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\(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=1-\sqrt{3}-\sqrt{3}-2\)
\(=-2\sqrt{3}-1\)
\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(4-2\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+4-2\sqrt{3}\)
\(=6-3\sqrt{3}\)
\(\frac{3}{\sqrt{7}-1}+\frac{3}{\sqrt{7}+1}=\frac{3\left[\sqrt{7}+1+\sqrt{7}-1\right]}{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}=\frac{6\sqrt{7}}{6}=\sqrt{7}\)
\(\frac{3}{\sqrt{X}-1}-\frac{2}{\sqrt{X}+1}+\frac{X-7}{X-1}=\frac{3\left(\sqrt{X}+1\right)-2\left(\sqrt{X}-1\right)+X-7}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{X+\sqrt{X}-2}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{\sqrt{X}+2}{\sqrt{X}+1}\)
TÍNH GIÁ TRỊ BIỂU THỨC:
\(\frac{3}{\sqrt{7}-1}\) + \(\frac{3}{\sqrt{7}+1}\)= \(\frac{3\left(\sqrt{7}+1\right)+3\left(\sqrt{7}-1\right)}{\left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right)}\)= \(\frac{3\sqrt{7}+3+3\sqrt{7}-3}{6}\)=\(\frac{6\sqrt{7}}{6}\)=\(\sqrt{7}\)
RÚT GỌN BIỂU THỨC:
\(\frac{3}{\sqrt{X}-1}\)-\(\frac{2}{\sqrt{X}+1}\)+\(\frac{X-7}{X-1}\)
= \(\frac{3\left(\sqrt{X}+1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)-\(\frac{2\left(\sqrt{X}-1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)+\(\frac{X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{3\sqrt{X}+3-2\sqrt{X}+2+X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{X+\sqrt{X}-2}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\left(\sqrt{X}+1\right)\left(\sqrt{X}-2\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\sqrt{X}-2}{\sqrt{X}-1}\)
CHÚC EM HỌC TỐT!
1. Trục căn thức ở mẫu:
\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\frac{1}{\sqrt{2005}+\sqrt{2009}}\)
=\(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+....+\frac{\sqrt{2005}-\sqrt{2001}}{4}+\frac{\sqrt{2009}-\sqrt{2005}}{4}\)
\(=\frac{\sqrt{2009}-1}{4}\)
2/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)
\(=6+3x\)
=> \(x^3-3x=6\)
=> \(B=x^3-3x+2000=6+2000=2006\)
\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
\(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}=\sqrt{2}+\sqrt{5}\)
\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-\dfrac{5}{\sqrt{3}-2\sqrt{2}}-\dfrac{5}{\sqrt{3}+\sqrt{8}}=\sqrt{\sqrt{3}^2+2\sqrt{3}.1+1^2}+\sqrt{\sqrt{3}^2-2\sqrt{3}.1+1^2}-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{\left(\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{3}+2\sqrt{2}\right)}-\dfrac{5\left(\sqrt{3}-2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\dfrac{5\sqrt{3}+10\sqrt{2}}{9-8}-\dfrac{5\sqrt{3}-10\sqrt{2}}{9-8}=\sqrt{3}+1+\sqrt{3}-1-5\sqrt{3}-10\sqrt{2}-5\sqrt{3}+10\sqrt{2}=-8\sqrt{3}\)\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}=2\sqrt{3}\)
a)\(\)https://www.cymath.com/answer?q=2sqrt(27)-6sqrt(4%2F3)%2B3%2F5sqrt(75)
\(M=2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}=2\sqrt{3^2.3}-6\sqrt{\frac{2^2.3}{3^2}}+\frac{3}{5}\sqrt{5^2.3}=.\)
\(=6\sqrt{3}-4\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
\(P=\frac{2}{x-1}\sqrt{\frac{x^2-2x+1}{4x^2}}.Với...0< x< 1\Leftrightarrow\) \(P=\frac{2}{x-1}\sqrt{\frac{\left(x-1\right)^2}{\left(2x\right)^2}}=\frac{2}{(x-1)}.\frac{\left(1-x\right)}{2x}=\frac{-1}{x}.\)
\(B=\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\) ĐK : \(x\ge0;x\ne1\)
\(=\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2}{\sqrt{x}+3}\)
\(=\frac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{1}{\sqrt{x}-1}\)
\(=\frac{3\sqrt{x}+1}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}-\frac{2}{\sqrt{x}+3}\)
\(=\frac{3\sqrt{x}+1-2\cdot\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}\)
\(=\frac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}-1}\)
Muốn rút gọn biểu thức này thì em cần trục căn thức ở mẫu, em nhé!
A = \(\frac{1}{3+\sqrt3}\) + \(\frac{1}{3-\sqrt3}\)
A = \(\frac{3-\sqrt3}{\left(3+\sqrt3\right).\left(3-\sqrt3\right)}\) + \(\frac{3+\sqrt3}{\left(3-\sqrt{3)}.\left(3+\sqrt3\right)\right.}\)
A = \(\frac{3-\sqrt3}{9-3}\) + \(\frac{3+\sqrt3}{9-3}\)
A = \(\frac{3-\sqrt3+3+\sqrt3}{6}\)
A = \(\frac{\left(3+3\right)+\left(\sqrt3-\sqrt3\right)}{6}\)
A = \(\frac{6+0}{6}\)
A = \(\frac66\)
A = 1
(1/3+căn bậc hai của 3 ) mũ 2 +(1/3-căn bặc hai của ba) mũ 2
= 1/9+3 +1/9-3
=1/12+1/6
=1/12+2/12
=3/12=1/4