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\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
cho\(\Delta ABC\)có 3 góc nhọn, đường cao BE, CF cắt nhau tại H. Qua A vẽ các đường thảng song song với BE và CF lần lượt cắt các đường thẳng CF và BE tại P và Q
1) CM: AH.AB=QA.BC
2)CM: BF.BA+CE.CA=BC2
3) Đường trung tuyến AM của tam giác ABC cắt PQ tại K. CM: 4 điểm A, K, E, Q cùng thuộc một đường tròn
Lời giải:
Từ điều kiện đề bài suy ra: \(\left\{\begin{matrix} x+y=\sqrt{7}\\ xy=1\end{matrix}\right.\)
\(A=x^7+y^7=(x^3+y^3)(x^4+y^4)-(x^3y^4+x^4y^3)\)
Có:
\(x^3+y^3=(x+y)^3-3xy(x+y)=(\sqrt{7})^3-3\sqrt{7}=4\sqrt{7}\)
\(x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2(xy)^2=(7-2)^2-2.1^2=23\)
\(x^3y^4+x^4y^4=(xy)^3(x+y)=1^3.\sqrt{7}=\sqrt{7}\)
Do đó:
\(A=4\sqrt{7}.23-\sqrt{7}=92\sqrt{7}-\sqrt{7}=91\sqrt{7}\)
3.
\(•x=3+\sqrt{2}\\ x^2=\left(3+\sqrt{2}\right)^2\\ x^2=9+2.3.\sqrt{2}+2\\ x^2=11+6\sqrt{2}\\• y=\sqrt{11+6\sqrt{2}}\\ y^2=\left(\sqrt{11+6\sqrt{2}}\right)^2\\ y^2=11+6\sqrt{2}\)
\(\Rightarrow x^2=y^2=11+6\sqrt{2}\)
1. ta có : \(4\sqrt{7}=\sqrt{112}\)
\(3\sqrt{3}=\sqrt{27}\)
ta thấy : \(\sqrt{112}>\sqrt{27}\) hay \(4\sqrt{7}>3\sqrt{3}\)
2. \(\dfrac{1}{4}\sqrt{82}=\sqrt{\dfrac{41}{8}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
ta thấy :\(\sqrt{\dfrac{41}{8}}< \sqrt{\dfrac{36}{7}}\) hay \(\dfrac{1}{4}\sqrt{82}< 6\sqrt{\dfrac{1}{7}}\)
3. \(x^2=\left(3+\sqrt{2}\right)^2\)
\(y^2=11+6\sqrt{2}\)=\(\left(3+\sqrt{2}\right)^2\)
ta thấy : \(x^2=y^2\Rightarrow x=y\)
Có \(x=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}=\frac{2}{\sqrt[3]{16}+\sqrt[3]{8}+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{4}\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right)}=\frac{\sqrt[3]{2}}{\sqrt[3]{4}+\sqrt[3]{2}+1}\)
\(y=\frac{2}{2\sqrt[3]{2}-2+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{16}-\sqrt[3]{8}+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{4}\left(\sqrt[3]{4}-\sqrt[3]{2}+1\right)}=\frac{\sqrt[3]{2}}{\sqrt[3]{4}-\sqrt[3]{2}+1}\)
Đặt \(\sqrt[3]{2}=a\)
=> \(x=\frac{a}{a^2+a+1}\) ,\(y=\frac{a}{a^2-a+1}\)
Có: \(x+y=\frac{a}{a^2+a+1}+\frac{a}{a^2-a+1}=\frac{a^3-a^2+a+a^3+a^2+a}{\left(a^2+a+1\right)\left(a^2-a+1\right)}=\frac{2a^3+2a}{a^4+a^2+1}\)
\(x-y=\frac{a}{a^2+a+1}-\frac{a}{a^2-a+1}=\frac{a^3-a^2+a-a^3-a^2-a}{\left(a^2+a+1\right)\left(a^2-a+1\right)}=\frac{-2a^2}{a^4+a^2+1}\)
Có x2-y2= (x-y)(x+y)=\(\frac{2a^3+2a}{a^4+a^2+1}.\frac{-2a^2}{a^4+a^2+1}=\frac{-2a^2.2a\left(a^2+1\right)}{\left(a^4+a^2+1\right)^2}=\frac{-4a^3\left(a^2+1\right)}{\left(a^4+a^2+1\right)^2}=\frac{-4.2\left(\sqrt[3]{4}+1\right)}{\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)^2}\)
=\(\frac{-8\left(\sqrt[3]{4}+1\right)}{\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)^2}\)
5.
ĐKXĐ: \(-\frac{1}{2}\le x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{2}+x+2\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=1\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
6.
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x^2-1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\left(vn\right)\end{matrix}\right.\)
2.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)
\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-x+1\\x+1=4x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-3=0\\4x^2-5x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)