x : 10 = 24 x + (-10)

x : 10 = 24x - 10

x = 240x - 100

240x - x = 100 

239  x = 100

x = 100: 239 

x = 100/239 

x : 10 = 24 x + (-10)

x : 10 = 24x - 10

x = 240x - 100

240x - x = 100 

239  x = 100

\(A=\left|2x-2\right|+\left|2x-13\right|=\left|2x-2\right|+\left|13-2x\right|\ge\left|2x-2+13-2x\right|=11\)

Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}x\ge1\\x\le\dfrac{13}{2}\end{matrix}\right.\Leftrightarrow1\le x\le\dfrac{13}{2}\)

=> x = 1 ; 2 ; 3 ; 4 ; 5 ; 6 

Ta có: \(A=\left|2x-2\right|+\left|2x-2013\right|=\left|2x-2\right|+\left|2013-2x\right|\)

\(\ge \left|2x-2+2013-2x\right|=2011\)

Dấu ''='' xảy ra khi và chỉ khi \(\left(2x-2\right)\left(2013-2x\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}2x-2\ge0\\2013-2x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}2x-2\le0\\2013-2x\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\x\le\dfrac{2013}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\x\ge\dfrac{2013}{2}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow1\le x\le\dfrac{2013}{2}\); \(x\in Z\)

Ta có ^AOC = 2^AOD ; ^BOC = 2^BOE 

Cộng vế với vế 

^AOC + ^BOC = 2(^AOD + ^BOE) 

180⁰ = 2(^AOD + ^BOE) 

<=> ^AOD + ^BOE = 90⁰

Theo bài ra ta có 

\(\left\{{}\begin{matrix}\widehat{AOD}-\widehat{BOE}=30^0\\\widehat{AOD}+\widehat{BOE}=90^0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\widehat{AOD}=120^0\\\widehat{BOE}=\widehat{AOD}-30^0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\widehat{AOD}=60^0\\\widehat{BOE}=30^0\end{matrix}\right.\)

-> ^AOC = 2^AOD = 120⁰ ; ^BOC = 2^BOD = 60⁰

Ta có: \(\widehat{AOD}-\widehat{BOE}=30^o\Leftrightarrow\widehat{AOD}=30^o+\widehat{BOE}\)

Vì OD và OE lần lượt là phân giác của góc AOC và góc BOC nên: 

\(\widehat{AOD}=\widehat{DOC};\widehat{COE}=\widehat{EOB}\)

\(\Rightarrow\widehat{DOC}=30^o+\widehat{COE}\left(1\right)\)

Lại có \(\widehat{AOC};\widehat{COB}\) là 2 góc kề nhau nên \(\widehat{DOC}+\widehat{COE}=90^o\Rightarrow\widehat{DOC}=90^o-\widehat{COE}\left(2\right)\)

Từ (1) và (2) suy ra: \(30^o+\widehat{COE}=90^o-\widehat{COE}\Leftrightarrow2\widehat{COE}=60^o\Leftrightarrow\widehat{BOC}=60^o\)

\(\Rightarrow\widehat{AOC}=180^o-60^o=120^o\)

A = \(\sqrt{x}\) + 2 

\(\sqrt{x}\) ≥ 0 ⇔ A = \(\sqrt{x}\) + 2 ≥ 2 ⇔ A(min) = 2 ⇔ x = 0

B = \(\sqrt{x+5}\) - 3

\(\sqrt{x+5}\) ≥ 0 ⇔ \(\sqrt{x+5}\) - 3 ≥ -3 ⇔ A(min)  =-3 ⇔ x = -5

a)ĐKXĐ: \(x\ge0\)

Ta có: \(\sqrt{x}+2\)

Ta thấy \(\sqrt{x}\ge0\forall x\in R\)

\(\Rightarrow\sqrt{x}+2\ge2\forall x\in R\)

Dấu ''='' xảy ra khi \(\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

b) ĐKXĐ: \(x\ge-5\)

Ta thấy \(\sqrt{x+5}\ge0\forall x\in R\)

\(\Rightarrow\sqrt{x+5}-3\ge-3\forall x\in R\)

Dấu ''='' xảy ra khi \(\sqrt{x+5}=0\Leftrightarrow x=-5\left(tm\right)\)

n² + 3 ⋮ n -1 ⇔ (n-1)(n+1) + 4 ⋮ n-1 ⇔ 4 ⋮ n- 1 ⇔ n-1 ϵ {-4; -1; 1; 4}

⇔  n ϵ { -3; 0; 2; 5}

`(0,5)^2 + 2x = (0,7)^2`

`0,25 + 2x = 0,49`

`2x=0,49-0,25`

`2x= 0,24`

`x=0,24 : 2`

`x=0,12`

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`1-2x = 9/8 + 7/5 : 2/5`

`1-2x = 9/8 + 7/5 xx 5/2`

`1-2 x= 9/8 +7/2`

`1-2x = 37/8`

`2x=1-37/8`

`2x = -29/8`

`x=-29/8 : 2`

`x=-29/8 . 1/2`

`x=-29/16`

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`c` Thiếu đề