tính gt bt
a/\(P=\sqrt{1^3+\dfrac{1^2}{3}}+\sqrt{2^3+\dfrac{3^2}{5}}+\sqrt{3^3+\dfrac{5^2}{7}}+...+\sqrt{75^3+\dfrac{149^2}{151}}\)
Tính:
1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{2+\sqrt{5}}\)
2) \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
3) \(\dfrac{1}{\sqrt{5}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
4) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\)
5) \(-\dfrac{1}{\sqrt{2}-\sqrt{3}}\)\(-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
1: ta có: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=3+2\sqrt{2}+\sqrt{5}-2\)
\(=2\sqrt{2}+\sqrt{5}+1\)
2: Ta có: \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
\(=3+2\sqrt{2}-3+2\sqrt{2}\)
\(=4\sqrt{2}\)
Thực hiện phép tính (rút gọn biểu thức)
a) \(\dfrac{1}{\sqrt{5}-2}+\dfrac{4}{\sqrt{5}+1}\)
b) \(\dfrac{4}{\sqrt{3}-1}+\dfrac{7}{3-\sqrt{2}}=-2\sqrt{3}\) c) \(\left(\dfrac{4}{3-\sqrt{5}}-\dfrac{1}{\sqrt{5}-2}\right)\dfrac{7}{3-\sqrt{2}}\)
Lời giải:
a.
\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)
$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.
$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$
$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.
$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$
$=1(3+\sqrt{2})=3+\sqrt{2}$
Bài 1: Rút gọn biểu thức:
a) \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}+\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}+\sqrt{5}}\right)\)
b) \(\dfrac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\dfrac{1}{3}}\)
c) \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
d) \(\sqrt{\dfrac{3}{4}}+\sqrt{\dfrac{1}{3}}+\sqrt{\dfrac{1}{12}}\)
Bài 2: Giải các phương trình sau:
a) \(x^2+4x+5=2\sqrt{2x+3}\)
b) \(x^2+9x+20=2\sqrt{3x+10}\)
c) \(x^2+7x+14=2\sqrt{x+4}\)
d) \(4\sqrt{x+1}=x^2-5x+14\)
e) \(\sqrt{6-x}=3x-4\)
f) \(\sqrt{5x-9}=9-2x\)
Mọi người làm ơn giúp mình với. Mình đang cần gấp ạ. Cảm ơn mọi người rất nhiều
Bài 1:
a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)
\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)
\(=\dfrac{5}{2}\sqrt{5}\)
c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)
\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)
Bài 2:
e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)
Ta có: \(\sqrt{6-x}=3x-4\)
\(\Leftrightarrow6-x=\left(3x-4\right)^2\)
\(\Leftrightarrow9x^2-24x+16+6-x=0\)
\(\Leftrightarrow9x^2-25x+22=0\)
\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)
Vậy: Phương trình vô nghiệm
a) \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}\)
b) \(\dfrac{3}{\sqrt{2}-1}+\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{3}+1}\)
c) \(\dfrac{3}{\sqrt{5}-2}-\dfrac{3}{\sqrt{5}+2}\)
a) \(=\dfrac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\dfrac{14}{49-48}=\dfrac{14}{1}=14\)
b) \(=\dfrac{3\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=3\sqrt{2}+3+\sqrt{2}=3+4\sqrt{2}\)
c) \(=\dfrac{3\left(\sqrt{5}+2\right)-3\left(\sqrt{5}-2\right)}{5-4}=3\sqrt{5}+6-3\sqrt{5}+6=12\)
Bài 1 : (2 điểm) : Thực hiện phép tính và rút gọn các biểu thức sau :
a)A\(=-\left(\dfrac{1}{3-\sqrt{5}}+\dfrac{1}{3+\sqrt{5}}\right):\sqrt{5}\)
b)\(B=\sqrt{48+\sqrt{5\dfrac{1}{3}+2\sqrt{75}-\sqrt[5]{1\dfrac{1}{3}}}}\)
\(A=-\dfrac{3+\sqrt{5}+3-\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\dfrac{\sqrt{5}}{5}\\ A=\dfrac{-6}{4}\cdot\dfrac{\sqrt{5}}{5}=\dfrac{-3\sqrt{5}}{10}\)
Thực hiến phép tính :
a, \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
b, \(\dfrac{2}{3\sqrt{2}-4}-\dfrac{2}{3\sqrt{2}+4}\)
c, \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)
d, \(\dfrac{3}{2\sqrt{2}-3\sqrt{3}}-\dfrac{3}{2\sqrt{2}+3\sqrt{3}}\)
e, \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
g, \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(a,=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\dfrac{6}{-1}=-6\\ b,=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}=\dfrac{16}{2}=8\\ c,=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}=\dfrac{16}{2}=8\)
\(d,=\dfrac{6\sqrt{2}+9\sqrt{3}-6\sqrt{2}+9\sqrt{3}}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=\dfrac{18\sqrt{3}}{-19}=\dfrac{-18\sqrt{3}}{19}\\ e,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
a \(\left(2\sqrt{6-}4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)5\sqrt{6}\)
b \(\left(\dfrac{1}{\sqrt{5-}\sqrt{2}}+\dfrac{1}{\sqrt{5}+\sqrt{2}}+1\right).\dfrac{1}{\sqrt{2}+1^2}\)
c \(\dfrac{3}{\sqrt{2+}\sqrt{7}}\dfrac{1}{\sqrt{2-}\sqrt{7}}-1\)
d \(2x\sqrt{300}-15\sqrt{75}+5\sqrt{75}.15\)
a: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)
\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)
\(=60-60\sqrt{2}+45\sqrt{3}\)
b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)
\(\dfrac{2\sqrt{30}}{\sqrt{5}+\sqrt{6}+\sqrt{7}} \)
\(\sqrt{24}+6\sqrt{\dfrac{2}{3}+\dfrac{10}{\sqrt{6}-1}}\)
\(\dfrac{2\sqrt{15}+\sqrt{16}}{\sqrt{84}+\sqrt{6}}\)
\(2\sqrt{40\sqrt{2}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(\dfrac{\left(2+\sqrt{3}\right)^2-1}{\left(\sqrt{3}+1\right)^2}:\dfrac{\left(3+\sqrt{5}\right)^2-4}{\left(\sqrt{5}+1\right)^2}\)
giúp em với ạ
\(2\sqrt{40\sqrt{3}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\cdot\sqrt{40\sqrt{3}}-2\cdot\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=2\cdot2\sqrt{10}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-6\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=4\sqrt{10}\sqrt{\sqrt{3}}-4\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
Rút gọn:
A = \(\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
B = \(\dfrac{3\sqrt{2}+\sqrt{11}}{\sqrt{2}+\sqrt{6+\sqrt{11}}}+\dfrac{3\sqrt{2}-\sqrt{11}}{\sqrt{2}-\sqrt{6-\sqrt{11}}}+18\)
C = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n+3}}\)với n thuộc N*
D = \(\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\left(\sqrt{15}-1\right)\left(7-2\sqrt{3}+\sqrt{5}\right)\)
E=\(\dfrac{\left(4+\sqrt{3}\right)}{\sqrt[]{1}+\sqrt{3}}+\dfrac{\left(8+\sqrt{15}\right)}{\sqrt{3}+\sqrt{5}}+...+\dfrac{2k+\sqrt{k^2-1}}{\sqrt{k-1}+\sqrt{k+1}}+...+\dfrac{240+\sqrt{14399}}{\sqrt{119}+\sqrt{121}}\)
F = \(\left(\dfrac{2a+1}{a\sqrt{a}-1}-\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\) với a >= 0 và a khác 1