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b: Xét ΔCFE vuông tại F và ΔCAB vuông tại A có
\(\widehat{C}\) chung
Do đó: ΔCFE\(\sim\)ΔCAB
Suy ra: \(\dfrac{CF}{CA}=\dfrac{CE}{CB}\)
\(\Leftrightarrow CF\cdot CB=CA\cdot CE\)
\(\Leftrightarrow CA\cdot CA\cdot\dfrac{1}{2}=CF\cdot CB\)
\(\Leftrightarrow CA^2=2\cdot CF\cdot CB\)
\(x=16\Rightarrow P=\dfrac{\sqrt{16}-2}{\sqrt{16}-3}=\dfrac{4-2}{4-3}=2\)
\(Q=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+3\sqrt{x}-6\sqrt{x}-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(A=P.Q=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}.\dfrac{\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{3\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}=\dfrac{5\sqrt{x}-2\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{5\sqrt{x}}{3\left(\sqrt{x}+3\right)}-\dfrac{2}{3}\)
Do \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}+3>0\end{matrix}\right.\) ; \(\forall x\ge0\Rightarrow\dfrac{5\sqrt{x}}{3\left(\sqrt{x}+3\right)}\ge0\)
\(\Rightarrow A\ge-\dfrac{2}{3}\)
\(A_{min}=-\dfrac{2}{3}\) khi \(x=0\)
a: Thay x=16 vào P, ta được:
\(P=\dfrac{4-2}{4-3}=2\)
b: Ta có: \(Q=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{6\sqrt{x}}{9-x}-\dfrac{3}{\sqrt{x}+3}\)
\(=\dfrac{x+3\sqrt{x}-6\sqrt{x}-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(a,=\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)\\ b,=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-2\left(\sqrt{x}+\sqrt{y}\right)\\ =\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}-2\right)\\ c,=x\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)=\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)\)
\(a,=\dfrac{\left(9-4\sqrt{5}\right)\left(5+2\sqrt{5}\right)}{4}+\dfrac{2\sqrt{5}}{5}\\ =\dfrac{5-2\sqrt{5}}{4}+\dfrac{2\sqrt{5}}{5}\\ =\dfrac{25-10\sqrt{5}+8\sqrt{5}}{20}=\dfrac{25-2\sqrt{5}}{20}\\ b,=\dfrac{\sqrt{x}+2-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\\ c,=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}-\dfrac{2}{\sqrt{x}-1}\\ =\dfrac{\sqrt{x}+1-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}-1}=1\\ d,=\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}+\dfrac{x+1}{1-x}\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1-x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-1}{\sqrt{x}+1}\)
Bài làm:
Δ ABC vuông tại A?
Ta có: \(\sin B=\frac{AC}{BC}=\frac{3}{5}\) <=> \(\frac{AC}{3}=\frac{BC}{5}=k\) \(\left(k\inℕ^∗\right)\)
=> \(AB^2=BC^2-CA^2=25k^2-9k^2=16k^2\)
=> \(AB=4k\)
Từ đây ta có thể dễ dàng tính được:
\(\cos B=\frac{AB}{BC}=\frac{4}{5}\) ; \(\tan B=\frac{AC}{AB}=\frac{3}{4}\) ; \(\cot B=\frac{AB}{AC}=\frac{4}{3}\)
\(sin^2b+cos^2b=1\)
\(\left(\frac{3}{5}\right)^2+cos^2b=1\)
\(\frac{9}{25}+cos^2b=1\)
\(cos^2b=\frac{16}{25}\)
\(cosb=\pm\sqrt{\frac{16}{25}}=\pm\frac{4}{5}\)
\(tanb=\frac{sinb}{cosb}=\orbr{\begin{cases}\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}\\\frac{\frac{3}{5}}{\frac{-4}{5}}=\frac{-3}{4}\end{cases}}\)
\(cotb=\frac{1}{tanb}=\orbr{\begin{cases}\frac{1}{\frac{3}{4}}=\frac{4}{3}\\\frac{1}{\frac{-3}{4}}=\frac{-4}{3}\end{cases}}\)
\(x+\sqrt{\left(x-1\right)^2}=x+\left|x-1\right|\)(1)
Với x < 1 (1) = x - ( x - 1 ) = x - x + 1 = 1
Với x >= 1 (1) = x + x - 1 = 2x - 1