\(a,\Rightarrow8x^2-16x-4x+5-8x^2+10x-5x=0\\ \Rightarrow-15x=-5\Rightarrow x=\dfrac{1}{3}\\ b,\Rightarrow9x^2-16-9x^2+12x-4=5\\ \Rightarrow12x=25\\ \Rightarrow x=\dfrac{25}{12}\)

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

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

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

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

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

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

4. ĐKXĐ: \(x\ge-7\)

\(\sqrt{x^2-49}-3\sqrt{x+7}=0\)

\(\Leftrightarrow\sqrt{x^2-49}=3\sqrt{x+7}\)

\(\Leftrightarrow x^2-49=9\left(x+7\right)\)

\(\Leftrightarrow x-7=9\)

\(\Leftrightarrow x=16\) ( thỏa mãn đk )

Vậy \(S=\left\{16\right\}\)

5. ĐKXĐ: \(x\ge0\)

\(x-14\sqrt{x}+45=0\)

\(\Leftrightarrow\left(\sqrt{x}-9\right)\left(\sqrt{x}-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-9=0\\\sqrt{x}-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=9\\\sqrt{x}=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=81\\x=25\end{matrix}\right.\)

Vậy \(S=\left\{81;25\right\}\)

6. ĐKXĐ: \(x\ge0\)

\(x-13\sqrt{x}+42=0\)

\(\Leftrightarrow\left(\sqrt{x}+6\right)\left(\sqrt{x}+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+7=0\\\sqrt{x}+6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-7\\\sqrt{x}=-6\end{matrix}\right.\) ( loại )

Vậy phương trình vô nghiệm

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 \(x+2>x-6\)