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mỗi lần đăng chỉ đc 1 hỏi bài thôi bạn đăng dài thế không ai trả lời đâu
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a: \(A=\left(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right):\dfrac{2}{\sqrt{5}-\sqrt{3}}\)
\(=\left(\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right):\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}\)
\(=\dfrac{\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}}=1\)
a:
\(A=\left(\dfrac{\sqrt{14}-\sqrt{7}}{2\sqrt{2}-2}+\dfrac{\sqrt{15}-\sqrt{5}}{2\sqrt{3}-2}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\dfrac{\sqrt{7}+\sqrt{5}}{2}\right)\cdot\dfrac{\sqrt{7}-\sqrt{5}}{1}=\dfrac{7-5}{2}=1\)
a:
\(A=\left(a-b\right)\cdot\sqrt{\dfrac{ab}{\left(a-b\right)^2}}\)
\(=\left(a-b\right)\cdot\dfrac{\sqrt{ab}}{\left|a-b\right|}\)
a<b<0
=>a-b<0
=>\(A=\left(a-b\right)\cdot\dfrac{\sqrt{ab}}{-\left(a-b\right)}=-\sqrt{ab}\)
a:
\(A=\sqrt{\dfrac{9+12a+4a^2}{b^2}}\)
\(=\sqrt{\dfrac{\left(2a+3\right)^2}{b^2}}=\left|\dfrac{2a+3}{b}\right|\)
a>=-3/2
=>2a+3>=0
b<0
=>\(\dfrac{2a+3}{b}< =0\)
=>\(A=\dfrac{-\left(2a+3\right)}{b}\)
b:
\(A=\left(\dfrac{1}{3-2\sqrt{2}}-\dfrac{6}{2+\sqrt{2}}\right)\left(3+5\sqrt{2}\right)\)
\(=\left(\dfrac{3+2\sqrt{2}}{1}-\dfrac{6\left(2-\sqrt{2}\right)}{2}\right)\left(3+5\sqrt{2}\right)\)
\(=\left(3+2\sqrt{2}-3\left(2-\sqrt{2}\right)\right)\cdot\left(3+5\sqrt{2}\right)\)
\(=\left(5\sqrt{2}-3\right)\left(5\sqrt{2}+3\right)\)
=50-9
=41
b:
\(A=\left(\dfrac{1}{\sqrt{5}-2}-\dfrac{59}{3\sqrt{7}-2}\right)\left(3\sqrt{7}+\sqrt{5}\right)\)
\(=\left(\dfrac{\sqrt{5}+2}{5-4}-\dfrac{59\left(3\sqrt{7}+2\right)}{63-4}\right)\left(3\sqrt{7}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}+2-3\sqrt{7}-2\right)\left(3\sqrt{7}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-3\sqrt{7}\right)\left(\sqrt{5}+3\sqrt{7}\right)\)
=5-63
=-58
b:
\(A=\dfrac{x}{\sqrt{x}-1}+\dfrac{2x-\sqrt{x}}{\sqrt{x}-x}\)
\(=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\sqrt{x}-1\)
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Em tách ra 1-2 bài/1 câu hỏi để mọi người hỗ trợ nhanh nhất nha!
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Đề bài không rõ ràng, không có điều kiện cụ thể. Bạn coi lại.
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a) Áp dụng BĐT AM-GM ta có:
\(x+y\ge2\sqrt{xy}\)
\(\Rightarrow\)\(\frac{x+y}{2}\ge\sqrt{xy}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y\)
b) Áp dụng BĐT AM-GM ta có:
\(\frac{\sqrt{x}}{\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{x}}\ge2\sqrt{\frac{\sqrt{x}}{\sqrt{y}}.\frac{\sqrt{y}}{\sqrt{x}}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y\)
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1. \(\left\{{}\begin{matrix}4x-2y=3.\\6x-3y=5.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}12x-6y=9.\\12x-6y=10.\end{matrix}\right.\)\(\Leftrightarrow x\in\phi.\)
2. \(\left\{{}\begin{matrix}2x+3y=5.\\4x+6y=10.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}8x+12y=20.\\8x+12y=20.\end{matrix}\right.\)\(\Leftrightarrow x\in\phi.\)
3. \(\left\{{}\begin{matrix}3x-4y+2=0.\\5x+2y=14.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-4y=-2.\\5x+2y=14.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-4y=-2.\\10x+4y=28.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-4y=-2.\\13x=26.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2.\\x=2.\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}2x+5y=3.\\3x-2y=14.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+15y=9.\\6x-4y=28.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x+5y=3.\\19y=-19.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4.\\y=-1.\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}.\\x+y-10=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-2y=0.\\x+y=10.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2y=0.\\3x+3y=30.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=10.\\-5y=-30.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4.\\y=6.\end{matrix}\right.\)
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a: ΔOCD cân tại O có OH là trung tuyến
nên OH vuông góc CD
góc OHS=góc OAS=90 độ
=>OHAS nội tiếp
b: góc SIA=1/2(sđ cung AC+sđ cung BD)
=1/2(sđ cung AC+sđ cung BA+sđ cung AD)
=1/2(sđ cung BC+sđ cung AD)
góc SAH=góc SAB+góc HAB
=1/2(sđ cung BC+sđ cung AD)
=>góc SIA=góc SAH
mà góc ISA chung
nên ΔSAH đồng dạng với ΔSIA
f: \(F=\dfrac{1-\sqrt{x-1}}{\sqrt{x-2\sqrt{x-1}}}\)
\(=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1}-1}\)
=-1