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\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{\sqrt{2}}{2}\Rightarrow B=45^0\)
\(cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{1}{2}\Rightarrow A=60^0\)
\(\Rightarrow C=180^0-\left(A+B\right)=75^0\)
\(h_a=\frac{bc.sinA}{a}=\frac{2.\left(\sqrt{3}+1\right)sin60^0}{\sqrt{6}}=\frac{\sqrt{6}+\sqrt{2}}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a=t\sqrt{3}\\b=t\sqrt{2}\\c=\frac{t\left(\sqrt{6}-\sqrt{2}\right)}{2}\end{matrix}\right.\)
\(cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{2t^2+\left(2-\sqrt{3}\right)t^2-3t^2}{t^2.\sqrt{2}\left(\sqrt{6}-\sqrt{2}\right)}=-\frac{1}{2}\)
\(\Rightarrow A=120^0\)
\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{\sqrt{2}}{2}\Rightarrow B=45^0\)
\(\Rightarrow C=180^0-\left(A+B\right)=15^0\)
\(R=\frac{a}{2sinA}=\frac{2\sqrt{3}}{2sin120^0}=2\)
b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{2}+2}-\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}+\sqrt{2}\right|\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\) (vì \(\sqrt{5}\ge\sqrt{2}\)
=0
c) \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+\sqrt{3+1}\) (vì \(\sqrt{3}\ge1\))
\(=2\sqrt{3}\)
a)\(\sqrt{5+2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}-\sqrt{3-2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\) (vì \(\sqrt{3}\ge\sqrt{2}\))
=0
\(cosA=\frac{AB^2+AC^2-BC^2}{2AB.AC}=\frac{a^2+b^2+a^2+c^2-b^2-c^2}{2AB.AC}=\frac{a^2}{AB.AC}>0\)
\(\Rightarrow A< 90^0\)
Tương tự ta có: \(cosB=\frac{b^2}{AB.BC}>0\Rightarrow B< 90^0\)
\(cosC=\frac{c^2}{AC.BC}>0\Rightarrow C< 90^0\)
\(\Rightarrow\Delta ABC\) là tam giác nhọn
a, \(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
= \(\sqrt{2+2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\)
= \(\sqrt{2}\) + 1 - 2 + \(\sqrt{2}\)
= 2\(\sqrt{2}\) - 1
b, \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
= \(\sqrt{5-4\sqrt{5}+4}-\sqrt{5}\)
= \(\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
= \(\sqrt{5}-2-\sqrt{5}\)
= - 2
c, \(\sqrt{28+8\sqrt{7}}-\sqrt{7}\)
= \(\sqrt{16+8\sqrt{7}+7}-\sqrt{7}\)
= \(\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)
= 4 + \(\sqrt{7}\) - \(\sqrt{7}\)
= 4
a/ \(\Leftrightarrow\sqrt{x^2+x+3}-\sqrt{x^2+2}+\sqrt{x^2+x+8}-\sqrt{x^2+7}=0\)
\(\Leftrightarrow\frac{x+1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{x+1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}\right)=0\)
\(\Leftrightarrow x+1=0\) (ngoặc to phía sau luôn dương)
\(\Rightarrow x=-1\)
b/
\(\sqrt{7-x^2+x\sqrt{x+5}}=\sqrt{3-2x-x^2}\) (1)
\(\Rightarrow7-x^2+x\sqrt{x+5}=3-2x-x^2\)
\(\Leftrightarrow x\sqrt{x+5}=-2x-4\)
\(\Rightarrow x^2\left(x+5\right)=4x^2+16x+16\)
\(\Rightarrow x^3+x^2-16\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
Do các phép biến đổi ko tương đương nên cần thay nghiệm vào (1) để kiểm tra
c/ ĐKXĐ: \(x\ge\frac{5}{3}\)
\(\Leftrightarrow\sqrt{10x+1}-\sqrt{9x+4}+\sqrt{3x-5}-\sqrt{2x-2}=0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{x-3}{\sqrt{3x-5}+\sqrt{2x-2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{1}{\sqrt{3x-5}+\sqrt{2x-2}}\right)=0\)
\(\Leftrightarrow x-3=0\) (ngoặc phía sau luôn dương)
d/ Đề bài là \(2\sqrt{2x+3}\) hay \(2\sqrt{2x-3}\) bạn?
e/ ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\sqrt{x+3+2\sqrt{x+3}+1}=x+4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+3}+1\right)^2}=x+4\)
\(\Leftrightarrow\sqrt{x+3}+1=x+4\)
\(\Leftrightarrow x+3-\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+3}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+3=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
M=\(\left(x_1+x_2\right)^2-2x_1.x_2+\left(y_1+y_2\right)^2-2y_1.y_2\)
Áp dụng định lý viettel :( :v )
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\);\(\left\{{}\begin{matrix}y_1+y_2=-\dfrac{b}{c}\\y_1y_2=\dfrac{a}{c}\end{matrix}\right.\)
\(M=\dfrac{b^2}{a^2}-\dfrac{2c}{a}+\dfrac{b^2}{c^2}-\dfrac{2a}{c}=\dfrac{b^2-4ac}{a^2}+\dfrac{b^2-4ac}{c^2}+2\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)
\(\ge2\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge4\)
Dấu = xảy ra: \(\left\{{}\begin{matrix}a=c\\b^2=4ac\end{matrix}\right.\)\(\Leftrightarrow b^2=4a^2=4c^2\)
@_@ 
đưa thẳng câu hỏi luôn đi ; nói như zầy chưa nghỉ ra câu trả lời ; chống mặt chết trước rồi
\(cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{1}{2}\Rightarrow A=60^0\)
\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{\sqrt{2}}{2}\Rightarrow B=45^0\)
\(\Rightarrow C=180^0-\left(A+B\right)=75^0\)
\(sinA=sin60^0=\frac{\sqrt{3}}{2}\)
\(\Rightarrow h_a=\frac{bc.sinA}{a}=\sqrt{3}+1\)