Question

[Linh Nguyen]

Mấy bạn giải giúp mik bài 7 và bài 9 vs ạ

Answer 1

[Lê Thị Thục Hiền]

Bài 7:

a)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge m+1\\x\ge\dfrac{m}{4}\end{matrix}\right.\)

TH1: \(m+1< \dfrac{m}{4}\Rightarrow m< -\dfrac{4}{3}\)

\(\Rightarrow x\ge\dfrac{m}{4}\)\(\Rightarrow x\in\)\([\dfrac{m}{4};+\)\(\infty\)\()\)

Để hàm số xác định với mọi x dương \(\Leftrightarrow\)\(\left(0;+\infty\right)\subset\)\([\dfrac{m}{4};+\)\(\infty\)\()\)

\(\Leftrightarrow\dfrac{m}{4}\ge0\Leftrightarrow m\ge0\) kết hợp với \(m< -\dfrac{4}{3}\Rightarrow m\in\varnothing\)

TH2:\(m+1\ge\dfrac{m}{4}\Rightarrow m\ge-\dfrac{4}{3}\)

\(\Rightarrow x\ge m+1\)\(\Rightarrow\)\(x\in\)\([m+1;+\)\(\infty\))

Để hàm số xác định với mọi x dương \(\Leftrightarrow\)\(\left(0;+\infty\right)\subset\)\([m+1;\)\(+\infty\)\()\)

\(\Leftrightarrow m+1\le0\Leftrightarrow m\le-1\) kết hợp với \(m\ge-\dfrac{4}{3}\)

\(\Rightarrow m\in\left[-\dfrac{4}{3};-1\right]\)

Vậy...

b)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge2-m\\x\ne-m\end{matrix}\right.\)\(\Rightarrow x\in\)\([2-m;+\)\(\infty\)) (vì \(-m< 2-m\))

Để hàm số xác ddingj với mọi x dương

\(\Leftrightarrow\left(0;+\infty\right)\subset\)\([2-m;+\)\(\infty\))

\(\Leftrightarrow2-m\le0\Leftrightarrow m\ge2\)

Vậy...

Answer 2

[Lê Thị Thục Hiền]

Bài 9:

a)Đặt \(f\left(x\right)=x^2+2x-2\)

TXĐ:\(D=R\)

TH1:\(x\in\left(-\infty;-1\right)\)

Lấy \(x_1;x_2\in\left(-\infty;-1\right)\)\(:x_1\ne x_2\) 

Xét \(I=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{x_1^2+2x_1-2-\left(x_2^2+2x_2-2\right)}{x_1-x_2}=x_1+x_2+2\)

Vì \(x_1;x_2\in\left(-\infty;-1\right)\Rightarrow x_1+x_2< -1+-1=-2\)\(\Leftrightarrow x_1+x_2+2< 0\)

\(\Rightarrow I< 0\)

Suy ra hàm nb trên \(\left(-\infty;-1\right)\)

TH2:\(x\in\left(-1;+\infty\right)\)

Lấy \(x_1;x_2\in\left(-1;+\infty\right)\)\(:x_1\ne x_2\) 

Xét \(I=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{x_1^2+2x_1-2-\left(x_2^2+2x_2-2\right)}{x_1-x_2}=x_1+x_2+2>0\)

Suy ra hàm đb trên \(\left(-1;+\infty\right)\)

Vậy...

b)Đặt \(f\left(x\right)=\dfrac{2}{x-3}\)

TXĐ:\(D=R\backslash\left\{3\right\}\)

TH1:\(x\in\left(-\infty;3\right)\)

Lấy \(x_1;x_2\in\left(-\infty;3\right)\)\(:x_1\ne x_2\) 

Xét \(I=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{\dfrac{2}{x_1-3}-\dfrac{2}{x_2-3}}{x_1-x_2}=\dfrac{-2}{\left(x_1-3\right)\left(x_2-3\right)}\)

Vì \(x_1;x_2\in\left(-\infty;3\right)\Rightarrow x_1-3< 0;x_2-3< 0\Rightarrow\left(x_1-3\right)\left(x_2-3\right)>0\)

\(\Rightarrow I< 0\)

Suy ra hàm nb trên \(\left(-\infty;3\right)\)

TH2:\(x\in\left(3;+\infty\right)\)

Lấy \(x_1;x_2\in\left(3;+\infty\right)\)\(:x_1\ne x_2\) 

Xét \(I=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{\dfrac{2}{x_1-3}-\dfrac{2}{x_2-3}}{x_1-x_2}=\dfrac{-2}{\left(x_1-3\right)\left(x_2-3\right)}\)

Vì \(x_1;x_2\in\left(3;+\infty\right)\Rightarrow x_1-3>0;x_2-3>0\Rightarrow\left(x_1-3\right)\left(x_2-3\right)>0\)

\(\Rightarrow I< 0\)

Suy ra hàm nb trên \(\left(3;+\infty\right)\)

Vậy hàm nb trên \(\left(-\infty;3\right)\) và \(\left(3;+\infty\right)\)