\(\sqrt{-1}\) cái này là số phức nha bạn
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A(frac{left(sqrt{x}-1right)left(x+sqrt{x}+1right)}{sqrt{x}left(sqrt{x}-1right)}-frac{left(sqrt{x}+1right)left(x-sqrt{x}+1right)}{sqrt{x}left(sqrt{x}+1right)}):(frac{2left(sqrt{x}-1right)^2}{x-1}left(frac{x+sqrt{x}+1-left(x-sqrt{x}+1right)}{sqrt{x}}right).frac{sqrt{x}+1}{2left(sqrt{x}-1right)}frac{2sqrt{x}}{sqrt{x}}.frac{sqrt{x}+1}{2left(sqrt{x}-1right)}frac{sqrt{x}+1}{sqrt{x}-1}
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A=(\(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)-\(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)):(\(\frac{2\left(\sqrt{x}-1\right)^2}{x-1}\)
=\(\left(\frac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}\right).\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
=\(\frac{2\sqrt{x}}{\sqrt{x}}.\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(y=\frac{1}{x^2+\sqrt{x}}\sqrt[]{}\sqrt{\sqrt[]{}\sqrt[]{}\frac{\frac{\frac{\frac{ }{ }}{ }}{ }}{ }}\)
Câu 10:
Cho \(\frac{1}{\sqrt{2}+x}\cdot\frac{1}{\sqrt{2}+y}=a\)
Và \(\frac{1}{\sqrt{2}+x}+\frac{1}{\sqrt{2+y}}=b\)
Tìm x + y ( theo a và b )
$\sqrt{4x^2+5x+1}+2\sqrt{x^2-x+1}=3-9x$
tinh gia tri bieu thuc a = \(\sqrt{4+\sqrt[3]{8}+\frac{2}{3}}+3^{10}\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)
tinh gia tri bieu thuc a = \(\sqrt{4+\sqrt[3]{8+\frac{2}{3}+3^{10}\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}}}\)
tinh gia tri bieu thuc a = \(\sqrt{4+\sqrt[3]{8+\frac{2}{3}+3^{10}\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}}}\)
ngoài sân có 1 tỉ con chó và ngoài vườn có \(\sqrt[23123232]{\sqrt{453}\frac{565}{099999}}\)con gà=???????????????????
sqrt{4+sqrt{6}}..........sqrt{4+sqrt{6}}..........sqrt{4+sqrt{6}}..........sqrt{4+sqrt{6}}..........sqrt{4+sqrt{6}}..........sqrt{4+sqrt{6}}..........sqrt{4+sqrt{6}}..........sqrt{4+sqrt{6}}..........sqrt{4+sqrt{6}}..........sqrt{4+sqrt{6}}..........
Đọc tiếp
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)
\(\sqrt{4+\sqrt{6}}=..........\)