrac{1}{100}=x\Rightarrow x=\dfrac{99}{100}\)

Vậy \(x=\dfrac{99}{100}\)

\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}=\dfrac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}+\dfrac{1}{2}=2\\x=-\dfrac{3}{2}+\dfrac{1}{2}=-1\end{matrix}\right.\)

\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

x = 2 hoặc x = -1