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a) A= \(\sqrt{2-\sqrt{3}}\) \(\left(\sqrt{6}-\sqrt{2}\right)\)\(\left(2+\sqrt{3}\right)\)
A= \(\sqrt{2-\sqrt{3}}\) . \(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{3}}\) .\(\left(\sqrt{6}-\sqrt{2}\right)\)
A= \(\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\) . \(\sqrt{2+\sqrt{3}}\) . \(\sqrt{2}\left(\sqrt{3}-1\right)\)
A= 1. \(\sqrt{2\left(2+\sqrt{3}\right)}\) \(\left(\sqrt{3}-1\right)\)
A=\(\sqrt{4+2\sqrt{3}}\) .\(\left(\sqrt{3}-1\right)\)
A=\(\sqrt{\left(\sqrt{3}+1\right)^2}\) \(\left(\sqrt{3}-1\right)\)
A=\(\left|\sqrt{3}+1\right|\)\(\left(\sqrt{3}-1\right)\)
A=\(\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
A=3-1
A=2
Vậy A=2
b)\(\frac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2}+\sqrt{3}}\) = \(\frac{\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}{\sqrt{2}+\sqrt{3}}\) = \(\frac{\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{\sqrt{2}+\sqrt{3}}\)=\(\frac{\sqrt{2+\sqrt{3}}.1}{\sqrt{2}+\sqrt{3}}\) = \(\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}+\sqrt{3}}\) .
Lần sau bạn chú ý viết rõ yêu cầu đề bài.
Lời giải:
\(P=3\left(1+\frac{1}{\sqrt{x}-1}\right):\left(\frac{x+2}{(\sqrt{x}-1)(\sqrt{x}+2)}-\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}+2)(\sqrt{x}-1)}\right)\)
\(=3.\frac{\sqrt{x}}{\sqrt{x}-1}:\frac{2+\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}=3.\frac{\sqrt{x}}{\sqrt{x}-1}.(\sqrt{x}-1)=3\sqrt{x}\)
Câu A=4
Cách giải:
\(\left(5\sqrt{3}+2\sqrt{12}-\sqrt{75}\right):\sqrt{3}\)
\(=\left(5\sqrt{3}+2\sqrt{4\cdot3}-\sqrt{25\cdot3}\right)\)\(:\sqrt{3}\)
\(=\left(5\sqrt{3}+4\sqrt{3}-5\sqrt{3}\right)\)\(:\sqrt{3}\)
\(1,\sqrt{\left(-0,3\right)^2}=\sqrt{0,09}=0,3\)
\(2,-\frac{1}{2}\sqrt{\left(0,3\right)^2}=-\frac{1}{2}.0,3=-0,15\)
\(3,\sqrt{a^{10}}=\sqrt{\left(a^5\right)^2}=a^5\left(a\ge0\right)\)
\(4,\sqrt{\left(2-x\right)^2}=\left|2-x\right|=2-x\left(x\le2\right)\)
\(5,\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)
\(6,\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)(Vì \(1< \sqrt{2}\))
\(7,\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
\(8,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)
\(=-2\)
\(9,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
a)
\(\sqrt[3]{(\sqrt{2}+1)(3+2\sqrt{2})}=\sqrt[3]{(\sqrt{2}+1)(2+2\sqrt{2}+1)}\)
\(=\sqrt[3]{(\sqrt{2}+1)(\sqrt{2}+1)^2}=\sqrt[3]{(\sqrt{2}+1)^3}=\sqrt{2}+1\)
b)
\(\sqrt[3]{(4-2\sqrt{3})(\sqrt{3}-1)}=\sqrt[3]{(3-2\sqrt{3}+1)(\sqrt{3}-1)}\)
\(=\sqrt[3]{(\sqrt{3}-1)^2(\sqrt{3}-1)}=\sqrt[3]{(\sqrt{3}-1)^3}=\sqrt{3}-1\)
c)
\((\sqrt[3]{4}+1)^3-(\sqrt[3]{4}-1)^3=[(\sqrt[3]{4}+1-(\sqrt[3]{4}-1)][(\sqrt[3]{4}+1)^2+(\sqrt[3]{4}+1)(\sqrt[3]{4}-1)+(\sqrt[3]{4}-1)^2]\)
\(=2[\sqrt[3]{16}+1+2\sqrt[3]{4}+\sqrt[3]{16}-1+\sqrt[3]{16}+1-2\sqrt[3]{4}]\)
\(=2(3\sqrt[3]{16}+1)\)
d)
\((\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}+\sqrt[3]{2})=[(\sqrt[3]{3})^2-\sqrt[3]{3}.\sqrt[3]{2}+(\sqrt[3]{2})^2](\sqrt[3]{3}+\sqrt[3]{2})\)
\(=(\sqrt[3]{3})^3+(\sqrt[3]{2})^3=3+2=5\)
e)
\(E=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
Áp dụng công thức $(a+b)^3=a^3+b^3+3ab(a+b)$ ta có:
\(E^3=20+14\sqrt{2}+20-14\sqrt{2}+3\sqrt[3]{(20+14\sqrt{2})(20-14\sqrt{2})}.E\)
\(E^3=40+3\sqrt[3]{20^2-(14\sqrt{2})^2}.E\)
\(E^3=40+3\sqrt[3]{8}.E=40+6E\)
\(\Leftrightarrow E^2(E-4)+4E(E-4)+10(E-4)=0\)
\(\Leftrightarrow (E-4)(E^2+4E+10)=0\)
Dễ thấy $E^2+4E+10=(E+2)^2+6\neq 0$ nên $E-4=0$ hay $E=4$
\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\) ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))
\(=\sqrt{\left(2+\sqrt{7}+2-\sqrt{7}\right)\left(2+\sqrt{7}-2+\sqrt{7}\right)}\)
\(=\sqrt{4\cdot\sqrt{7}}\)
\(2,\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}-\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}=\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2=\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2-\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(=\left(3\sqrt{5}-5\sqrt{2}+5\sqrt{2}+3\sqrt{5}\right)\left(3\sqrt{5}-5\sqrt{2}-5\sqrt{2}-3\sqrt{5}\right)\)
\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)
\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}\)
cuối lười tính nên thôi nhá :>
\(a.\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\dfrac{3}{3-\sqrt{6}}=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}=\sqrt{6}-\dfrac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{3\sqrt{2}-3\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{-3\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=-3\) \(b.\left(2\sqrt{2}-\sqrt{3}\right)^2-2\sqrt{3}\left(\sqrt{3}-2\sqrt{2}\right)=\left(2\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)=8-3=5\) \(c.\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{9-5}:\sqrt{5}=\dfrac{2\sqrt{5}}{4}.\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{1}{2}\) \(d.\left(3-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3+\dfrac{\sqrt{ab}-3\sqrt{a}}{\sqrt{b}-3}\right)=\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)=9-a\)
a: \(=\sqrt{2}-1\)
b: \(=\sqrt{3}+1+2-\sqrt{3}=3\)