\(1997.2003=\left(2000-3\right)\left(2000+3\right)=2000^2-3^2=4000000-9=3999991\)

\(1997\cdot2003=\left(2000-3\right)\left(2000+3\right)=3999991\)

Bài 1:

1. ĐKXĐ: \(x\ge0\) 
\(x-7\sqrt{x}+10=0\)

\(\Leftrightarrow x-2\sqrt{x}-5\sqrt{x}+10=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=25\end{matrix}\right.\) ( thỏa mãn đk )

Vậy \(S=\left\{4;25\right\}\)

2. ĐKXĐ: \(x\ge4\) 
\(\sqrt{x^2-16}-5\sqrt{x-4}=0\)

\(\Leftrightarrow\sqrt{x^2-16}=5\sqrt{x-4}\)

\(\Leftrightarrow x^2-16=25\left(x-4\right)\)

\(\Leftrightarrow x+4=25\)

\(\Leftrightarrow x=21\) ( thỏa mãn đk )

Vậy \(S=\left\{21\right\}\)

3. ĐKXĐ: \(x\ge-4\)
\(\sqrt{x^2-16}-3\sqrt{x+4}=0\)

\(\Leftrightarrow\sqrt{x^2-16}=3\sqrt{x+4}\)

\(\Leftrightarrow x^2-16=9\left(x+4\right)\)

\(\Leftrightarrow x-4=9\)

\(\Leftrightarrow x=13\) ( thỏa mãn đk )

Vậy \(S=\left\{13\right\}\)

Bài 1:
a) ĐKXĐ: \(x\ge0\)

\(x-7\sqrt{x}+10=0\)

\(\Rightarrow x+10=7\sqrt{x}\)

\(\Rightarrow x^2+20x+100=49x\)

\(\Rightarrow x^2-29x+100=0\)

\(\Rightarrow\left(x^2-4x\right)-\left(25x-100\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x-25\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\x-25=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=25\end{matrix}\right.\)

b) ĐKXĐ:\(\left\{{}\begin{matrix}x^2-16\ge0\\x-4\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-4\right)\left(x+4\right)\ge0\\x-4\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x+4\ge0\\x-4\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge-4\\x\ge4\end{matrix}\right.\Rightarrow x\ge4\)

\(\sqrt{x^2-16}-5\sqrt{x-4}=0\)

\(\Rightarrow\sqrt{\left(x-4\right)\left(x+4\right)}-5\sqrt{x-4}=0\\ \Rightarrow\sqrt{x-4}\left(\sqrt{x+4}-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-4}=0\\\sqrt{x+4}-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\\sqrt{x+4}=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x+4=25\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=21\end{matrix}\right.\)

Bài 1:

a: Bảng giá trị:

Vẽ đồ thị:

b: Tọa độ A là:

\(\begin{cases}y=0\\ -x-1=0\end{cases}\Rightarrow\begin{cases}y=0\\ x=-1\end{cases}\)

=>A(-1;0)

\(OA=\sqrt{\left(-1-0\right)^2+\left(0-0\right)^2}=1\)

Tọa độ B là:

\(\begin{cases}x=0\\ y=-0-1=-1\end{cases}\)

=>B(0;-1)

=>\(OB=\sqrt{\left(0-0\right)^2+\left(-1-0\right)^2}=1\)

ΔOAB vuông tại O

=>\(S_{OAB}=\frac12\cdot OA\cdot OB=\frac12\cdot1\cdot1=\frac12\)

c:

(d2): y=-2x+1

Tọa độ C là:

\(\begin{cases}y=0\\ -2x+1=0\end{cases}\Rightarrow\begin{cases}y=0\\ -2x=-1\end{cases}\Rightarrow\begin{cases}y=0\\ x=\frac12\end{cases}\)

=>\(OC=\sqrt{\left(\frac12-0\right)^2+\left(0-0\right)^2}=\frac12\)

Tọa độ D là:

\(\begin{cases}x=0\\ y=-2\cdot0+1=1\end{cases}\)

=>D(0;1)

=>\(OD=\sqrt{\left(0-0\right)^2+\left(1-0\right)^2}=1\)

\(CD=\sqrt{\left(\frac12-0\right)^2+\left(1-0\right)^2}=\sqrt{\frac14+1}=\sqrt{\frac54}=\frac{\sqrt5}{2}\)

Chu vi tam giác OCD là:

OC+OD+CD

\(=0,5+1+\frac{\sqrt5}{2}=\frac{3+\sqrt5}{2}\)

d: Gọi α là góc tạo bởi (d1): y=-x-1 với trục Ox

tan α=a=-1

=>α=135 độ

Gọi β là góc tạo bởi (d2): y=-2x+1 với trục Ox

tan β=a=-2

=>β≃117 độ

ĐKXĐ:

a.

\(1-sinx\ne0\)

\(\Leftrightarrow sinx\ne1\)

\(\Leftrightarrow x\ne\dfrac{\pi}{2}+k2\pi\)

b.

\(sin\left(2-x\right)\ne0\)

\(\Leftrightarrow2-x\ne k\pi\)

\(\Leftrightarrow x\ne2+k\pi\)

c.

\(1-cos5x\ne0\)

\(\Leftrightarrow cos5x\ne1\)

\(\Leftrightarrow5x\ne k2\pi\)

\(\Leftrightarrow x\ne\dfrac{k2\pi}{5}\)

d.

\(3+x>0\)

\(\Leftrightarrow x>-3\)

e.

\(2x-3\ne0\)

\(\Leftrightarrow x\ne\dfrac{3}{2}\)

a) \(198+789+502+311\)

\(=\left(198+502\right)+\left(789+311\right)\)

\(=700+1100\)

\(=1800\)

b) \(15.6.4.125.8\)

\(=\left(15.4\right).6.\left(125.8\right)\)

\(=60.6.1000\)

\(=360000\)

c) \(367+129+133+371+17\)

\(=\left(367+133\right)+\left(129+371\right)+17\)

\(=500+500+17\)

\(=1000+17\)

\(=1017\)

d) \(37.64+37.35+37\)

\(=37.\left(64+35+1\right)\)

\(=37.100\)

\(=3700\)

e) \(3.25.8+4.37.6+2.38.12\)

\(=\left(3.8\right).25+\left(4.6\right).37+\left(2.12\right).38\)

\(=24.25+24.37+24.38\)

\(=24.\left(25+37+38\right)\)

\(=24.100\)

\(=2400\)

f) \(136.48+16.272+68.20.2\)

\(=136.48+16.2.136+68.2.20\)

\(=136.48+32.136+136.20\)

\(=136.\left(48+32+20\right)\)

\(=136.100\)

\(=13600\)

g) \(1+6+11+16+...+46+51\)

Số số hạng là : 

\(\left(51-1\right):5+1=11\) ( số )

Tổng là : 

\(\left(1+51\right)\times11:2=286\)

Chúc bạn học tốt

Đề sai :) 

\(x+2>x-6\)  

\(x-x>-2-6\)  

\(0>-8\left(llđ\right)\)   

Vậy \(\forall x\in R\) thì \(x+2>x-6\)

\(4x+12=-20\)

\(\Rightarrow4x=-20-12\)

\(\Rightarrow4x=-32\)

\(\Rightarrow x=-32\div4\)

\(\Rightarrow x=-8\)

\(4x+12=-20\)

\(\Rightarrow4x=(-20)-12\)

\(\Rightarrow4\text{x}=-32\)

\(\Rightarrow\text{x}=-\frac{32}{4}\Rightarrow\text{x}=-8\)

Chúc bạn học tốt :>