? cho a,b,c tìm x,y,z là seo?

a)\(3\sqrt{5}\left(\sqrt{45}+2\sqrt{125}-4\sqrt{80}\right)=3\sqrt{5}\left(3\sqrt{5}+10\sqrt{5}-16\sqrt{5}\right)=45+150-240=-45\)b) \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2-\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\sqrt{4-3}=1\)

c) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}=\left|4-\sqrt{15}\right|+\left|3-\sqrt{15}\right|=4-\sqrt{15}+\sqrt{15}-3=1\)

d) \(\sqrt{\dfrac{6\sqrt{3}-8}{2\sqrt{3}+1}}=\sqrt{\dfrac{\left(6\sqrt{3}-8\right)\left(2\sqrt{3}-1\right)}{11}}=\sqrt{\dfrac{36-6\sqrt{3}-16\sqrt{3}+8}{11}}=\sqrt{\dfrac{44-22\sqrt{3}}{11}}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

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B

D

Ta có \(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

           0,2---->0,4----->0,2------->0,2(mol)

\(V_{H_2\left(đktc\right)}=0,2\cdot22,4=4,48\left(l\right)\)

\(m_{ZnCl_2}=0,2\cdot\left(65+35,5\cdot2\right)=27,2\left(g\right)\)

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\(B=3\sqrt{\dfrac{1}{2}}+\sqrt{4,5}-\sqrt{12,5}=\dfrac{3\sqrt{2}}{2}+\dfrac{3\sqrt{2}}{2}-\dfrac{5\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}\)

\(C=\sqrt{\dfrac{1}{3}}+\sqrt{1\dfrac{1}{5}}+4\sqrt{3}=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{30}}{5}+4\sqrt{3}=\dfrac{5\sqrt{3}+3\sqrt{30}+4\sqrt{3}}{15}=\dfrac{9\sqrt{3}+3\sqrt{30}}{15}=\dfrac{\sqrt{30}+3\sqrt{3}}{5}\)\(D=2\sqrt{\dfrac{8y}{5}}+\sqrt{\dfrac{45y}{2}}-\sqrt{10y}=\dfrac{20\sqrt{2y}}{5}+\dfrac{3\sqrt{10y}}{2}-\sqrt{10y}=\dfrac{40\sqrt{2y}+15\sqrt{10y}-10\sqrt{10y}}{10}=\dfrac{40\sqrt{2y}+5\sqrt{10y}}{10}=\dfrac{8\sqrt{2y}+\sqrt{10y}}{2}\)

1. \(P=\left(\dfrac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\left(x>0\right)\)

\(P=\dfrac{1-x}{\sqrt{x}}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(P=\dfrac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1+1-\sqrt{x}}\)

\(P=\dfrac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(1+\sqrt{x}\right)^2}{-\sqrt{x}}\)

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

\(P=\dfrac{\left(1+\sqrt{\left(\sqrt{3}+1\right)^2}\right)^2}{-\sqrt{\left(\sqrt{3}+1\right)^2}}=\dfrac{\left(2+\sqrt{3}\right)^2}{-\sqrt{3}-1}=\dfrac{-7-4\sqrt{3}}{\sqrt{3}+1}\)

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Xét \(\Delta ABC\) có AD=DB;DE//BC nên AE=EC hay E là trung điểm AC

Xét \(\Delta ADE\) và \(\Delta EFC\) có:

\(\left\{{}\begin{matrix}\widehat{DAE}=\widehat{FEC}\\AE=EC\left(cmt\right)\\\widehat{AED}=\widehat{ECF}\end{matrix}\right.\)

\(\Rightarrow\)\(\Delta ADE\) \(=\) \(\Delta EFC\)\(\left(g.c.g\right)\) 

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1. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)\left(x>0\right)\)

\(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

2. Để \(P>0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\sqrt{2}\\x>\sqrt{-1}\left(L\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< \sqrt{2}\\x< \sqrt{-1}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< \sqrt{2}\end{matrix}\right.\)

Vậy \(P>0\Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< \sqrt{2}\end{matrix}\right.\)

a) \(A=\sqrt{x-2}+\sqrt{6-x}\)

\(\Rightarrow A^2=x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}\)

Ta có \(\sqrt{\left(x-2\right)\left(6-x\right)}\ge0,\forall x\)

Do đó \(A^2=4+2\sqrt{\left(x-2\right)\left(6-x\right)}\ge4\)

Mà A không âm \(\Leftrightarrow A\ge2\)

Dấu "=" \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

Áp dụng BĐT Bunhiacopxky:

\(A^2=\left(\sqrt{x-2}+\sqrt{6-x}\right)^2\le\left(x-2+6-x\right)\left(1+1\right)=4\cdot2=8\)

\(\Leftrightarrow A\le\sqrt{8}\)

Dấu "=" \(\Leftrightarrow x-2=6-x\Leftrightarrow x=4\)

Mấy bài còn lại y chang nha 

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