Сtg(2arcctg(-1/√3)=ctg(2*2π/3))=ctg(4π/3)=ctgπ/3=1/√3
-х^2+4=0
-x^2=0-4
-x^2=-4
x=-4:(-2)
х=2
![\frac{\sqrt{72}*14} {\sqrt{63}}= \frac{\sqrt{36*2}*14} {\sqrt{9*7}} = \frac{6\sqrt{2}*14} {3\sqrt{7}}= \frac{2\sqrt{2}*7*2} {\sqrt{7}}= \frac{4\sqrt{2}*(\sqrt{7})^{2}}{\sqrt{7}}=4 \sqrt{14}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B72%7D%2A14%7D+%7B%5Csqrt%7B63%7D%7D%3D+%5Cfrac%7B%5Csqrt%7B36%2A2%7D%2A14%7D+%7B%5Csqrt%7B9%2A7%7D%7D+%3D+%5Cfrac%7B6%5Csqrt%7B2%7D%2A14%7D+%7B3%5Csqrt%7B7%7D%7D%3D+%5Cfrac%7B2%5Csqrt%7B2%7D%2A7%2A2%7D+%7B%5Csqrt%7B7%7D%7D%3D+%5Cfrac%7B4%5Csqrt%7B2%7D%2A%28%5Csqrt%7B7%7D%29%5E%7B2%7D%7D%7B%5Csqrt%7B7%7D%7D%3D4+%5Csqrt%7B14%7D)
Если 14 тоже под корнем, то решение такое :
![\frac{\sqrt{72*14}} {\sqrt{63}} = \sqrt{\frac{36*2*2*7}{9*7}} = \frac{6*2}{3}=4](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B72%2A14%7D%7D+%7B%5Csqrt%7B63%7D%7D+%3D+%5Csqrt%7B%5Cfrac%7B36%2A2%2A2%2A7%7D%7B9%2A7%7D%7D+%3D+%5Cfrac%7B6%2A2%7D%7B3%7D%3D4)
Сtg75°=ctg(45°+30°)=cos(45°+30°)/sin(45°+30°)=
=(cos45°·cos30°-sin45°·sin30°)/(sin45°cos30°+cos45°sin30°)=
= (так как sin45°=cos45°=√2/2)=
=(cos30°-sin30°)/(cos30°+sin30°)=(√3-1)/(√3+1)=(√3-1)²/2=2-√3