Cho tam giác ABC vuông tại A và đường cao AH. Tính tỉ số lượng giác của góc B và góc C khi BC = 25, AH = 12
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\(\left(\frac{1}{1-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1-\sqrt{a}}\right).\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=\left(\frac{0}{1-\sqrt{a}}\right).\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=0.\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)
\(=0\)
\(A=\left(\frac{1}{1-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\) đkxđ:\(a>0;a\ne1\)
\(A=\left(\frac{1}{1-\sqrt{a}}-\frac{1}{1-\sqrt{a}}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}}\)\
\(A=0\)
a ) \(A=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{3}\right)-\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)
\(=\frac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{5-3}\)
\(=\frac{-2\sqrt{3}}{2}\)
\(=-\sqrt{3}\)
c ) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{\sqrt{3}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+1\right)+\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)-2\left(2+\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2\sqrt{3}+4}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2\left(\sqrt{3}+2\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2.\sqrt{3}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3.\left(3-1\right)}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{3}\)
\(=\frac{3-\sqrt{3}}{3}\)
\(=1-\frac{\sqrt{3}}{3}\)
\(E=\left(\frac{\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}}+\frac{\sqrt{\sqrt{x}+1}}{\sqrt{\sqrt{x}-1}}\right):\sqrt{\frac{1}{x-1}}\) \(ĐKXĐ:x>1\)
\(E=\left(\frac{\left(\sqrt{\sqrt{x}-1}\right)^2}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}+\frac{\left(\sqrt{\sqrt{x}+1}\right)^2}{\left(\sqrt{\sqrt{x}-1}\right)\left(\sqrt{\sqrt{x}+1}\right)}\right)\cdot\sqrt{\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{1}}\)
\(E=\left(\frac{\sqrt{x}-1}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{\sqrt{x}-1}\right)\left(\sqrt{\sqrt{x}+1}\right)}\right)\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(E=\frac{\sqrt{x}-1+\sqrt{x}+1}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(E=\frac{2\sqrt{x}}{\sqrt{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}}\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=2\sqrt{x}\)
Ta có:\(x=19-8\sqrt{3}=16-2.4\sqrt{3}+3=\left(4-\sqrt{3}\right)^2\)
\(\Rightarrow2\sqrt{x}=2.\sqrt{\left(4-\sqrt{3}\right)^2}=2.\left(4-\sqrt{3}\right)=8-2\sqrt{3}\)
\(C=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x-\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\) (tự tìm ĐKXĐ)
\(=\frac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}-1\right)+2\left(\sqrt{x}+1\right)\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}+1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+3\)
GTNN:\(x-\sqrt{x}+3=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
\(\Rightarrow Min\left(C\right)=\frac{11}{4}khi..\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)
TL:
\(\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\)
\(=\frac{8-3\sqrt{7}-8-3\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)
\(=\frac{-6\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)
Cho \(A=\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\)
CACH 1 : \(\Rightarrow A\sqrt{2}=\sqrt{16-6\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(\Rightarrow A\sqrt{2}=\sqrt{9-2.3.\sqrt{7}+7}-\sqrt{9+2.3.\sqrt{7}+7}\)
\(\Rightarrow A\sqrt{2}=\sqrt{\left(3-\sqrt{7}\right)^2}-\sqrt{\left(3+\sqrt{7}\right)^2}\)
\(\Rightarrow A\sqrt{2}=|3-\sqrt{7}|-|3+\sqrt{7}|\)
\(\Rightarrow A\sqrt{2}=3-\sqrt{7}-3-\sqrt{7}=-2\sqrt{7}=-\sqrt{28}\)
\(\Rightarrow A=-\sqrt{14}\)
CACH 2 : \(A^2=8-3\sqrt{7}+8+3\sqrt{7}-2.\sqrt{8^2-\left(3\sqrt{7}\right)^2}\)
\(\Rightarrow A^2=16-2\sqrt{64-63}=16-2=14\)
\(\Rightarrow A=\sqrt{14}\) hoặc \(A=-\sqrt{14}\)
Mà \(8+3\sqrt{7}>8-3\sqrt{7}\) \(\Rightarrow\sqrt{8+3\sqrt{7}}>\sqrt{8-3\sqrt{7}}\)
Vây A âm \(\Rightarrow A=-\sqrt{14}\)
a,ĐKXĐ:\(\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\)
\(\sqrt{x-2}.\sqrt{x+2}-\sqrt{x-2}=0\)
\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=4-\sqrt{x}-\sqrt{y}\left(đk:x;y>0\right)\)
\(\Leftrightarrow\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{y}}+\sqrt{y}=4\)
Do x,y là các số thực dương nên sử dụng BĐT AM-GM cho 2 số không âm ta có :
\(\frac{1}{\sqrt{x}}+\sqrt{x}\ge2\sqrt{\frac{1}{\sqrt{x}}.\sqrt{x}}=2\)
\(\frac{1}{\sqrt{y}}+\sqrt{y}\ge2\sqrt{\frac{1}{\sqrt{y}}.\sqrt{y}}=2\)
Cộng theo vế các bất đẳng thức cùng chiều ta được :
\(\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{y}}+\sqrt{y}\ge2+2=4\)
Dấu = xảy ra khi và chỉ khi \(\hept{\begin{cases}\frac{1}{\sqrt{x}}=\sqrt{x}\Leftrightarrow x=1\\\frac{1}{\sqrt{y}}=\sqrt{y}\Leftrightarrow y=1\end{cases}\Leftrightarrow}x=y=1\)
Vậy nghiệm của phương trình trên là \(x=y=1\)