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a)
ĐKXĐ: \(2x\geq 0\Leftrightarrow x\geq 0\)
Vậy TXĐ của $x$ là \(D= [0;+\infty)\)
b)
ĐK: \((2x-1)(x+3)\neq 0\Leftrightarrow \left\{\begin{matrix} 2x-1\neq 0\\ x+3\neq 0\end{matrix}\right.\Leftrightarrow \Leftrightarrow \left\{\begin{matrix} x\neq \frac{1}{2}\\ x\neq -3\end{matrix}\right.\)
Vậy TXĐ \(D=\mathbb{R}\setminus \left\{\frac{1}{2}; -3\right\}\)
c)
ĐK: \(8x^3+1\neq 0\Leftrightarrow x^3\neq \frac{-1}{8}\Leftrightarrow x\neq \frac{-1}{2}\)
Vậy TXĐ \(D=\mathbb{R}\setminus \left\{\frac{-1}{2}\right\}\)
d)
ĐK:
\(|x-2015|+1\neq 0\Leftrightarrow |x-2015|\neq -1\Leftrightarrow x\in\mathbb{R}\)
Vậy TXĐ \(D=\mathbb{R}\)
e)
ĐK: \(\left\{\begin{matrix} |x-1,2|\neq 0\\ 2x-5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 1,2\\ x\neq 2,5\end{matrix}\right.\)
Vậy TXĐ: \(D=\mathbb{R}\setminus \left\{1,2; 2,5\right\}\)
f)
ĐK: \(x^2-4\neq 0\Leftrightarrow (x-2)(x+2)\neq 0\Leftrightarrow x\neq \pm 2\)
Vậy TXĐ: \(D=\mathbb{R}\setminus \left\{\pm 2\right\}\)
a) \(x+5\ge0\Leftrightarrow x\ge-5\)
b) \(2x+5\ne0\Leftrightarrow2x\ne-5\Leftrightarrow x\ne-\frac{5}{2}\)
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
Bài easy quá mà!
4. a) Áp dụng tỉ dãy số bằng nhau:
\(\frac{a_1-1}{100}=\frac{a_2-2}{99}=...=\frac{a_{100}-100}{1}\)
\(=\frac{\left(a_1+a_2+...+a_{100}\right)-\left(1+2+...+100\right)}{100+99+...+2+1}=\frac{5050}{5050}=1\)
Suy ra: \(a_1-1=100\Leftrightarrow a_1=101\)
\(a_2-2=99\Leftrightarrow a_2=101\)
.......v.v...
\(a_{100}-100=1\Leftrightarrow a_{100}=101\)
Do đó: \(a_1=a_2=a_3=...=a_{100}=101\)
Bài 5/
Theo t/c dãy tỉ số bằng nhau,ta có: \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)\(=\frac{2x}{x}\)
Suy ra:
\(\frac{y+z-x}{x}=\frac{2x}{x}\Leftrightarrow y+z-x=2x\Rightarrow x=y=z\) (vì nếu \(x\ne y\ne z\Rightarrow y+z-x\ne2x\) "không thỏa mãn")
Thay vào A,ta có: \(A=\left(1+\frac{x}{x}\right)\left(1+\frac{y}{y}\right)\left(1+\frac{z}{z}\right)=2.2.2=8\)