Rút gọn các căn thức sau
\(\sqrt{12-3\sqrt{7}}\)
\(\sqrt{2-\sqrt{3}}\)
\(\sqrt{3-\sqrt{5}}\)
\(\sqrt{7-3\sqrt{5}}\)
Rút gọn các biểu thức sau:
9, A = \(\sqrt{4+\sqrt{15}}-\sqrt{7-3\sqrt{5}}\)
10, A = \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
11, A = \(\text{}\text{}\text{}\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
12, A = \(\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{6-3\sqrt{3}}\)
13, A = \(\sqrt{9-4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3=6
13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
Rút gọn các biểu thức :
a) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
b) \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
c)\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)
Rút gọn các biểu thức:
\(\frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7+\sqrt{3}}}}{\sqrt{\sqrt{7-2}}}\)
Tính giá trị biểu thức:
\(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
Rút gọn các căn thức sau:
\(\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7-2\sqrt{10}}}\)
\(\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(=\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{6\left(2+\sqrt{3}\right)}-4+\sqrt{2\left(2+\sqrt{3}\right)}}{2\sqrt{6}}\right)\)
\(=\frac{\sqrt{2+\sqrt{3}}}{2}.\left(\frac{2\sqrt{6}}{\sqrt{12+6\sqrt{3}}-4+\sqrt{4+2\sqrt{3}}}\right)\)
\(=\frac{\sqrt{6\left(2+\sqrt{3}\right)}}{\left|\sqrt{3}+3\right|-4+\left|\sqrt{3}+1\right|}\)
\(=\frac{\left|\sqrt{3}+3\right|}{\sqrt{3}+3-4+\sqrt{3}+1}\)
\(=\frac{\sqrt{3}+3}{2\sqrt{3}}\)
\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7-2\sqrt{10}}}\)
\(=\frac{\sqrt{3}+\sqrt{\left(\sqrt{2}\right)^2+6\sqrt{2}+9}-\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{6}+\left(\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}+1}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{10}+\left(\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}+3-\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}+\sqrt{2}}\)
\(=\frac{3}{2\sqrt{2}+1}\)
B 5. Rút gọn các biểu thức sau:
a)\(\sqrt{7+4\sqrt{3}}\) b)\(\sqrt{9-4\sqrt{5}}\)
c)\(\sqrt{14+6\sqrt{5}}\) d)\(\sqrt{17-12\sqrt{2}}\)
a.\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\)
b.\(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2\)
c.\(\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3\)
d.\(\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)
Rút gọn các biểu thức sau:
a \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
b \(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
c \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
d \(\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)
\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)
\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)
\(\Leftrightarrow B^3+9B-10=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)
\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))
c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)
\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)
\(\Rightarrow C=1\)
d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)
\(=\sqrt[3]{3}+\sqrt[3]{2}\)
Vậy...
rút gọn các biểu thức sau:
\(\dfrac{1}{2}\sqrt{20}+5\)
\(\sqrt{16}+\sqrt{64}\)
\(\sqrt{20}-\sqrt{45}+3\sqrt{18}\)
\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{2}\)
A= \(\dfrac{2}{\sqrt{7}-5}-\dfrac{2}{\sqrt{7}+5}\)
B=\(\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
mình cần gấp á. tại vì mình khá là ngu toán nên giúp mik vs
a) \(\dfrac{1}{2}\sqrt{20}+5=\dfrac{1}{2}\cdot2\sqrt{5}+5=5+\sqrt{5}\)
b) \(\sqrt{16}+\sqrt{64}=4+8=12\)
c) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}=9\sqrt{2}-\sqrt{5}\)
d) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{2}=2-\sqrt{2}+\sqrt{2}=2\)
Rút gọn căn thức sau:
\(B= \dfrac{\sqrt{7-\sqrt{5}}+\sqrt{7+\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
Ta có:\(\left(\sqrt{7-\sqrt{5}}+\sqrt{7+\sqrt{5}}\right)^2=7-\sqrt{5}+7+\sqrt{5}+2\sqrt{\left(7-\sqrt{5}\right)\left(7+\sqrt{5}\right)}=14+2\sqrt{44}=14+4\sqrt{11}\)
=>\(\sqrt{7-\sqrt{5}}+\sqrt{7+\sqrt{5}}=\sqrt{14+4\sqrt{11}}=\sqrt{2}.\sqrt{7+2\sqrt{11}}\)
=>B=\(\dfrac{\sqrt{2}.\sqrt{7+2\sqrt{11}}}{\sqrt{7+2\sqrt{11}}}\cdot\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
=\(\sqrt{2}\cdot\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)(mình làm tắt tách 4=2+2=\(\sqrt{4}+\sqrt{4}\))
=\(\sqrt{2}\)\(\cdot\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}\cdot\left(1+\sqrt{2}\right)=2+\sqrt{2}\)
\(B=\dfrac{\sqrt{7-\sqrt{5}}+\sqrt{7+\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(B=\dfrac{\sqrt{14-2\sqrt{5}}+\sqrt{14+2\sqrt{5}}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{6}+\sqrt{8}+2}{\sqrt{2}+\sqrt{3}+2}\)
\(B=\dfrac{\sqrt{\left(\left(\sqrt{7+2\sqrt{11}}\right)-\left(\sqrt{7-2\sqrt{11}}\right)\right)^2}+\sqrt{\left(\left(\sqrt{7+2\sqrt{11}}\right)+\left(7-2\sqrt{11}\right)\right)^2}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{2}\left(\sqrt{3}+2+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(B=\dfrac{\sqrt{7+2\sqrt{11}}-\sqrt{7-2\sqrt{11}}+\sqrt{7+2\sqrt{11}}+\sqrt{7-2\sqrt{11}}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(B=\dfrac{2.\sqrt{7+2\sqrt{11}}}{\sqrt{2}.\sqrt{7+2\sqrt{11}}}.\left(1+\sqrt{2}\right)\)
\(B=\sqrt{2}.\left(1+\sqrt{2}\right)=\sqrt{2}+2\)
Rút gọn căn thức
\(\sqrt{5+5\sqrt{3}}\cdot\sqrt{48-10\sqrt{7+4\sqrt{3}}}\)
\(=\sqrt{5.\left(\sqrt{3}+1\right)}.\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}\)
\(=\sqrt{5}.\left(\sqrt{3}+1\right).\sqrt{48-10.\left(2+\sqrt{3}\right)}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{28-10\sqrt{3}}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\sqrt{\left(5-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{15}+\sqrt{5}\right).\left(5-\sqrt{3}\right)\)
Vậy...
~ Chắc chắn đúng cậu nhé ~ Tiếc gì 1 tk cho tớ nào?
Rút gọn các biểu thức sau:
a. \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}\) - \(\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
b.\(\dfrac{1}{4-3\sqrt{2}}\) - \(\dfrac{1}{4+3\sqrt{2}}\)
c.\(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right)\): \(\sqrt{28}\)
d.\(\dfrac{3}{\sqrt{6}-\sqrt{3}}\)+\(\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)
\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)
\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)
b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)
\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)
b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)