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những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
a, Xét \(M^2=4-\sqrt{10-2\sqrt{5}}+4+\sqrt{10-2\sqrt{5}}-2\sqrt{\left(4-\sqrt{10-2\sqrt{5}}\right)\left(4+\sqrt{10-2\sqrt{5}}\right)}\)
\(=8-2\sqrt{4^2-10+2\sqrt{5}}\\ =8-2\sqrt{16-10+2\sqrt{5}}\\ =8-2\sqrt{6+2\sqrt{5}}\\ =8-2\sqrt{\left(\sqrt{5}+1\right)^2}\\ =8-2\left(\sqrt{5}+1\right)\\ =8-2\sqrt{5}-2=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\\ \Rightarrow M=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
b,
\(P\sqrt{2}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\\ =\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}\\ =\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}\\ =\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}\\ =\frac{\sqrt{2}\left[\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)+\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)\right]}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ =\frac{\sqrt{2}\left(2\sqrt{3}+3-2-\sqrt{3}+2\sqrt{3}-3+2-\sqrt{3}\right)}{\sqrt{3}\left(3-1\right)}\\ =\frac{\sqrt{2}\left(2\sqrt{3}\right)}{\sqrt{3}\cdot2}=\sqrt{2}\\ \Rightarrow P=1\)
c,
\(Q\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\\ \Rightarrow Q=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{2}\cdot\sqrt{3}=6\)
Chúc bạn học tốt nha
.
ĐK: \(x-9\ne0\Rightarrow x\ne9\)
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)
\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)
2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)