Câu 1:
a) ĐKXĐ: \(x\ne25;x\ne9\)
b) Ta có:
\(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)\(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)\(A=\left(\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)\(A=\left(\dfrac{-5}{\sqrt{x}+5}\right).\left(\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{25-x-x+9+x-25}\right)\)\(A=\dfrac{-5\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{5}{\sqrt{x}+3}\)
c) Để A nguyên thì \(\dfrac{5}{\sqrt{x}+3}\in Z\)
\(\Rightarrow5⋮\left(\sqrt{x}+3\right)\)
\(\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(5\right)\)
\(\Rightarrow\sqrt{x}+3\in\left\{\pm1;\pm5\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{-2;-4;-8;2\right\}\)
Vì \(\sqrt{x}\ge0\) nên \(\sqrt{x}=2\) \(\Leftrightarrow x=4\)
Vậy \(x=4\) thì A đạt giá trị nguyên.
d) ĐKXĐ: \(x\ne25;x\ne9\)
\(B=\dfrac{A\left(x+16\right)}{5}=\left(\dfrac{5}{\sqrt{x}+3}\right)\left(x+16\right).\dfrac{1}{5}\)
\(B=\dfrac{5\left(x+16\right)}{\left(\sqrt{x}+3\right).5}=\dfrac{x+16}{\sqrt{x}+3}\)
\(B=\dfrac{x-9+25}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+25}{\sqrt{x}+3}\)
\(B=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\)=\(\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)
Theo bất đẳng thức Cô-si ta có:
\(\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\left(\dfrac{25}{\sqrt{x}+3}\right)}-6\)\(=2.5-6=4\)
Dấu "=" xảy ra khi và chỉ khi \(\sqrt{x}+3=\dfrac{25}{\sqrt{x}+3}\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+3=5\\\sqrt{x}+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=-8\end{matrix}\right.\)(loại \(\sqrt{x}=-8\) vì \(\sqrt{x}\ge0\))
\(\Rightarrow\sqrt{x}=2\Leftrightarrow x=4\)
Vậy GTNN của \(B=4\) khi \(x=4\)
Câu 3:
a)\(\dfrac{\sqrt{\dfrac{abc+4}{a}-4\sqrt{\dfrac{bc}{a}}}}{\sqrt{abc}-2}=\dfrac{1}{\sqrt{a}}\) với \(\left\{{}\begin{matrix}a>0\\b>0\\\sqrt{abc}>2\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\dfrac{abc+4}{a}-4\sqrt{\dfrac{bc}{a}}}.\sqrt{a}=\sqrt{abc}-2\)
Biến đổi VT của đẳng thức ta có:
\(\sqrt{\dfrac{abc+4}{a}-4\sqrt{\dfrac{bc}{a}}}.\sqrt{a}=\sqrt{\left(\dfrac{abc+4}{a}-4\sqrt{\dfrac{bc}{a}}\right).a}\)
\(=\sqrt{\dfrac{abc+4-4\sqrt{abc}}{a}.a}\)\(=\sqrt{\left(\sqrt{abc-2}\right)^2}\)
\(=\sqrt{abc}-2=VP\)
Vậy \(VT=VP\Rightarrowđpcm\)
b) Đăt \(P=\dfrac{1}{2a+2bc+1}+\dfrac{1}{2b+2ca+1}+\dfrac{1}{2c+2ab+1}\)
Ta có:
\(a^2+b^2\ge2ab\) ; \(b^2+c^2\ge2bc\) ; \(a^2+c^2\ge2ac\)
\(\Rightarrow a^2+b^2+b^2+c^2+c^2+a^2\ge2ab+2bc+2ac\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow a^2+b^2+c^2\ge1\) (1)
Mặt khác:
\(\left(a-1\right)^2\ge0\Leftrightarrow a^2+1\ge2a\) (2)
\(\left(b-1\right)^2\ge0\Leftrightarrow b^2+1\ge2b\) (3)
\(\left(c-1\right)^2\ge0\Leftrightarrow c^2+1\ge2c\) (4)
Kết hợp (1) ; (2) ; (3) ; (4) ta có:
\(P\ge\dfrac{1}{\left(a^2+1\right)+\left(b^2+c^2\right)+1}+\dfrac{1}{\left(b^2+1\right)+\left(c^2+a^2\right)+1}+\dfrac{1}{\left(c^2+1\right)+\left(a^2+b^2\right)+1}\)
\(P\ge\dfrac{1}{a^2+b^2+c^2+2}+\dfrac{1}{a^2+b^2+c^2+2}+\dfrac{1}{a^2+b^2+c^2+2}\)
Vì \(a^2+b^2+c^2\ge1\) nên ta có:
\(P\ge\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)
hay: \(\dfrac{1}{2a+2bc+1}+\dfrac{1}{2b+2ac+1}+\dfrac{1}{2c+2ab+1}\ge1\)\(\left(đpcm\right)\)
Câu 5:
\(Q=x\sqrt{\dfrac{\left(2011+y^2\right)\left(2011+z^2\right)}{\left(2011+x^2\right)}}+y\sqrt{\dfrac{\left(2011+x^2\right)\left(2011+z^2\right)}{\left(2011+y^2\right)}}+z\sqrt{\dfrac{\left(2011+y^2\right)\left(2011+x^2\right)}{\left(2011+z^2\right)}}\left(1\right)\)
Vì \(xy+yz+zx=2011\) nên ta có:
\(2011+x^2=xy+yz+zx+x^2=\left(x+z\right)\left(x+y\right)\)
\(2011+y^2=xy+xz+yz+y^2=\left(x+y\right)\left(y+z\right)\)
\(2011+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào (1) ta có:
\(Q=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+z\right)\left(x+y\right)}}+y\sqrt{\dfrac{\left(x+y\right)\left(x+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(x+z\right)\left(y+z\right)}}\)
\(Q=x.\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(Q=xy+xz+yx+yz+zx+zy\)
\(Q=2\left(xy+yz+zx\right)=2.2011=4022\)
Vậy giá trị của biểu thức \(Q=4022\)
Câu 2:
a) \(\left(x^2-x+1\right)\left(x^2+4x+1\right)=6x^2\)
\(\Leftrightarrow x^4+4x^3+x^2-x^3-4x^2-x+x^2+4x+1-6x^2=0\)
\(\Leftrightarrow x^4+3x^3-8x^2+3x+1=0\)
\(\Leftrightarrow x^4-2x^3+x^2+5x^3-10x^2+5x+x^2-2x+1=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(x^2+5x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2+5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-5+\sqrt{21}}{2}\\x=\dfrac{-5-\sqrt{21}}{2}\end{matrix}\right.\)
Vậy \(S=\left\{1;\dfrac{-5+\sqrt{21}}{2};\dfrac{-5-\sqrt{21}}{2}\right\}\)
b)\(3x^2+2x=2\sqrt{x^2+x}+1-x\)
ĐKXĐ: \(\) thỏa mãn với \(\forall x\)
\(\Leftrightarrow3x^2+2x-2\sqrt{x^2+x}-1+x=0\)
\(\Leftrightarrow3x^2+3x-2\sqrt{x^2+x}-1=0\)
\(\Leftrightarrow3\left(x^2+x\right)-2\sqrt{x^2+x}-1=0\) (1)
Đặt \(\sqrt{x^2+x}=a\left(a\ge0\right)\). Thay vào (1) ta có:
\(pt\Leftrightarrow3a^2-2a-1=0\)
\(\Leftrightarrow3a^2-3a+a-1=0\)
\(\Leftrightarrow\left(3a+1\right)\left(a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3a+1=0\\a-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=\sqrt{x^2+x}=\dfrac{-1}{3}\left(loại\right)\\a=\sqrt{x^2+x}=1\left(TM\right)\end{matrix}\right.\)
Ta có:
\(\sqrt{x^2+x}=1\Rightarrow x^2+x-1=0\)
Giải phương trình tìm được \(x=\pm\dfrac{\sqrt{5}-1}{2}\)
Vậy \(S=\left\{\pm\dfrac{\sqrt{5}-1}{2}\right\}\)
c) \(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
ĐKXĐ: \(x\ge-3;x\ge-1\)
\(\Leftrightarrow\sqrt{x+3}+2x\sqrt{x+1}-2x-\sqrt{\left(x+1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x+3}\right)\left(1-\sqrt{x+1}\right)+2x\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left(1-\sqrt{x+1}\right)\left(\sqrt{x+3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-\sqrt{x+1}=0\\\sqrt{x+3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{0;1\right\}\)
d) \(x^2+9x+20=2\sqrt{3x+10}\)
ĐKXĐ: \(x\ge\dfrac{-10}{3}\)
\(\Leftrightarrow x^2+9x+20-2\sqrt{3x+10}=0\)
\(\Leftrightarrow x^2+6x+9+3x+10-2\sqrt{3x+10}+1=0\)
\(\Leftrightarrow\left(x+3\right)^2+\left(\sqrt{3x+10}-1\right)^2=0\)
Vì \(\left(x+3\right)^2\ge0;\left(\sqrt{3x+10}-1\right)^2\ge0\) nên:
\(\left\{{}\begin{matrix}\left(x+3\right)^2=0\\\left(\sqrt{3x+10}-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow x=3\)
Vậy \(S=\left\{-3\right\}\)
Câu 4:
a)Gọi O là giao điểm của DH và AH.
Xét tứ giác ADHE có:
\(DAE=ADH=AEH=90^o\) (D, E là hình chiếu của H lên AB, AC)
\(\Rightarrow ADHE\) là hình chữ nhật.
Ta có: \(AHD+BHD=90^o\)
\(AHD+AHE=90^o\)
\(\Rightarrow BHD=AHE\) (1)
Xét \(\Delta AHE\) có: \(AHE+EAH=90^o\)
Xét \(\Delta BHD\) có: \(BHD+HBD=90^o\)
Theo (1), \(BHD=AHE\Rightarrow EAH=HBD\) (*)
Mặt khác, xét \(\Delta OAE\) có: \(OA=OE\) (tính chất hai đường chéo của hình chữ nhật).\(\Rightarrow\Delta OAE\) cân tại O
\(\Rightarrow OAE=OEA\) hay \(EAH=DEA\) (**)
Từ (*) và (**) ta có:\(DEA=HBD\left(=EAH\right)\) hay:\(DEA=CBA\)
Xét \(\Delta AED\left(A=90^o\right)\) và \(\Delta ABC\left(A=90^o\right)\) có:
\(DEA=CBA\left(cmt\right)\) \(\Rightarrow\Delta AED\) \(\sim\Delta ABC\left(g-g\right)\)
Vậy \(\Delta AED\sim\Delta ABC\left(đpcm\right)\)
b) Ta có:
\(S_{AHE}=\dfrac{AE.EH}{2}=\dfrac{1}{2}AE.EH=\dfrac{1}{2}S_{ADHE}\)
Mà: \(S_{ADHE}=\dfrac{1}{2}S_{ABC}\Rightarrow S_{AHE}=\dfrac{1}{4}S_{ABC}\)
Xét \(\Delta AHE\left(E=90^o\right)\) và \(\Delta BHD\left(D=90^o\right)\) có:
\(EAH=DBH\) (theo (*),a)
\(AE=BD\) (tính chất hình chữ nhật)
\(\Rightarrow\Delta AHE=\Delta BHD\)(cạnh góc vuông-góc nhọn)
\(\Rightarrow S_{AHE}=S_{BHD}=\dfrac{1}{4}S_{ABC}\)
Lại có:
\(S_{ADHE}+S_{BHD}+S_{CHE}=S_{ABC}\)
\(\Rightarrow S_{CHE}=\left(1-\dfrac{1}{2}-\dfrac{1}{4}\right)S_{ABC}=\dfrac{1}{4}S_{ABC}\)
\(\Rightarrow S_{CHE}=S_{BHD}\)\(\Rightarrow\Delta CHE=\Delta BHD\)
\(\Rightarrow HC=HB\)(hai cạnh tương ứng).
\(\Rightarrow AH\) là đường trung tuyến của \(\Delta ABC\)
Vì AH vừa là đường trung tuyến vừa là đường cao của \(\Delta ABC\) nên \(\Delta ABC\) vuông cân tại A.
Vậy \(\Delta ABC\) vuông cân tại A (đpcm).