a: \(f\left(-x\right)=\left|-2x-5\right|-\left|-2x+5\right|\)

\(=\left|2x+5\right|-\left|2x-5\right|=-f\left(x\right)\)

=>Hàm số lẻ

b: \(f\left(-x\right)=\dfrac{\left|9-3\cdot\left(-x\right)\right|-\left|3\cdot\left(-x\right)+9\right|}{-x}\)

\(=\dfrac{\left|3x+9\right|-\left|3x-9\right|}{-x}=\dfrac{\left|3x-9\right|-\left|3x+9\right|}{x}=f\left(x\right)\)

=>Hàm số chẵn

1.

ĐK: \(-1\le x\le4\)

Đặt \(\sqrt{x+1}+\sqrt{4-x}=t\left(t\ge0\right)\)

\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{t^2-5}{2}\)

\(PT\Leftrightarrow t+\frac{t^2-5}{2}=5\Rightarrow t^2+2t-15=0\) \(\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)

\(t=3\Rightarrow\sqrt{-x^2+3x+4}=2\) \(\Leftrightarrow-x^2+3x+4=4\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (tm)

2.

ĐK:\(x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t\ge0\right)\)

\(\Rightarrow2\sqrt{x^2-16}=t^2-2x\)

\(PT\Leftrightarrow t=2x-12+t^2-2x\)

\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\) Giải tiếp như trên.

@tran duc huy Bình phương rồi chuyển vế nha.

a/ \(x^4+y^4=1\Rightarrow\left\{{}\begin{matrix}x^4\le1\\y^4\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|x\right|\le1\\\left|y\right|\le1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^4\ge x^6\\y^4\ge y^6\end{matrix}\right.\) \(\Rightarrow x^6+y^6\le x^4+y^4\le1\)

Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}x^4=x^6\\y^4=y^6\\x^4+y^4=1\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y\right)=\left(1;0\right);\left(0;1\right);\left(-1;0\right);\left(0;-1\right)\)

b/ \(\Rightarrow x^9+y^4=1.\left(x^4+y^4\right)\)

\(\Rightarrow x^9+y^9=\left(x^5+y^5\right)\left(x^4+y^4\right)\)

\(\Rightarrow x^9+y^9=x^9+y^9+x^5y^4+x^4y^5\)

\(\Rightarrow x^4y^4\left(x+y\right)=0\Rightarrow\left[{}\begin{matrix}xy=0\\x=-y\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(0;1\right);\left(1;0\right)\)

5. \(y=\dfrac{-3x}{x+2}\)

xác định khi: \(x+2\ne0\Leftrightarrow x\ne-2\)

vậy D= (\(-\infty;+\infty\))\{-2}

6. \(y=\sqrt{-2x-3}\)

xác định khi: \(-2x-3\ge0\Leftrightarrow x\le\dfrac{-3}{2}\)

vậy D= (\(-\infty;\dfrac{-3}{2}\)]

7. \(y=\dfrac{3-x}{\sqrt{x-4}}\)

xác định khi: x-4 >0 <=> x>4

vậy D= (\(4;+\infty\))

8. \(y=\dfrac{2x-5}{\left(3-x\right)\sqrt{5-x}}\)

xác định khi: \(\left\{{}\begin{matrix}3-x\ne0\\5-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x< 5\end{matrix}\right.\)

vậy D= (\(-\infty;5\))\ {3}

9.\(y=\sqrt{2x+1}+\sqrt{4-3x}\)

xác định khi: \(\left\{{}\begin{matrix}2x+1\ge0\\4-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-1}{2}\\x\le\dfrac{4}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{-1}{2}\le x\le\dfrac{4}{3}\)

vậy D= [\(\dfrac{-1}{2};\dfrac{4}{3}\)]

1. \(y=\dfrac{3x-2}{x^2-4x+3}\)

xác định khi : \(x^2-4x+3\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne1\end{matrix}\right.\)

vậy tập xác định là: D = \(\left(-\infty;+\infty\right)\backslash\left\{3;1\right\}\)

2.\(y=2\sqrt{5-4x}\)

xác định khi \(5-4x\ge0\Leftrightarrow x\le\dfrac{5}{4}\)

vậy D= (\(-\infty;\dfrac{5}{4}\)]

3. \(y=\dfrac{2}{\sqrt{x+3}}+\sqrt{5-2x}\)

xác định khi: \(\left\{{}\begin{matrix}x+3>0\\5-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow-3< x\le\dfrac{5}{2}\)

vậy D= (\(-3;\dfrac{5}{2}\)]

4.\(\sqrt{9-x}+\dfrac{1}{\sqrt{x+2}-2}\)

xác định khi: \(\left\{{}\begin{matrix}9-x\ge0\\x+2\ge0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x\ge-2\\x\ne2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le9\\x\ne2\end{matrix}\right.\)

Vậy D= [\(-2;9\)]\{2}

a)\(\frac{x-1}{9}=\frac{8}{3}\)

(x-1)3=8.9

3x-3=72

3x=72+3

3x=75

x=75:3

x=25

b)\(\frac{-x}{4}=\frac{-9}{x}\)

(\(\frac{-x}{4}=\frac{-9}{x}\))=(\(\frac{x}{4}=\frac{9}{x}\))

x.x=9.4

^{x^2=36}

^{x^2=6^2}

^➤x=6

^{c)\(\frac{x}{4}=\frac{18}{x+1}\)}

^x(x+1)=18.4

^x(x+1)=72

^{(Ta có x và x+1 là hai số tự nhiên liên tiếp và một chẵn và một số lẻ)}

Ta có:Ư(72)=1;2;3;4;6;8;9;72;36;12;18;24

Và vì x và x+1 là 2 số tự nhiên liên tiếp, nên:

Ta có bảng sau

d)\(\frac{5}{12}=\frac{-x}{72}\)

(-x).12=5.72

(-x).12=360

(-x)=360:12

(-x)=30

➤x=-30

e)\(\frac{x+3}{-15}=\frac{1}{3}\)

(x+3)3=1.(-15)

3x+9=-15

3x=(-15)-9

3x=-24

x=(-24):3

➤x=-8

a: \(f\left(-x\right)=\dfrac{-x^5+x}{\sqrt{\left(-x\right)^2+\left|-x\right|}}=-f\left(x\right)\)

=>f(x) lẻ

b: \(f\left(-x\right)=\left(\left|9+2x\right|-\left|9-2x\right|\right)\left(-x+5x^3\right)\)

\(=f\left(x\right)\)

=>f(x) chẵn

c: \(f\left(-x\right)=\dfrac{\left|3+x\right|-\left|3-x\right|}{\left(-x\right)^4+1}=-f\left(x\right)\)

=>f(x) lẻ