Ta có :

\(x-y=10\)

\(\Rightarrow\left(x-y\right)^2=100\left(x>y\right)\)

\(\Rightarrow\left(x+y\right)^2-4xy=100\)

\(\Rightarrow\left(x+y\right)^2=100+4xy\)

mà \(x.y=24\)

\(\Rightarrow\left(x+y\right)^2=100+4.24=196\)

\(\Rightarrow\left(x+y\right)^2=14^2\)

\(\Rightarrow\left[{}\begin{matrix}x+y=4\\x+y=-4\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}D=x+y=4\\D=x+y=-4\end{matrix}\right.\)

Đính Chính

\(x+y=\pm14\)

Anh đang trên xe đi chơi nên xin phép gõ không latex

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(2x+1)^2 - 4x^2 + 4x -1 =0

<=> (2x+1)^2 - (2x-1)^2=0

<=> (2x + 1 + 2x -1). (2x+1 - 2x +1)=0

<=> 4x. 2= 0

<=> 8x=0

<=> x =0

`@` `\text {Ans}`

`\downarrow`

`(2x + 1)^2 - 4x^2 + 4x - 1 = 0`

`<=> 4x^2 + 4x + 1 - 4x^2 + 4x - 1 = 0`

`<=> (4x^2 - 4x^2) + (4x + 4x) + (1 - 1) = 0`

`<=> 8x = 0`

`<=> x = 0`

Vậy, `x = 0.`

2/5= 6/15

1/3= 5/15

Ta thấy 6/15 > 5/15 do 6>5 (cùng mẫu)

Nên 2/5 không bằng 1/3

sai vì \(\dfrac{2}{5}và\dfrac{1}{3}\) là 2 phân số tối giản

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

\(\Leftrightarrow\left(16x^2-8x+1\right)-4\left(4x^2+4x+1\right)-x-4=0\)

\(\Leftrightarrow16x^2-8x+1-16x^2-16x-4-x-4=0\)

\(\Leftrightarrow25x-7=0\)

\(\Leftrightarrow25x=7\)

\(\Leftrightarrow x=\dfrac{7}{25}\)

`@` `\text {Ans}`

`\downarrow`

`(4x - 1)^2 - 4(2x + 1)^2 - x - 4 = 0`

`<=> 16x^2 - 8x + 1 - 4(4x^2 + 4x + 1) - x - 4 = 0`

`<=> 16x^2 - 8x + 1 - 16x^2 - 16x - 4 - x - 4 = 0`

`<=> -25x - 7 = 0`

`<=> -25x = 7`

`<=> x =`\(\dfrac{-7}{25}\)

Vậy, \(x= \dfrac{-7}{25}\)

x¹⁰ = x

x¹⁰ - x = 0

x(x⁹ - 1) = 0

x = 0 hoặc x⁹ - 1 = 0

*) x⁹ - 1 = 0

x⁹ = 1

x = 1

Vậy x = 0; x = 1

\(x^{10}=x\)

\(\Rightarrow x=1\)

mik cần gấp giúp vs ạ

Gọi số cần tìm là ab

Theo bài ra ta có:

\(ab3-ab=750\)

\(\Rightarrow abx10+3-abx1=750\)

\(\Rightarrow abx9+3=750\)

\(\Rightarrow abx9=750-3=747\)

\(\Rightarrow ab=747:9=83\)

Vậy số cần tìm là 83

\(C=4x^2+y^2-4x+8y+12\)

\(C=4x^2-4x+1+y^2+8y+16-5\)
\(C=\left(4x^2-4x+1\right)+\left(y^2+8y+16\right)-5\)

\(C=\left(2x-1\right)^2+\left(y+4\right)^2-5\)

Mà: \(\left\{{}\begin{matrix}\left(2x-1\right)^2\ge0\forall x\\\left(y+4\right)^2\ge0\forall x\end{matrix}\right.\)

Nên: \(C=\left(2x-1\right)^2+\left(y+4\right)^2-5\ge-5\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}2x-1=0\\y+4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-4\end{matrix}\right.\)

Vậy: \(C_{min}=-5\) khi \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-4\end{matrix}\right.\)

D = \(\dfrac{1}{1\times1981}\) + \(\dfrac{1}{2\times1982}\)+...+ \(\dfrac{1}{25\times2005}\)

D =\(\dfrac{1}{1980}\times\)( \(\dfrac{1980}{1\times1981}\)+ \(\dfrac{1980}{2\times1982}\)+....+ \(\dfrac{1980}{25\times2005}\))

D = \(\dfrac{1}{1980}\) \(\times\)(\(\dfrac{1}{1}\) - \(\dfrac{1}{1981}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{1982}\)+....+ \(\dfrac{1}{25}\) \(\times\) \(\dfrac{1}{2005}\))

D= \(\dfrac{1}{1980}\)[( \(\dfrac{1}{1}\) + \(\dfrac{1}{2}\) +....+ \(\dfrac{1}{25}\)) - ( \(\dfrac{1}{1981}\)+ \(\dfrac{1}{1982}\)+...+ \(\dfrac{1}{2005}\))]

E =\(\dfrac{1}{25}\times\)( \(\dfrac{1}{1\times26}\)+ \(\dfrac{1}{2\times27}\)+...+ \(\dfrac{1}{1980\times2005}\))

E =  \(\dfrac{1}{25}\). (\(\dfrac{25}{1\times26}\) + \(\dfrac{25}{2\times27}\)+....+ \(\dfrac{25}{1980\times2005}\))

E = \(\dfrac{1}{25}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{26}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{27}\)+...+\(\dfrac{1}{1980}\)-\(\dfrac{1}{2005}\))

E=\(\dfrac{1}{25}\)[\(\dfrac{1}{1}\)+...+ \(\dfrac{1}{25}\)+ (\(\dfrac{1}{26}\)+...+\(\dfrac{1}{1980}\)) - (\(\dfrac{1}{26}\)+...+\(\dfrac{1}{1980}\)) - (\(\dfrac{1}{1981}\)+..\(\dfrac{1}{2005}\))]

E = \(\dfrac{1}{25}\) .[\(\dfrac{1}{1}\)+\(\dfrac{1}{2}\)+...+\(\dfrac{1}{25}\) - (\(\dfrac{1}{1981}\)+\(\dfrac{1}{1982}\)+...+ \(\dfrac{1}{2005}\))]

\(\dfrac{D}{E}\) = \(\dfrac{\dfrac{1}{1980}}{\dfrac{1}{25}}\) = \(\dfrac{5}{396}\)