Sử dụng hằng đẳng thức bình phương của một tổng, bình phương của một hiểu để khai phương
a) \(\sqrt{17-3\sqrt{32}}\) +\(\sqrt{17+3\sqrt{32}}\)
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\(\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{6}+\sqrt{2}}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}+\frac{4\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}\)
\(=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\frac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}\)
\(=\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}\)
\(=\sqrt{5}+\sqrt{6}\)
ĐKXĐ: Bạn tự làm nha
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}+1\)
\(=\frac{x^2-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{x^2+x+1}{x+\sqrt{x}+1}\)
\(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{1\left(\sqrt{a}-1\right)-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}.\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-1-2}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(a-1\right)}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
\(\left(5\sqrt{7}+7\sqrt{5}\right):\sqrt{35}=\left(\sqrt{5^2.7}+\sqrt{7^2.5}\right):\sqrt{35}\)
\(=\left(\sqrt{35.5}+\sqrt{35.7}\right):\sqrt{35}\)
\(=\sqrt{35}\left(\sqrt{5}+\sqrt{7}\right):\sqrt{35}\)
\(=\sqrt{5}+\sqrt{7}\)
Toán Học Team
ĐKXĐ: \(x\ge0;x\ne1;\)
\(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}\)
\(=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(1-x\right)^2}{2}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{1-2x+x^2}{2}\)
\(=\frac{x\sqrt{x}+2x+\sqrt{x}-2x-4\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^1}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{x\sqrt{x}-x-4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}\)
\(=\frac{\sqrt{x}\left(x-\sqrt{x}-4\right)\left(x-1\right)}{2\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(x-\sqrt{x}-4\right)\left(\sqrt{x}-1\right)}{2}\)
Ta có: \(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\times\frac{\left(1-x\right)^2}{2}\)
\(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(1-x\right)^2}{2}\)
\(P=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(1-x\right)^2}{2}\)
\(P=\frac{x\sqrt{x}-4\sqrt{x}-x}{-\left(1-x\right)\left(\sqrt{x}+1\right)}.\frac{\left(1-x\right)^2}{2}\)
\(P=\frac{\sqrt{x}\left(x-4-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{2}\)
Có \(\left(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\right)^2\)
\(=\left(\sqrt{17-3\sqrt{32}}\right)^2+2\left(\sqrt{17-3\sqrt{32}}\right)\left(\sqrt{17+3\sqrt{32}}\right)\)\(+\left(\sqrt{17=3\sqrt{32}}\right)^2\)
\(=17-3\sqrt{32}+2\sqrt{\left(17-3\sqrt{32}\right)\left(17+3\sqrt{32}\right)}\)\(+17+3\sqrt{32}\)
\(=34+2\sqrt{17^2-9.32}\)
\(=34+2\sqrt{289-288}\)
\(=34+2\sqrt{1}=34+2=36\)
\(\Rightarrow\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=\sqrt{36}=6\)
(Vì có \(\hept{\begin{cases}\sqrt{17-3\sqrt{32}}\ge0\\\sqrt{17+3\sqrt{32}}\ge0\end{cases}}\)nên \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\ge0\))
Ở cuối dòng 2 mình nhầm dấu + thành dấu = nghe mọi người