f)\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)- \(\dfrac{\sqrt{6}-3}{\sqrt{2}-\sqrt{3}}\)
g)\(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0,4}\right)\)
giải chi tiết cụ thể giúp mk với ạ
Câu1:\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)- \(\dfrac{\sqrt{6}-3}{\sqrt{2}-\sqrt{3}}\)
Câu2:\(\dfrac{\sqrt{2}}{\sqrt{3}-1}-\sqrt{\dfrac{3}{2}}\)
giải cụ thể giúp mk vớiiiii ạ
Câu 1:
\(=\sqrt{3}-\sqrt{2}-\sqrt{2}=3-2\sqrt{2}\)
1.)\(\sqrt{\left(\sqrt{3}-3\right)^2}\)+\(\sqrt{4-2\sqrt{3}}\)
2)\(\dfrac{1}{\sqrt{5}-2}\) +\(\dfrac{\sqrt{10}-\sqrt{5}}{1-\sqrt{2}}\)
3)\(\dfrac{\sqrt{2}}{\sqrt{3}-1}\) -\(\sqrt{\dfrac{3}{2}}\)
giải chi tiết giúp mk vớiiiii ạ
1)\(=\left|\sqrt{3}-3\right|+\sqrt{\left(\sqrt{3}-1\right)^2}=3-\sqrt{3}+\left|\sqrt{3}-1\right|=3-\sqrt{3}+\sqrt{3}-1=2\)
2: \(=\sqrt{5}+2-\sqrt{5}=2\)
rÚT GỌN: G=\(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{6}}-\sqrt{2}\)
chững minh : a) \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt[]{6}=9\)
b)\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
c)\(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
giúp mk với tối mai mk nạp rồi
a) \(VT=2\sqrt{6}-4\sqrt{2}+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
\(=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)
\(=-4\sqrt{2}+1+4\sqrt{2}+8\)
\(=1+8\)
\(=9\)
\(\Rightarrow VT=VP\) (đpcm).
b) \(VT=\left(3\sqrt{10}-3\sqrt{2}+\sqrt{50}-\sqrt{10}\right)\sqrt{3-\sqrt{5}}\)
\(=\left(3\sqrt{10}-3\sqrt{2}+5\sqrt{2}-\sqrt{10}\right)\sqrt{3-\sqrt{5}}\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{10}+2\sqrt{2}\right)^2\cdot\left(3-\sqrt{5}\right)}\)
\(=\sqrt{\left(40+8\sqrt{20}+8\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{\left(48+16\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{16\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{16\left(9-5\right)}\)
\(=\sqrt{64}\)
\(=8\)
\(\Rightarrow VT=VP\) (đpcm).
c) \(VT=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{2+\sqrt{5}}\)
\(=2\left(\sqrt{5}+2\right)-\dfrac{2\left(2-\sqrt{5}\right)}{-1}\)
\(=2\sqrt{5}+4+2\left(2-\sqrt{5}\right)\)
\(=2\sqrt{5}+4+4-2\sqrt{5}\)
\(=4+4\)
\(=8\)
\(\Rightarrow VT=VP\) (đpcm).
mn cs thể giải chi tiết đc ko ạ
\(A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 4$
\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-1\right]:\left[\frac{(3-\sqrt{x})(3+\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+3)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right]\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-(\sqrt{x}-2)}{\sqrt{x}+3}=\frac{-3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{-(\sqrt{x}-2)}=\frac{3}{\sqrt{x}-2}\)
Tính (rút gọn)
a) \(\left(\sqrt{3+\sqrt{5}}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-}\sqrt{15}\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3+2}}\)
d) \(\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
f)\(\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\)
g)\(\dfrac{\sqrt{2+\sqrt{3}}}{3+\sqrt{3}}\)
h)\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)
\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)
\(d,\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)\\ =\left(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right)\sqrt{2}\left(\sqrt{5}-1\right)\\ =\left(2\sqrt{4+\sqrt{5}-1}\right)\sqrt{2}\left(\sqrt{5}-1\right)\\ =\sqrt{24+8\sqrt{5}}\left(\sqrt{5}-1\right)\\ =\sqrt{\left(2\sqrt{5}+2\right)^2}\left(\sqrt{5}-1\right)\\ =2\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\\ =2\left(5-1\right)\\ =8\)
rút gọn
a) \(\left(-7\sqrt{7}\right)\left(-2\sqrt{8}\right)\)
b) \(-\sqrt{33}.3\sqrt{3}\)
c) \(\left(3\sqrt{5}\right).\left(-10\sqrt{3}\right)\)
d) \(\dfrac{1}{2}\sqrt{5}.\left(-6\sqrt{2}\right)\)
e) \(\dfrac{2}{3}\sqrt{7}.\left(-\dfrac{9}{16}\sqrt{3}\right)\)
f) \(15\sqrt{6}:5\sqrt{3}\)
g) \(-25\sqrt{12}:\left(-5\sqrt{6}\right)\)
h) \(36\sqrt{8}:12\sqrt{2}\)
i) \(4\sqrt{27}:\left(-2\sqrt{3}\right)\)
i: =-12*căn 3/2căn 3=-6
h: =72căn 2/12căn 2=6
g: =25căn 12/5căn 6=5căn 2
f: =(15:5)*căn 6:3=3căn 2
d: =-1/2*6*căn 10=-3căn 10
Tính :
a) \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\)
b) \(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21+12\sqrt{3}}}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
d) \(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
e) \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
f) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
g) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)-\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
Rút gọn:
1) \(\dfrac{16-6\sqrt{7}}{\sqrt{7}-3}\)
2) \(\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2+4\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
3) \(\dfrac{\left(\sqrt{3}+2\sqrt{5}\right)^2-8\sqrt{15}}{\sqrt{6}-2\sqrt{10}}\)
Giúp em với ạ. Help mee !!!
Câu 1,2 bạn đã đăng và có lời giải rồi
Câu 3:
\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)
Thực hiện các phép tính (không được ghi mỗi kết quả không, phải giải chi tiết)
A = \(2\sqrt{10}.3\sqrt{8}.2\)
B = \(\sqrt{20}\left(2\sqrt{3}-\sqrt{5}\right)\)
C = \(\left(2\sqrt{5}-3\right)\left(2\sqrt{5}+3\right)\)
a: \(=12\sqrt{80}=48\sqrt{5}\)
b: \(=2\sqrt{5}\cdot2\sqrt{3}-10=4\sqrt{15}-10\)
c: =20-9=11