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1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)
\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)
\(=6-8=-2\)
2) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=3^2-\left(\sqrt{5}\right)^2\)
\(=9-5=4\)
3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)
=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn
\(a,\dfrac{-3}{5}.\sqrt{\left(-0.5\right)^2}\\ =\dfrac{-3}{5}.0,5\\ =\dfrac{-3}{5}.\dfrac{1}{2}\\ =-\dfrac{3}{10}\)
Câu (b) nhìn hơi lạ lạ á :v
\(c,\sqrt{\left(1-\sqrt{7}\right)^2}+\sqrt{7}\\ =\sqrt{7}-1+\sqrt{7}\\ =2\sqrt{7}-1\)
\(d,\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\\ =3+\sqrt{2}-\left(3-\sqrt{2}\right)\\ =3+\sqrt{2}-3+\sqrt{2}\\ =2\sqrt{2}\)
A = \(\sqrt{2}\left(\sqrt{8}-\sqrt{32}-2\sqrt{18}\right)=\sqrt{16}-\sqrt{64}-2\sqrt{36}=4-8-2\cdot6=-4-12=-16\)
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\(B=\sqrt{2}-\sqrt{3-\sqrt{5}}=\dfrac{2-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{2-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{2-\sqrt{5}+1}{\sqrt{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}\)
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\(C=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)
còn lại lúc nx mk lm nốt nhé, h bận
Lời giải:
\(B=(\sqrt{2}-\sqrt{3-\sqrt{5}})\sqrt{2}=2-\sqrt{6-2\sqrt{5}}\)
\(=2-\sqrt{5+1-2\sqrt{5}}=2-\sqrt{(\sqrt{5}-1)^2}=2-(\sqrt{5}-1)=3-\sqrt{5}\)
\(C=\sqrt{4-\sqrt{7}}-\sqrt{4}+\sqrt{7}=\sqrt{\frac{8-2\sqrt{7}}{2}}-2+\sqrt{7}\)
\(=\sqrt{\frac{7+1-2\sqrt{7}}{2}}-2+\sqrt{7}\)
\(=\sqrt{\frac{(\sqrt{7}-1)^2}{2}}-2+\sqrt{7}\)
\(=\frac{|\sqrt{7}-1|}{\sqrt{2}}-2+\sqrt{7}=\frac{\sqrt{7}-1}{\sqrt{2}}-2+\sqrt{7}\)
$D$: bạn xem lại đề, mình thấy biểu thức không rút gọn được nữa.
\(E=\sqrt{4+2\sqrt{2}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-(2+\sqrt{2})}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}.\sqrt{2^2-(\sqrt{2})^2}=\sqrt{2}.\sqrt{2}=2\)
\(F=(\sqrt{2}-\sqrt{3+\sqrt{5}})\sqrt{2}+2\sqrt{5}\)
\(=2-\sqrt{6+2\sqrt{5}}+2\sqrt{5}\)
\(=2-\sqrt{5+1-2\sqrt{5}}+2\sqrt{5}\)
\(=2-\sqrt{(\sqrt{5}-1)^2}+2\sqrt{5}\)
\(=2-(\sqrt{5}-1)+2\sqrt{5}=3+\sqrt{5}\)
\(G=(\sqrt{14}-\sqrt{10}).\sqrt{6+\sqrt{35}}=\sqrt{2}(\sqrt{7}-\sqrt{5})\sqrt{6+\sqrt{35}}\)
\(=(\sqrt{7}-\sqrt{5})\sqrt{12+2\sqrt{35}}=(\sqrt{7}-\sqrt{5}).\sqrt{7+5+2\sqrt{7.5}}\)
\(=(\sqrt{7}-\sqrt{5}).\sqrt{(\sqrt{7}+\sqrt{5})^2}=(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})=7-5=2\)
\(H=\sqrt{11-4\sqrt{7}}-\sqrt{2}.\sqrt{8+3\sqrt{7}}\)
\(=\sqrt{2^2+7-2.2.\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(=\sqrt{(2-\sqrt{7})^2}-\sqrt{3^2+7+2.3\sqrt{7}}=\sqrt{(2-\sqrt{7})^2}-\sqrt{(3+\sqrt{7})^2}\)
\(=|2-\sqrt{7}|-|3+\sqrt{7}|=\sqrt{7}-2-(3+\sqrt{7})=-5\)
a) \(\sqrt{3^2}-\sqrt{7^2}+\sqrt{\left(-1\right)^2}=|3|-|7|+|-1|=3-7+1=-3\)
b) \(-2\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}+\sqrt{3^2}=-2|2|+|-5|+\left|3\right|=-4+5+3=4\)
c) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2-\sqrt{2}\right|+\left|2+\sqrt{2}\right|=2-\sqrt{2}+2+\sqrt{2}=4\)
d) \(\sqrt{\left(3\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}=\left|3\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3\sqrt{2}-\sqrt{2}+1=2\sqrt{2}+1\)
e) \(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}-1\right|+\left|\sqrt{2}+1\right|=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}\)
f) \(\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{5}+2\right)^2}=\left|\sqrt{5}-2\right|+\left|\sqrt{5}+2\right|=\sqrt{5}-2+\sqrt{5}+2=2\sqrt{5}\)
g) \(\sqrt{9-4\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\sqrt{9-2\sqrt{8}}+\sqrt{2-2\sqrt{2}.3+9}=\sqrt{\left(\sqrt{8}-1\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}=\sqrt{8}-1+3-\sqrt{2}=2-\sqrt{2}+\sqrt{8}\)
h) \(\sqrt{12+8\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{12+2\sqrt{4}\sqrt{8}}+\sqrt{6-2\sqrt{2}\sqrt{4}}=\sqrt{\left(\sqrt{4}+\sqrt{8}\right)^2}+\sqrt{\left(\sqrt{4}-\sqrt{2}\right)^2}=\sqrt{4}+\sqrt{8}+\sqrt{4}-\sqrt{2}\)
k) \(\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{\left(\sqrt{3}+2\right)^2}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)
a)\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{10\left(4-\sqrt{15}\right)}+\sqrt{6\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
b) \(2\left(\sqrt{10}-\sqrt{2}\right)\left(4+\sqrt{6-2\sqrt{5}}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{\left(1-\sqrt{5}\right)^2}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{5}-1\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=6\sqrt{10}+2\sqrt{50}-6\sqrt{2}-2\sqrt{10}\)
\(=6\sqrt{10}+10\sqrt{2}-6\sqrt{2}-2\sqrt{10}\)
\(=4\sqrt{10}+4\sqrt{2}\)
c) \(\left(\sqrt{7}+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=7\sqrt{7}-7\sqrt{2}+\sqrt{98}-\sqrt{28}\)
\(=7\sqrt{7}-7\sqrt{2}+7\sqrt{2}-2\sqrt{7}\)
\(=5\sqrt{7}\)
d) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)
\(=\sqrt{\dfrac{289+42\sqrt{2}}{16}}\)
\(=\dfrac{\sqrt{289+42\sqrt{2}}}{\sqrt{4^2}}\)
\(=\dfrac{\sqrt{\left(1+12\sqrt{2}\right)^2}}{4}\)
\(=\dfrac{1+12\sqrt{2}}{4}\)
e) \(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{147}-\sqrt{63}+7-\sqrt{21}\)
\(=7\sqrt{3}-\sqrt{63}+7-\sqrt{21}\)
f) bạn xem đề lại nhé
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)
\(=33-3\sqrt{22}-11+3\sqrt{22}\)
\(=22\)
\(a=\left(1+\sqrt{3}\right)^2-\sqrt{2}^2=4+2\sqrt{3}-2=2+2\sqrt{3}\)
\(b=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=6+2\sqrt{9-5}=6+4=10\)
\(c=\sqrt{11}+\sqrt{7}+\sqrt{11}-\sqrt{7}-2\sqrt{\left(\sqrt{11}+\sqrt{7}\right)\left(\sqrt{11}-\sqrt{7}\right)}\)
\(=2\sqrt{11}-2\sqrt{11-7}=2\sqrt{11}-4\)
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 + \(6\sqrt{2}=11+6\sqrt{2}\left(đpcm\right)\)
\(B=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2+\sqrt{7}\right)^2}\)
\(B=\left|3-\sqrt{7}\right|+\left|2+\sqrt{7}\right|\)
\(B=3-\sqrt{7}+2+\sqrt{7}\)
\(B=5\)
b) B= (2 + sqrt(7)) * sqrt(11 - 4sqrt(7)) - sqrt 20+5 sqrt 5 ...2