Trường hợp 1: x - 1 ≥ 0 → x ≥ 1

→ x - 1 = 2x - 5

→ x - 2x = -5 + 1

→ - x = - 4

→ x = 4

Trường hợp 2: x - 1 ≤ 1 → x ≤ 1

→ - ( x - 1) = 2x - 5

→ - x + 1 = 2x - 5

→ -x - 2x = -5 - 1

→  -3x = 6

→  x = 2 (loại)

Vậy, x = 4

Để (d)//(d') thì \(\left\{{}\begin{matrix}3-m=2\\m-5\ne3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m=1\\m\ne8\end{matrix}\right.\)

=>m=1

=>(d): y=(3-1)x+1-5=2x-4

Ta có: (d): y=2x-4; (d'): y=2x+3

Lấy A(3;2) thuộc (d)

=>KHoảng cách từ (d) đến (d') sẽ là khoảng cách từ A đến (d')

(d'): y=2x+3

=>2x-y+3=0

Khoảng cách từ A đến (d') là:

\(\dfrac{\left|2\cdot3+\left(-1\right)\cdot2+3\right|}{\sqrt{2^2+\left(-1\right)^2}}=\dfrac{7}{\sqrt{5}}=\dfrac{7\sqrt{5}}{5}\)

=>\(d\left(\left(d\right);\left(d'\right)\right)=\dfrac{7\sqrt{5}}{5}\)

a: =>\(x^2\cdot2\sqrt{2}+x\left(2+2\sqrt{2}\right)+4=0\)

\(\text{Δ}=\left(2\sqrt{2}+2\right)^2-4\cdot2\sqrt{2}\cdot4=12-24\sqrt{2}< 0\)

=>PTVN

b: 

\(\Leftrightarrow2x^2+2x+\sqrt{3}-x^2+2\sqrt{3}x+\sqrt{3}=0\)

=>\(x^2+x\left(2\sqrt{3}+2\right)+2\sqrt{3}=0\)

\(\text{Δ}=\left(2\sqrt{3}+2\right)^2-4\cdot2\sqrt{3}=16>0\)

PT có hai nghiệm là;

\(\left\{{}\begin{matrix}x_1=\dfrac{-2\sqrt{3}-2-4}{2}=-\sqrt{3}-3\\x=\dfrac{-2\sqrt{3}-2+4}{2}=-\sqrt{3}+1\end{matrix}\right.\)

Ta có: \(3x-2⋮x-5\)

\(\Leftrightarrow3x-15+13⋮x-5\)

mà \(3x-15⋮x-5\)

nên \(13⋮x-5\)

\(\Leftrightarrow x-5\inƯ\left(13\right)\)

\(\Leftrightarrow x-5\in\left\{1;-1;13;-13\right\}\)

hay \(x\in\left\{6;4;18;-8\right\}\)

Vậy: \(x\in\left\{6;4;18;-8\right\}\)

Ta có : \(3x-2⋮x-5\)

\(\Leftrightarrow3x-15+13⋮x-5\)

Thấy \(3x-15=3\left(x-5\right)⋮x-5\)

Nên để \(3x-2⋮x-5\)

\(\Leftrightarrow13⋮x-5\)

\(\Leftrightarrow x-5\inƯ_{\left(13\right)}\)

\(\Leftrightarrow x-5\in\left\{1;-1;13;-13\right\}\)

\(\Leftrightarrow x\in\left\{5;4;18;-8\right\}\)

Vậy ...

\(3x-2⋮x-5\)

\(\Leftrightarrow\left(3x-2\right)-\left(3x-15\right)⋮x-5\)

\(\Leftrightarrow7⋮x-5\)

\(\Rightarrow\) x-5 là ước của 7

\(\Rightarrow x-5\in\left\{1;-1;7;-7\right\}\)

\(\Rightarrow x\in\left\{6;4;12;-2\right\}\)

Min của A là 99 khi (x;y)=(2;3).

Chúc abh học tốt.

\(A=\left(x+\dfrac{4}{x}\right)+5\left(\dfrac{y}{3}+\dfrac{3}{y}\right)+17\left(x+y\right)\)

\(A\ge2\sqrt{\dfrac{4x}{x}}+5.2\sqrt{\dfrac{3y}{3y}}+17.5=99\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;3\right)\)

a/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}\sqrt{x^2+1}+\dfrac{2x}{x}+\dfrac{1}{x}}{\dfrac{x}{x}\sqrt[3]{\dfrac{2x^3}{x^3}+\dfrac{x}{x^3}+\dfrac{1}{x^3}}+\dfrac{x}{x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+2}{\sqrt[3]{2}+1}=+\infty\)

b/ \(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2.1^2-1+1}-\sqrt[3]{2.1+3}}{3.1^2-2}=...\)

c/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+x\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{x\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)

\(\text{#ID07 - DNfil}\)

`A = -(x + 1)^2 + 5`

Ta có: `(x + 1)^2 \ge 0` `AA` `x`

`=> -(x + 1)^2 \le 0` `AA` `x`

`=> -(x + 1)^2 + 5 \le 5` `AA` `x`

Vậy, GTLN của A là `5` khi `(x + 1)^2 = 0 => x + 1 = 0 => x = -1`

________

2.

`2x - 0,7 = 1,3`

`=> 2x = 1,3 + 0,7`

`=> 2x = 2`

`=> x = 1`

Vậy, `x = 1`

__

`x - \sqrt{25} = (2/5 - 6/5)`

`=> x - \sqrt{25} = -3/5`

`=> x = -3/5 + \sqrt{25}`

`=> x = -3/5 + 5`

`=> x = 22/5`

Vậy, `x = 22/5`

__

`3/4 + 1/4 \div x = 2/5`

`=> 1/4 \div x = 2/5 - 3/4`

`=> 1/4 \div x = -7/20`

`=> x = 1/4 \div (-7/20)`

`=> x = -5/7`

Vậy, `x = -5/7.`