cho 2 số nguyên dương lẻ m,n nguyên tố cùng nhau và

\(\left\{{}\begin{matrix}m^2+2⋮n\\n^2+2⋮m\end{matrix}\right.\)

chứng minh rằng \(m^2+n^2+2⋮4mn\)

\(\left(\sqrt{4+\sqrt{15}}\right)^2\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}=\left(\sqrt{4+\sqrt{15}}\sqrt{4-\sqrt{15}}\right)\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4+\sqrt{15}}=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)

đề sai rồi nha bạn

\(=\sqrt{13+\sqrt{30+\sqrt{2+\sqrt{8+2\cdot2\sqrt{2}+1}}}}=\sqrt{13+\sqrt{30+\sqrt{2+\sqrt{\left(2\sqrt{2}\right)^2+2\cdot2\sqrt{2}+1}}}}=\sqrt{13+\sqrt{30+\sqrt{2+\sqrt{\left(2\sqrt{2}\text{+}1\right)^2}}}}=\sqrt{13+\sqrt{30\text{+}\sqrt{2\text{+}2\sqrt{2}+1}}}=\sqrt{13+\sqrt{30+\sqrt{\left(\sqrt{2}+1\right)^2}}}=\sqrt{13+\sqrt{30+\sqrt{2}+1}}=\sqrt{13+\sqrt{31+\sqrt{2}}}\)

\(=\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+1=x+\sqrt{x}\)đk \(x>1\)

\(A=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x^2++2xy+y^2\right)}{4}}=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x-y\right)^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\sqrt{3}\left(x-y\right)}{2}=\dfrac{\sqrt{3}}{x+y}\)

\(B=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\left(1-4a+4a^2\right)}=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\left(2a-1\right)^2}=\dfrac{1}{2a-1}\cdot\sqrt{5}a^2\left(2a-1\right)=\sqrt{5}\cdot a^2\)

a)\(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right)\div\sqrt{15}=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right)\div\sqrt{3}\sqrt{5}=10\sqrt{3}\div\sqrt{3}\sqrt{5}=\sqrt{2}\sqrt{5}\div\sqrt{5}=\sqrt{2}\)b)\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{4}\sqrt{9}\sqrt{7}-\sqrt{100}\sqrt{7}+\sqrt{16}\sqrt{9}\sqrt{7}-\sqrt{64}\sqrt{7}=2\cdot3\cdot\sqrt{7}-10\cdot\sqrt{7}+4\cdot3\cdot\sqrt{7}-8\sqrt{7}=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}=0\)

c)\(\sqrt{27^2-23^2}+\sqrt{37^2-35^2}=\sqrt{\left(27-23\right)\left(27+23\right)}+\sqrt{\left(37-35\right)\left(37+35\right)}=\sqrt{4\cdot50}\cdot\sqrt{2\cdot72}=\sqrt{4\cdot50\cdot2\cdot72}=\sqrt{2^2\cdot2\cdot25\cdot2\cdot36\cdot2}=\sqrt{16}\cdot\sqrt{25}\cdot\sqrt{36}=4\cdot5\cdot6=120\)

d)\(\left(\sqrt{\dfrac{1}{7}}+\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right)\div\sqrt{7}=\left(\dfrac{1}{\sqrt{7}}+\dfrac{4}{\sqrt{7}}+\dfrac{3}{\sqrt{7}}\right)\cdot\dfrac{1}{\sqrt{7}}=\dfrac{7}{\sqrt{7}}\cdot\dfrac{1}{\sqrt{7}}=1\)

Đặt \(23-n=x^2;n-3=y^2\)

\(\Rightarrow x^2-y^2=23-n+n-3=20\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=20=1\cdot20=2\cdot10=4\cdot5\)\(=-1\cdot\left(-20\right)=-4\cdot\left(-5\right)=-2\cdot\left(-10\right)\)

bạn kẻ bảng tìm x,y

rồi tính x²,y² . tìm n

\(\dfrac{1}{D}=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1+2}{\sqrt{x}+1}=1+\dfrac{2}{\sqrt{x}+1}\)

Để \(\dfrac{1}{D}\) nguyên thì \(\left(\sqrt{x}+1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\Leftrightarrow\sqrt{x}\in\left\{0;2;-1;3\right\}\Leftrightarrow x\in\left\{0;4;9\right\}\)