đưa nhân tử ra ngoài dấu căn:
a, \(\sqrt{5\left(1-\sqrt{2}\right)^2}\)
b, \(\sqrt{27\left(2-\sqrt{5}\right)^2}\)
c, \(\sqrt{\dfrac{2}{\left(3-\sqrt{10}\right)^2}}\)
d, \(\sqrt{\dfrac{5\left(1-\sqrt{3}\right)^2}{4}}\)
Tính:
\(A=\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}\)
\(B=\dfrac{1}{\sqrt{2}-1}+\dfrac{14}{3+\sqrt{2}}\)
\(C=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)
\(D=\sqrt{\left(1-\sqrt{2}\right)^2}-3\sqrt{18}+4\sqrt{\dfrac{1}{2}}\)
Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
Các số sau đây có căn bậc hai không?
a) A = \(\left(1-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+2\right)\)
b) B = \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)
a) \(A=\left(1-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+2\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+\dfrac{4}{2}\right)\)
\(=\dfrac{2-\left(\sqrt{3}-1\right)}{2}:\dfrac{\left(\sqrt{3}-1\right)+4}{2}\)
\(=\dfrac{3-\sqrt{3}}{2}.\dfrac{2}{\sqrt{3}+3}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(1+\sqrt{3}\right)}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2}{2}\)
Vì \(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)^2>0\\2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\left(\sqrt{3}-1\right)^2}{2}>0\) hay A>0
=> A có căn bậc 2
Vậy......
b)\(B=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)\left(1+\sqrt{3}\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\sqrt{5}\right):\dfrac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\left(\dfrac{\sqrt{2}\left(3-1\right)}{1-3}-\sqrt{5}\right).\dfrac{5-2}{\sqrt{5}+\sqrt{2}}\)
\(=\left(-\sqrt{2}-\sqrt{5}\right).\dfrac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-\left(\sqrt{2}+\sqrt{5}\right).\dfrac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-3\)
Vì -3 < 0 hay B < 0
=> B không có căn bậc 2
Vậy.....
\(a:\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{5+2\sqrt{6}}\)
b : \(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}\)
c : \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right).\left(2+\dfrac{5-3\sqrt{5}}{3-\sqrt{5}}\right)\)
d : \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
a) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{5+2\sqrt{6}}\)
\(=\left|\sqrt{3}-2\right|+\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{3}\)
b) \(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\sqrt{2}\)
\(=\sqrt{2}-\sqrt{2}\)
\(=0\)
c) \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\cdot\left(2+\dfrac{5-3\sqrt{5}}{3-\sqrt{5}}\right)\)
\(=\left[2-\dfrac{\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}\right]\cdot\left[2-\dfrac{\sqrt{5}\left(3-\sqrt{5}\right)}{3-\sqrt{5}}\right]\)
\(=\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)\)
\(=4-4\sqrt{5}+5\)
\(=9-4\sqrt{5}\)
d) \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(=6-121\)
\(=-115\)
Tính:
\(A=\sqrt{27}-2\sqrt{48}+3\sqrt{75}\)
\(B=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(C=\sqrt{\left(2\sqrt{3}+1\right)^2}+\sqrt{\left(2\sqrt{3}-5\right)^2}\)
\(D=\sqrt{9-4\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(E=\dfrac{4}{\sqrt{5}-2}-\dfrac{32}{\sqrt{5}+1}\)
\(M=\dfrac{10}{3\sqrt{2}-4}+\dfrac{28}{3\sqrt{2}+2}\)
please help ;-;
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
B1: tính : a, \(\sqrt{400.0,81}\)
b, \(\sqrt{\dfrac{5}{27}.\dfrac{3}{20}}\)
c, \(\sqrt{\left(-5\right)^2.3^2}\)
d, \(\sqrt{\left(2-\sqrt{5}\right)^2.\left(2+\sqrt{5}\right)^2}\)
a)\(\sqrt{400.0,81}=\sqrt{4.81}=\sqrt{2^2.9^2}=2.9=18\)
b)\(\sqrt{\dfrac{5}{27}.\dfrac{3}{20}}=\sqrt{\dfrac{5}{3^3}.\dfrac{3}{2^2.5}}=\sqrt{\dfrac{1}{3^2.2^2}}=\dfrac{1}{3.2}=\dfrac{1}{6}\)
c)\(\sqrt{\left(-5\right)^2.3^2}=\sqrt{5^2.3^2}=5.3=15\)
d)\(\sqrt{\left(2-\sqrt{5}\right)^2\left(2+\sqrt{5}\right)^2}=\sqrt{\left[2^2-\left(\sqrt{5}\right)^2\right]^2}=\sqrt{\left(-1\right)^2}=1\)
Bài 1: Tính
a) \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
b) \(1\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
c) \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
d) \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
Bài 2: Cho (d₁): y = \(\dfrac{1}{2}x-4\) và (d₂): y = \(-3x+3\) . Vẽ (d₁) và (d₂) trên cùng một hệ trục tọa độ. Tìm tọa độ giao điểm A của 2 đường thẳng trên.
Helpp!!
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)
Tính
\(A=\sqrt{20}-3\sqrt{8}+5\sqrt{45}\)
\(B=\dfrac{30}{\sqrt{7}-1}+\dfrac{15}{\sqrt{7}+2}\)
\(C=\left(3-\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3+\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
\(D=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(E=\sqrt{7-4\sqrt{3}}-\sqrt{3+2\sqrt{3}}\)
1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)
2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)
3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)
4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)