giúp mình với ạ
a,\(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
b,\(\sqrt{31-12\sqrt{3}}-\sqrt{31+12\sqrt{3}}\)
mình cảm ơn nhiều .
Những câu hỏi liên quan
giúp em với ạ
\(\sqrt{5
+2\sqrt{ }6}\)
\(\sqrt{12+2\sqrt{ }35}-\sqrt{12-2\sqrt{ }35}\)
\(\sqrt{16+6\sqrt{ }7}\)
\(\sqrt{31-12\sqrt{ }3}\)
\(\sqrt{27+10\sqrt{ }2}\)
\(\sqrt{14+6\sqrt{ }5}\)
a: \(\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
b: \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)
c: \(\sqrt{16+6\sqrt{7}}=4+\sqrt{7}\)
d: \(\sqrt{31-12\sqrt{3}}=3\sqrt{3}-2\)
e: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)
f: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)
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Cho mình hỏi:
a.\(\sqrt{15-6\sqrt{6}}+\sqrt{42-12\sqrt{6}}\)
b.1\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{24}+\sqrt{25}}\)
Trả lời giúp mình với ạ! mình cảm ơn!
tính giá trị của biểu thức \(A=\left(x-y\right)^3+3\left(x-y\right)\left(xy+1\right)\)
biết \(x=\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}};\) \(y=\sqrt[3]{17+12\sqrt{2}}-\sqrt[3]{17-12\sqrt{2}}\)
giúp mình với, thanks nhiều
Giúp mình 2 câu này với ạ. Mình xin cảm ơn.
a) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
b) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(a)\)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\)\(\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}\)
\(=\)\(\sqrt{\left(\sqrt{6}+3\right)}+\sqrt{\left(\sqrt{24}+3\right)}\)
\(=\)\(\left|\sqrt{6}+3\right|+\left|\sqrt{24}+3\right|\)
\(=\)\(\sqrt{6}+3+\sqrt{24}+3\)
\(=\)\(\sqrt{6}\left(1+\sqrt{4}\right)+9\)
\(=\)\(3\sqrt{6}+9\)
Chúc bạn học tốt ~
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\(b)\)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=\)\(\left|2-\sqrt{3}\right|+\sqrt{3-2\sqrt{3}+1}\)
\(=\)\(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\) ( vì \(2=\sqrt{4}>\sqrt{3}\) )
\(=\)\(2-\sqrt{3}+\left|\sqrt{3}-1\right|\)
\(=\)\(2-\sqrt{3}+\sqrt{3}-1\) ( vì \(\sqrt{3}>\sqrt{1}=1\) )
\(=\)\(1\)
Chúc bạn học tốt ~
PS : mới lớp 8 sai thì thông cảm >.<
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\(\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}\) + \(\sqrt{\left(\sqrt{24}-\sqrt{9}\right)^2}\)
\(\sqrt{9}-\sqrt{6}\) + \(\sqrt{24}-\sqrt{9}\) vì \(\sqrt{9}>\sqrt{6}\),\(\sqrt{24}>\sqrt{9}\)
\(3-\sqrt{6}\) + \(2\sqrt{6}-3\)
= \(\sqrt{6}\)
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Xem thêm câu trả lời
18 tháng 9 2021 lúc 22:07
Giúp vs, làm câu nào cx đc, làm hết thì tốta) sqrt{17+12sqrt{2}}+sqrt{17-12sqrt{2}}b) sqrt{27-10sqrt{2}}+sqrt{18-8sqrt{2}}c) sqrt{3-sqrt{5}}.sqrt{8}d) dfrac{sqrt{2-sqrt{3}}}{sqrt{2}}.sqrt{8}e) dfrac{sqrt{15}-sqrt{5}}{sqrt{3}-1}+dfrac{5-2sqrt{5}}{2sqrt{5}-4}g) dfrac{3+2sqrt{3}}{sqrt{3}}+dfrac{2-sqrt{2}}{sqrt{2}-1}-left(sqrt{2}+3right)h) dfrac{sqrt[3]{135}}{sqrt[3]{5}}-sqrt[3]{54}.sqrt[3]{4}i) left(sqrt[3]{25}-sqrt[3]{10}+sqrt[3]{4}right).left(sqrt[3]{5}+sqrt[3]{2}right)k) sqrt[3]{left(4-2sqrt{3}r...
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Giúp vs, làm câu nào cx đc, làm hết thì tốt
a) \(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
b) \(\sqrt{27-10\sqrt{2}}+\sqrt{18-8\sqrt{2}}\)
c) \(\sqrt{3-\sqrt{5}}.\sqrt{8}\)
d) \(\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}.\sqrt{8}\)
e) \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
g) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2-\sqrt{2}}{\sqrt{2}-1}-\left(\sqrt{2}+3\right)\)
h) \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
i) \(\left(\sqrt[3]{25}-\sqrt[3]{10}+\sqrt[3]{4}\right).\left(\sqrt[3]{5}+\sqrt[3]{2}\right)\)
k) \(\sqrt[3]{\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)
L) \(A=\sqrt[3]{10+14\sqrt{2}}+\sqrt[3]{10-14\sqrt{2}}\)
k: \(\sqrt[3]{\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)
\(=\sqrt[3]{\left(\sqrt{3}-1\right)^3}\)
\(=\sqrt{3}-1\)
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Tính các biểu thức sau:
a)\(\sqrt{3+2\sqrt{2}}-\sqrt{17-12\sqrt{2}}\)
b)\(\sqrt{5-2\sqrt{6}}-\sqrt{14-4\sqrt{6}}-\sqrt{48}\)
c)\(\sqrt{11+3\sqrt{8}}-\sqrt{17-12\sqrt{2}}-4\sqrt{8}\)
Mọi người ơi,em đang bí những bài này,nếu ai biết thì giải giúp em nhé!
Em xin cảm ơn ạ!!!
a) \(\sqrt{3+2\sqrt{2}}-\sqrt{17-12\sqrt{2}}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(3-2\sqrt{2}\right)^2}\)
= \(\left|\sqrt{2}+1\right|-\left|3-2\sqrt{2}\right|\)
= \(\sqrt{2}+1-3+2\sqrt{2}\)
= \(3\sqrt{2}-2\)
b) \(\sqrt{5-2\sqrt{6}}-\sqrt{14-4\sqrt{6}}-\sqrt{48}\)
= \(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{3}-\sqrt{2}\right)^2}-4\sqrt{3}\)
= \(\left|\sqrt{3}-\sqrt{2}\right|-\left|2\sqrt{3}-\sqrt{2}\right|-4\sqrt{3}\)
= \(\sqrt{3}-\sqrt{2}-2\sqrt{3}+\sqrt{2}-4\sqrt{3}\)
= \(-5\sqrt{3}\)
c) \(\sqrt{11+3\sqrt{8}}-\sqrt{17-12\sqrt{2}}-4\sqrt{8}\)
= \(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-2\sqrt{2}\right)^2}-8\sqrt{2}\)
= \(\left|3+\sqrt{2}\right|-\left|3-2\sqrt{2}\right|-8\sqrt{2}\)
= \(3+\sqrt{2}-3+2\sqrt{2}-8\sqrt{2}\)
= \(-5\sqrt{2}\)
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1,\(\sqrt{49-12\sqrt{ }5}+\sqrt{49+12\sqrt{ }5}\)
2,\(\sqrt{29+12\sqrt{ }5}\)+\(\sqrt{29-12\sqrt{ }5}\)
3,\(\sqrt{31-12\sqrt{ }3}-\sqrt{31+12\sqrt{ }3}\)
4,\(\sqrt{39-12\sqrt{ }3}-\sqrt{39+12\sqrt{ }3}\)
Bài 1 : \(\sqrt{49-12\sqrt{5}}+\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-4\sqrt{45}+4}+\sqrt{45+4\sqrt{45}+4}\)
\(=\sqrt{\left(\sqrt{45}-2\right)^2}+\sqrt{\left(\sqrt{45}+2\right)^2}\)
\(=\sqrt{45}-2+\sqrt{45}+2=2\sqrt{45}\)
Bài 2 : \(\sqrt{29+12\sqrt{5}}+\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{20+6\sqrt{20}+9}+\sqrt{20-6\sqrt{20}+9}\)
\(=\sqrt{\left(\sqrt{20}+3\right)^2}+\sqrt{\left(\sqrt{20}-3\right)^2}\)
\(=\sqrt{20}+3+\sqrt{20}-3=2\sqrt{20}\)
Bài 3 : \(\sqrt{31-12\sqrt{3}}+\sqrt{31+12\sqrt{3}}\)
\(=\sqrt{27-4\sqrt{27}+4}+\sqrt{27+4\sqrt{27}+4}\)
\(=\sqrt{\left(\sqrt{27}-2\right)^2}+\sqrt{\left(\sqrt{27}+2\right)^2}\)
\(=\sqrt{27}-2+\sqrt{27}+2=2\sqrt{27}\)
Chúc bạn học tốt
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4 , Ta có :
\(\sqrt{39-12\sqrt{3}}-\sqrt{39+12\sqrt{3}}\)
\(=\sqrt{3-2.6.\sqrt{3}+6^2}-\sqrt{3+2.6.\sqrt{3}+6^2}\)
\(=\sqrt{\left(\sqrt{3}-6\right)^2}-\sqrt{\left(\sqrt{3}+6\right)^2}\)
\(=\left|\sqrt{3}-6\right|-\left|\sqrt{3}+6\right|\)
\(=6-\sqrt{3}-\sqrt{3}-6\)
\(=-2\sqrt{3}\)
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1,\(\sqrt{4-2.2.3\sqrt{5}+45}+\sqrt{4+2.2.3\sqrt{5}+45}\)
\(=\sqrt{\left(2-3\sqrt{5}\right)^2}+\sqrt{\left(2+3\sqrt{5}\right)^2}\)
\(=\left|2-3\sqrt{5}\right|+\left|2+3\sqrt{5}\right|\)
\(=3\sqrt{5}-2+2+3\sqrt{5}\)
\(=6\sqrt{5}\)
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Tính thu gọn :
a , sqrt{17-12sqrt{2}}-sqrt{17+12sqrt{2}}
b , sqrt{27+12sqrt{5}}-sqrt{27-12sqrt{5}}
c , sqrt{15-6sqrt{6}}+sqrt{15+sqrt{6sqrt{6}}}
d , sqrt{4+sqrt{15}}+sqrt{4-sqrt{15}}-2sqrt{3-sqrt{5}}
e , sqrt{17-3sqrt{32}}+sqrt{17+3sqrt{32}}
f , sqrt{5+sqrt{3-sqrt{29-12sqrt{5}}}}
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Tính thu gọn :
a , \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
b , \(\sqrt{27+12\sqrt{5}}-\sqrt{27-12\sqrt{5}}\)
c , \(\sqrt{15-6\sqrt{6}}+\sqrt{15+\sqrt{6\sqrt{6}}}\)
d , \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
e , \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
f , \(\sqrt{5+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)
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mik sửa lại câu f , tí nhé :
f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
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a,\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}\) \(=3-2\sqrt{2}-3-2\sqrt{2}=-4\sqrt{2}\)
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Xem thêm câu trả lời
Rút gọn :
\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}}{\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}}{\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)^2}+\dfrac{\sqrt{2}+1}{\left(\sqrt{2}+1\right)^2}=\dfrac{1}{\sqrt{2}-1}+\dfrac{1}{\sqrt{2}+1}\)
\(=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2}+1-\sqrt{2}+1=2\)
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