Theo tính chất của 2 đường thẳng song song : để Ax // By 

<=> 2 góc trong cùng phía bù nhau 

Hay \(\widehat{B\text{Ax}}+\widehat{ABy}=180^o\)

\(\Leftrightarrow a+3a=180^o\)

\(\Leftrightarrow4a=180^o\Leftrightarrow a=45^o\)

\(\dfrac{3}{1^2x2^2}+\dfrac{5}{2^2x3^2}+\dfrac{7}{3^2x4^2}+...+\dfrac{19}{9^2x10^2}\\ =\dfrac{2^2}{1^2x2^2}-\dfrac{1^2}{1^2x2^2}+\dfrac{3^2}{2^2x3^2}-\dfrac{2^2}{2^2x3^2}+\dfrac{4^2}{3^2x4^2}-\dfrac{3^2}{3^2x4^2}+...+\dfrac{10^2}{9^2x10^2}-\dfrac{9^2}{9^2x10^2}\\ =\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\\ =1-\dfrac{1}{10^2}< 1\\ =>DPCM\)

helppppp

b, \(\dfrac{2a+9-5a-17-3a}{a+3}=\dfrac{-6a-8}{a+3}=\dfrac{-6\left(a+3\right)+10}{a+3}=-6+\dfrac{10}{a+3}\)

\(\Rightarrow a+3\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

viết lại đề nhé bạn khó hiểu quá

\(\left(x-4\right)^7=\left(x-4\right)^3\)

\(\Rightarrow\left(x-4\right)^3-\left(x-4\right)^7=0\)

\(\Rightarrow\left(x-4\right)^3\left[1-\left(x-4\right)^4\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-4\right)^3=0\\1-\left(x-4\right)^4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\end{matrix}\right.\)

\(\left(x-4\right)^7=\left(x-4\right)^3\)

\(\Leftrightarrow\left(x-4\right)^7-\left(x-4\right)^3=0\)

\(\Leftrightarrow\left(x-4\right)^3\left[\left(x-4\right)^4-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\\left(x-4\right)^4-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\\left(x-4\right)^4=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\\left[{}\begin{matrix}x-4=1\\x-4=-1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\\x=3\end{matrix}\right.\)

Lời giải:

$x^3=x$

$x^3-x=0$

$x(x^2-1)=0$

$x(x-1)(x+1)=0$

$\Rightarrow x=0$ hoặc $x-1=0$ hoặc $x+1=0$

$\Rightarrow x=0$ hoặc $x=1$ hoặc $x=-1$

x³ = x

<=> x = 1 ; x = -1 ; x = 0

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 \(\sqrt{x+5}=0\Leftrightarrow x=-5\left(tm\right)\)