Vẽ (A;AC) và (B;BC). BT cắt (A) tại Z khác T, AK cắt (B) tại Y khác K. E đối xứng với C qua AB

Vì CA,CB vuông góc nhau nên CA tiếp xúc (B) và CB tiếp xúc (A)

Suy ra \(AC^2=AT^2=AK.AY\). Suy ra \(\widehat{ATK}=\widehat{AYT}\). Tương tự \(\widehat{BKT}=\widehat{BZK}\)

Dễ thấy AC=AT=AZ=AE, BC=BK=BY=BE suy ra CE là trục đẳng phương của (A) và (B)

Do đó \(P_{X/\left(A\right)}=\overline{XZ}.\overline{XT}=P_{X/\left(B\right)}=\overline{XY}.\overline{XK}\), suy ra (K,T,Y,Z)_cyc

Suy ra \(\widehat{ATK}=\widehat{AYT}=\widehat{BZK}=\widehat{BKT}\). Vậy tam giác MKT cân tại M hay MK = MT.

\(\sqrt{1+\frac{\sqrt{3}}{2}}=\sqrt{\frac{2+\sqrt{3}}{2}}=\sqrt{\frac{4+2\sqrt{3}}{4}}=\sqrt{\frac{3+2\sqrt{3}+1}{4}}\)

\(=\sqrt{\frac{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}{4}}=\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{4}}=\left|\frac{\sqrt{3}+1}{2}\right|=\frac{1+\sqrt{3}}{2}\)

\(\left(\frac{1+\sqrt{3}}{2}\right)^2=\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)^2\)

\(=\frac{1}{4}+2.\frac{1}{2}.\frac{\sqrt{3}}{2}+\frac{3}{4}\)

\(=1+\frac{\sqrt{3}}{2}\)

\(\RightarrowĐPCM\)

Ta có: \(\Delta=\left(-m\right)^2+4.3=m^2+12>0\)

=> pt luôn có 2 nghiệm phân biệt

Theo hệ thức vi-et, ta có: \(\hept{\begin{cases}x_1+x_2=m\\x_1x_2=-3\end{cases}}\)

Theo bài ra, ta có: x₁² + x₂² = 5m

<=> (x₁ + x₂)² - 2x₁x₂ = 5m

<=> m² + 6 = 5m

<=> x² - 5m + 6 = 0

<=> x² - 2m - 3m + 6 = 0

<=> (m - 2)(m - 3)= 0

<=> \(\orbr{\begin{cases}m=2\\m=3\end{cases}}\)

*có giải mà hỏi làm gì=))*

a) Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

\(\Rightarrow\sqrt{2}A=\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

\(\Leftrightarrow\sqrt{2}A=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}+1\right|+\left|\sqrt{5}-1\right|\)

\(\Leftrightarrow\sqrt{2}A=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\Rightarrow A=\sqrt{10}\left(đpcm\right)\)

b) \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(9+4\sqrt{2}\right)=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}=9+2\sqrt{6}\left(đpcm\right)\)

bn có mà bn che mà

hok tok

kkk

Bài 3

\(A=\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

\(=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\sqrt{3+\sqrt{5}}\cdot\sqrt{2}\left(\sqrt{5}-1\right)\)

\(=2\cdot\sqrt{6+2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)=2\cdot\sqrt{\left(\sqrt{5}+1\right)^2}\cdot\left(\sqrt{5}-1\right)\)

\(=2\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=2\cdot4=8\left(đpcm\right)\)

\(B=\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{2-\sqrt{3}}\right)=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)

\(=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\left(đpcm\right)\)

Bài 4

\(P=\frac{3\sqrt{10}+\sqrt{20}-3\sqrt{6}-\sqrt{12}}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{10}\left(\sqrt{2}+1\right)-\sqrt{6}\left(\sqrt{2}+1\right)}{\sqrt{5}-\sqrt{3}}\)

\(=\frac{\sqrt{2}\left(\sqrt{2}+1\right)\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=2+\sqrt{2}\)

\(Q=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+2\sqrt{2}+2+2}{\sqrt{2}+\sqrt{3}+2}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)

\(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

\(\left|3-\sqrt{5}\right|+\left|3+\sqrt{5}\right|+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(3-\sqrt{5}+3+\sqrt{5}+2\sqrt{3^2-\sqrt{5}^2}\)

\(6+2\sqrt{9-5}\)

\(6+2\sqrt{4}\)

\(=10\)

\(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=\left(1+\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)

\(=4+2\sqrt{3}-2=2+2\sqrt{3}\)

\(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)

\(1+\sqrt{3}+\sqrt{2}+\sqrt{3}+3+\sqrt{6}-\sqrt{2}-\sqrt{6}-2\)

\(2+2\sqrt{3}\)