thế điểm A(-4;1) vào đó bạn nên 1=2(-4)+b 

Vẽ đường kính AK \(\Rightarrow\angle ACK=90\Rightarrow AC\bot CK\)

mà \(BD\bot AC\Rightarrow\) \(BD\parallel CK\) mà BCKD nội tiếp nên BCKD là hình thang cân

\(\Rightarrow BK=CD\Rightarrow AB^2+CD^2=AB^2+BK^2=AK^2=4R^2\)

Gọi pt đường thẳng đề cho là \(y=ax+b\)

Vì song song với đường thẳng \(y=2x+7\Rightarrow\left\{{}\begin{matrix}a=2\\b\ne7\end{matrix}\right.\Rightarrow y=2x+b\)

Vì đường thẳng đề cho đi qua điểm \(A\left(-4;1\right)\Rightarrow1=-8+b\Rightarrow b=9\)

\(\Rightarrow y=2x+9\)

a) tam giác AHB vuông tại H có đường cao HE nên áp dụng hệ thức lượng

\(\Rightarrow AE.AB=AH^2\)

tam giác AHC vuông tại H có đường cao HF nên áp dụng hệ thức lượng

\(\Rightarrow AF.AC=AH^2=AE.AB\)

b) \(AE.AB=AF.AC\Rightarrow\dfrac{AE}{AC}=\dfrac{AF}{AB}\)

Xét \(\Delta AEF\) và \(\Delta ABC:\) Ta có: \(\left\{{}\begin{matrix}\dfrac{AE}{AC}=\dfrac{AF}{AB}\\\angle BACchung\end{matrix}\right.\)

\(\Rightarrow\Delta AEF\sim\Delta ABC\left(c-g-c\right)\)

c) Ta có: \(AH^4=AH^2.AH^2=AE.AB.AF.AC\)

tam giác ABC vuông tại A có đường cao AH nên áp dụng hệ thức lượng

\(\Rightarrow AB.AC=AH.BC\)

\(\Rightarrow AH^4=AE.AF.BC.AH\Rightarrow AH^3=AE.AF.BC\)

a) ĐKXĐ: \(x\ne3;-2;-\dfrac{5}{3}\)

\(B=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\left(x^2-2x+1\right)\)

\(=\left(\dfrac{x}{\left(x+2\right)\left(x-3\right)}-\dfrac{x-1}{\left(3x+5\right)\left(x-3\right)}\right):\dfrac{\left(x^2-1\right)^2}{\left(x+2\right)\left(3x+5\right)}\left(x-1\right)^2\)

\(=\dfrac{x\left(3x+5\right)-\left(x-1\right)\left(x+2\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}.\dfrac{\left(x+2\right)\left(3x+5\right)}{\left(x-1\right)^2\left(x+1\right)^2}.\left(x-1\right)^2\)

\(=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}.\dfrac{\left(x+2\right)\left(3x+5\right)}{\left(x+1\right)^2}\)

\(=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}.\dfrac{\left(x+2\right)\left(3x+5\right)}{\left(x+1\right)^2}=\dfrac{2}{x-3}\)

b) \(B=\dfrac{2}{2003-3}=\dfrac{2}{2000}=\dfrac{1}{1000}\)

c) \(x>3\Rightarrow x-3>0\Rightarrow\dfrac{2}{x-3}>0\)

ĐKXĐ: \(x\ge0\)

a) \(A=H+\sqrt{x}=1-\sqrt{x}+\sqrt{x}=1\) thì sao mà tìm được GTNN

b) \(A=H-4\sqrt{x}=1-\sqrt{x}-4\sqrt{x}=1-5\sqrt{x}\)

Ta có: \(5\sqrt{x}\ge0\Rightarrow-5\sqrt{x}\le0\Rightarrow1-5\sqrt{x}\le1\)

\(\Rightarrow A_{max}=1\) khi \(x=0\)

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}=2x+1\Rightarrow\left|x-3\right|=2x+1\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

a) \(C=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(a>0.a\ne1\right)\)

\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1-\sqrt{a}-2}{\sqrt{a}-1}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{-1}{\sqrt{a}-1}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(1-\sqrt{a}\right)=-\dfrac{1}{\sqrt{a}}\)

b) \(C=\dfrac{1}{4}\Rightarrow-\dfrac{1}{\sqrt{a}}=\dfrac{1}{4}\Rightarrow\sqrt{a}=-4\) (vô lý) \(\Rightarrow\) không có a thỏa đề

a) \(A=\sqrt{9a}-\sqrt{16a}-\sqrt{49a}=3\sqrt{a}-4\sqrt{a}-7\sqrt{a}=-8\sqrt{a}\)

b) \(B=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)

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

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

a) \(\sqrt{\dfrac{3}{2a}}=\dfrac{\sqrt{3}}{\sqrt{2a}}=\dfrac{\sqrt{3}.\sqrt{2a}}{\sqrt{2a}.\sqrt{2a}}=\dfrac{\sqrt{6a}}{2a}\)

b) \(\sqrt{\dfrac{3ab}{2}}=\dfrac{\sqrt{3ab}}{\sqrt{2}}=\dfrac{\sqrt{3ab}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6ab}}{2}\)