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1) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{10}-10-5\sqrt{10}\)
\(=-10\)
2) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(=14-2\sqrt{21}-7+2\sqrt{21}\)
\(=7\)
3) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\) (hẳn đề là như thế này)
\(=33-3\sqrt{22}-11+3\sqrt{22}\)
\(=22\)
a: \(=2\cdot3+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
b: \(=5\sqrt{10}+2\cdot5-5\sqrt{10}=10\)
c: \(=2\sqrt{7}\cdot\sqrt{7}-\sqrt{12}\cdot\sqrt{7}-\sqrt{7}\cdot\sqrt{7}+2\sqrt{21}=2\cdot7-7=7\)
d: \(=\left(2\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}+3\sqrt{22}=2\cdot11=22\)
a,\(\sqrt{2\left(11+6\sqrt{2}\right)}\)=\(\sqrt{2\left(9+2.3.\sqrt{2}+2\right)}\)=\(\sqrt{2\left(3+\sqrt{2}\right)^2}\)=\(\sqrt{2}\)(3+\(\sqrt{2}\))
\(a.\sqrt{22+12\sqrt{2}}=\sqrt{18+2.3\sqrt{2}.2+4}=3\sqrt{2}+2\)
\(b.\sqrt{\dfrac{5+2\sqrt{6}}{2}}=\sqrt{\dfrac{3+2\sqrt{3}.\sqrt{2}+2}{2}}=\dfrac{\sqrt{3}+\sqrt{2}}{2}\)
\(c.\sqrt{30+4\sqrt{2}.\sqrt{7}}=\sqrt{28+2.\sqrt{2}.2\sqrt{7}+2}=2\sqrt{7}+\sqrt{2}\)
\(d.\sqrt{5+2\sqrt{2-\sqrt{9-4\sqrt{2}}}}=\sqrt{5+2\sqrt{2-\sqrt{8-2.2\sqrt{2}+1}}}=\sqrt{5+2\sqrt{2-2\sqrt{2}+1}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{2}+1\) \(e.\sqrt{1+2\sqrt{\sqrt{2+\sqrt{11+6\sqrt{2}}}}}=\sqrt{1+2\sqrt{\sqrt{2+\sqrt{9+2.3\sqrt{2}+2}}}}=\sqrt{1+2\sqrt{\sqrt{5+\sqrt{2}}}}\)
\(f.\sqrt{1+\dfrac{\sqrt{3}}{2}+\sqrt{1-\dfrac{\sqrt{3}}{2}}}=\sqrt{1+\dfrac{\sqrt{3}}{2}+\sqrt{\dfrac{3}{4}-2.\dfrac{\sqrt{3}}{2}.\dfrac{1}{2}+\dfrac{1}{4}}}=\sqrt{\sqrt{3}+\dfrac{1}{2}}=\)
\(g.\sqrt{10-2\sqrt{21}}+\sqrt{4+2\sqrt{3}}=\sqrt{7-2\sqrt{7}.\sqrt{3}+3}+\sqrt{3+2\sqrt{3}+1}=\sqrt{7}-\sqrt{3}+\sqrt{3}+1=\sqrt{7}+1\)
\(2.3+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
\(5\sqrt{10}+2.5-5\sqrt{10}=10\)
\(14-2\sqrt{21}-7+2\sqrt{21}=7\)
\(33-3\sqrt{22}-11+3\sqrt{22}=22\)
1: Chứng minh
a) Ta có: \(VT=11+6\sqrt{2}\)
\(=9+2\cdot3\cdot\sqrt{2}+2\)
\(=\left(3+\sqrt{2}\right)^2=VP\)(đpcm)
b) Ta có: \(VP=\left(\sqrt{7}-1\right)^2\)
\(=7-2\cdot\sqrt{7}\cdot1+1\)
\(=8-2\sqrt{7}=VT\)(đpcm)
c) Ta có: \(VT=\left(5-\sqrt{3}\right)^2\)
\(=25-2\cdot5\cdot\sqrt{3}+3\)
\(=28-10\sqrt{3}=VP\)(đpcm)
d) Ta có: \(VP=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}-\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left|\sqrt{3}+1\right|-\left|\sqrt{3}-1\right|\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-\sqrt{3}+1\)
\(=2=VT\)(đpcm)
thêm dòng này nữa :33
⇔ 11 + \(6\sqrt{2}=11+6\sqrt{2}\left(đpcm\right)\)
a) Ta có: \(\sqrt{11-2\sqrt{10}}\)
\(=\sqrt{10-2\cdot\sqrt{10}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}\)
\(=\left|\sqrt{10}-1\right|=\sqrt{10}-1\)
b) Ta có: \(\sqrt{9-2\sqrt{14}}\)
\(=\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{2}\right|\)
\(=\sqrt{7}-\sqrt{2}\)
c) Ta có: \(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\)
\(=\sqrt{3}+1+\sqrt{3}-1\)
\(=2\sqrt{3}\)
d) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5+2\cdot\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\left(\sqrt{5}+2\right)\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
e) Ta có: \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}\right)-\sqrt{2}\cdot\left(\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}-\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\left(\sqrt{7}+1\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)
\(=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
g) Ta có: \(\sqrt{3}+\sqrt{11+6\sqrt{2}}+\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{3}+\sqrt{9+2\cdot3\cdot\sqrt{2}+2}+\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|3+\sqrt{2}\right|+\left|\sqrt{2}+\sqrt{3}\right|\)
\(=\sqrt{3}+3+\sqrt{2}+\sqrt{2}+\sqrt{3}\)
\(=3+2\sqrt{3}+2\sqrt{2}\)
h) Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{\left(\sqrt{3}+2\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\cdot\left(\sqrt{3}+2\right)}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\sqrt{3}-20}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\left(5-\sqrt{3}\right)}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
k) Ta có: \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
\(=\sqrt{49-2\cdot7\cdot\sqrt{45}+45}-\sqrt{49+2\cdot7\cdot\sqrt{45}+45}\)
\(=\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}\)
\(=\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|\)
\(=7-\sqrt{45}-\left(7+\sqrt{45}\right)\)
\(=7-\sqrt{45}-7-\sqrt{45}\)
\(=-2\sqrt{45}=-6\sqrt{5}\)
i) Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow A^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\cdot\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\cdot\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\cdot\sqrt{6-2\sqrt{5}}\)
\(=8+2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\cdot\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2\)
\(=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Leftrightarrow A=\sqrt{5}+1\)
mình làm mẫu 2 bài nhé 2 bài kia bạn làm tương tự
1)a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{3}=\sqrt{3}+1-\sqrt{3}=1\)
\(\sqrt{10-2\sqrt{21}}+\sqrt{7}=\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}+\sqrt{7}=\sqrt{7}+\sqrt{3}+\sqrt{7}=2\sqrt{7}+\sqrt{3}\)
2)a) \(\sqrt{12-6\sqrt{3}}-\sqrt{3}=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{3}=3-\sqrt{3}-\sqrt{3}=3-2\sqrt{3}\)
b) \(\sqrt{7+2\sqrt{6}}-\sqrt{3}=\sqrt{\left(1+\sqrt{6}\right)^2}-\sqrt{3}=1+\sqrt{6}-\sqrt{3}\)