\(a,x^{15}=x\\ \Rightarrow x^{15}-x=0\\ \Rightarrow x\left(x^{14}-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^{14}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(thoaman\right)\\x=1\left(thoaman\right)\\x=-1\left(loại\right)\end{matrix}\right.\)

\(b,\left(2x+1\right)^3=125\\ \Rightarrow\left(2x+1\right)^3=5^3\\ \Rightarrow2x+1=5\\ \Rightarrow2x=4\\ \Rightarrow x=2\left(thoaman\right)\)

\(c,\left(x-5\right)^4=\left(x-5\right)^6\\ \Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\\ \Rightarrow\left(x-5\right)^4\left(1-\left(x-5\right)^2\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\\x-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\left(thoaman\right)\)

\(d,x^{10}=x^1\\ \Rightarrow x^{10}-x^1=0\\ \Rightarrow x\left(x^9-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^9-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\left(thoaman\right)\)

\(e,2^x-15=17\\ \Rightarrow2^x=17+15\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\left(thoaman\right)\)

\(f,\left(7x-11\right)^3=2^5.5^2+200\\ \Rightarrow\left(7x-11\right)^3=32.25+200\\ \Rightarrow\left(7x-11\right)^3=800+200\\ \Rightarrow\left(7x-11\right)^3=1000\\ \Rightarrow\left(7x-11\right)^3=10^3\\ \Rightarrow7x-11=10\\ \Rightarrow7x=21\\ \Rightarrow x=21:7=3\left(thoaman\right)\)

@seven

\(a,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\\ =\sqrt{3}+\sqrt{2}-\left(\sqrt{3}-\sqrt{2}\right)\\ =\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\\=2\sqrt{2} \)

\(b,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)

\(c,=x-4+\sqrt{\left(4^2-2.4.x+x^2\right)}\\ =x-4+\sqrt{\left(4-x\right)^2}\\ =x-4+\left|4-x\right|\\ =x-4+x-4=2x-8\)    (vì \(x>4\) )

@seven 

\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)

\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)

@seven

\(a,3^x.3=243\\ \Rightarrow3^x=243:3\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\\ b,x^{20}=x\\ \Rightarrow x^{20}-x=0\\ \Rightarrow x\left(x^{19}-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^{19}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(c,2^x.16^2=1024\\ \Rightarrow2^x=1024:16^2\\ \Rightarrow2x=4\\ \Rightarrow x=4:2=2\)

\(d,64.4^x=16^8\\ \Rightarrow2^6.4^x=\left(2^4\right)^8\\\Rightarrow 4^x=2^{32}:2^6\\ \Rightarrow4^x=2^{26}\\ \Rightarrow4^x=\left(2^2\right)^{13}\\ \Rightarrow4^x=4^{13}\\ \Rightarrow x=13\\ e,2^x.4=128\\ \Rightarrow2^x=128:4\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\)

@seven

\(\dfrac{4}{\sqrt{7}+\sqrt{3}}+\dfrac{4}{\sqrt{7}-\sqrt{3}}\\ =\dfrac{4\left(\sqrt{7}-\sqrt{3}\right)}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}+\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}\\ =\dfrac{4\left(\sqrt{7}-\sqrt{3}\right)}{7-3}+\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{7-3}\\ =\dfrac{4\left(\sqrt{7}-\sqrt{3}\right)}{4}+\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{4}\\ =\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\\ =2\sqrt{7}\)

@seven

\(720:\left[41-\left(2x-5\right)\right]=2^3.5\\ \Rightarrow720:\left[41-\left(2x-5\right)\right]=8.5\\ \Rightarrow720:\left[41-\left(2x-5\right)\right]=40\\ \Rightarrow\left[41-\left(2x-5\right)\right]=720:40\\ \Rightarrow41-\left(2x-5\right)=18\\ \Rightarrow2x-5=41-18\\ \Rightarrow2x-5=23\\ \Rightarrow2x=23+5\\ \Rightarrow2x=28\\ \Rightarrow x=28:2=14\)

@seven

Xét tam giác \(ABH\) vuông tại H có

\(AH^2+HB^2=AB^2\left(Pytago\right)\)

\(\Leftrightarrow AH=\sqrt{AB^2-BH^2}=\sqrt{15^2-9^2}=12\left(cm\right)\)

Xét tam giác ABC vuông tại A

\(AB^2=HB.BC\\ \Rightarrow BC=\dfrac{AB^2}{HB}=\dfrac{15^2}{9}=25\left(cm\right)\\ HB+HC=BC\\ \Rightarrow HC=BC-BH=25-9=16\left(cm\right)\\ AB.AC=AH.BC\\ \Rightarrow AC=\dfrac{AH.BC}{AB}=\dfrac{12.25}{15}=20\left(cm\right)\)

\(f,\left(2x+1\right)^3=343\\\Rightarrow \left(2x+1\right)^3=7^3\\ \Rightarrow2x+1=7\\ \Rightarrow2x=6\\ \Rightarrow x=3\\ g,\left(x-1\right)^3=125\\ \Rightarrow\left(x-1\right)^3=5^3\\ \Rightarrow x-1=5\\ \Rightarrow x=6\\ h,2^{x+2}-2^x=96\\ \Rightarrow2^x.2^2-2^x=96\\ \Rightarrow2^x.\left(2^2-1\right)=96\\ \Rightarrow2^x=96:\left(2^2-1\right)\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\)

\(i,\left(x-5\right)^4=\left(x-5\right)^6\\ \Rightarrow\left(x-5\right)^4\left(1-\left(x-5\right)^2\right)=0\\\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{matrix}\right. \\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\\x-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(t/m\right)\\x=6\left(t|m\right)\\x=4\left(loại\right)\end{matrix}\right.\)

\(j,720:\left[41-\left(2x-5\right)\right]=2^3.5\\ \Rightarrow720:\left(41-2x+5\right)=40\\ \Rightarrow\left(46-2x\right)=720:40\\ \Rightarrow46-2x=18\\ \Rightarrow2x=46-18=28\\ \Rightarrow x=28:2=14\)

@seven

\(2x-5\sqrt{xy}+3y\\ =2x-2\sqrt{xy}-3\sqrt{xy}+3y\\ =2\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-3\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)\\ =\left(2\sqrt{x}-3\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)

\(x+\sqrt{x}-12=x+4\sqrt{x}-3\sqrt{x}-12\\ =\sqrt{x}\left(\sqrt{x}+4\right)-3\left(\sqrt{x}+4\right)\\ =\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)\)