1.Tính :
\(\sqrt{\left(4-\sqrt{11}\right)^2+\sqrt{11}}\)
\(\sqrt{\left(\sqrt{8}-7\right)^2-\sqrt{8}}\)
a) 11+6\(\sqrt{2}\) = \(\left(3+\sqrt{2}\right)^2\)
b) 8-2\(\sqrt{7}\)=\(\left(\sqrt{7}-1\right)^2\)
c)\(\sqrt{11+6\sqrt{2}}=\sqrt{11-6\sqrt{2}}=6\)
d) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=-2\)
1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right).\)
2)\(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
3)\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
4)\(\sqrt{2\sqrt{3}-4}+\sqrt{2\sqrt{3}+4}\)
5)\(\sqrt{4\sqrt{6}+11}-\sqrt{11-4\sqrt{6}}\)
6)\(\sqrt{10+2\sqrt{11}}-\sqrt{10-2\sqrt{11}}\)
7)\(\sqrt{5-2\sqrt{7-2\sqrt{6}}}\)
AI ĐÓ TỐT BỤNG GIÚP MK ZỚI:((
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
5) Đặt \(A=\sqrt{4\sqrt{6}+11}-\sqrt{11-4\sqrt{6}}\)
\(\Rightarrow A^2=\left(\sqrt{11+4\sqrt{6}}-\sqrt{11-4\sqrt{6}}\right)^2\)
\(=11+4\sqrt{6}-2\sqrt{\left(11+4\sqrt{6}\right)\left(11-4\sqrt{6}\right)}+11-4\sqrt{6}\)
\(=22-2\sqrt{121-96}\)
\(=22-2\sqrt{5}\)
=> \(A=\sqrt{22-2\sqrt{5}}\)
6) Đặt \(B=\sqrt{10+2\sqrt{11}}-\sqrt{10-2\sqrt{11}}\)
\(\Leftrightarrow B^2=\left(\sqrt{10+2\sqrt{11}}-\sqrt{10-2\sqrt{11}}\right)^2\)
\(=10+2\sqrt{11}-2\sqrt{\left(10+2\sqrt{11}\right)\left(10-2\sqrt{11}\right)}+10-2\sqrt{11}\)
\(=20-2\sqrt{100-44}\)
\(=20-4\sqrt{14}\)
=> \(B=\sqrt{20-4\sqrt{14}}\)
Giải phương trình:
e) \(\sqrt{x^2}=\left|-8\right|\)
Tính:
e) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}\)
f) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)
e) \(\sqrt{x^2}=\left|-8\right|\Rightarrow\left|x\right|=8\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
e) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}=\sqrt{\dfrac{8-2\sqrt{7}}{2}}-\sqrt{\dfrac{8+2\sqrt{7}}{2}}+\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}+\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}+\sqrt{2}\)
\(=\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}+\sqrt{2}=\dfrac{\sqrt{7}-1}{\sqrt{2}}-\dfrac{\sqrt{7}+1}{\sqrt{2}}+\sqrt{2}\)
\(=-\dfrac{2}{\sqrt{2}}+\sqrt{2}=-\sqrt{2}+\sqrt{2}=0\)
f) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)
\(=\sqrt{\dfrac{12+2\sqrt{11}}{2}}-\sqrt{\dfrac{12-2\sqrt{11}}{2}}+3\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{11}\right)^2+2.\sqrt{11}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{11}\right)^2-2.\sqrt{11}.1+1^2}{2}}+3\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{11}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{11}-1\right)^2}{2}}+3\sqrt{2}\)
\(=\dfrac{\left|\sqrt{11}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{11}-1\right|}{\sqrt{2}}+3\sqrt{2}=\dfrac{\sqrt{11}+1}{\sqrt{2}}-\dfrac{\sqrt{11}-1}{\sqrt{2}}+3\sqrt{2}\)
\(=\dfrac{2}{\sqrt{2}}+3\sqrt{2}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}\)
Rút gọn: (Giải chi tiết từng bước)
7) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
8) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
`a, (sqrt 28 - sqrt 12 - sqrt 7) sqrt 7 + 2 sqrt 21`.
`= sqrt(28.7) - sqrt(12.7) - sqrt(7.7) + 2 sqrt 21`.
`= sqrt(4. 7.7) - sqrt (12.7) - 7 + 2 sqrt 21`.
`= 14 - sqrt(4.3.7) - 7 + 2 sqrt 21`.
`= 7`.
`b, (sqrt99-sqrt18-sqrt11)sqrt11+3sqrt22`
`= sqrt(99.11)- sqrt(18.11)-sqrt(11.11) +3sqrt22`
`= sqrt(9.11.11)-sqrt(2.9.11)-11+3sqrt22`
`= 33 - 11 = 22`.
Rút gọn:
A = \(\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
B = \(\dfrac{3\sqrt{2}+\sqrt{11}}{\sqrt{2}+\sqrt{6+\sqrt{11}}}+\dfrac{3\sqrt{2}-\sqrt{11}}{\sqrt{2}-\sqrt{6-\sqrt{11}}}+18\)
C = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n+3}}\)với n thuộc N*
D = \(\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\left(\sqrt{15}-1\right)\left(7-2\sqrt{3}+\sqrt{5}\right)\)
E=\(\dfrac{\left(4+\sqrt{3}\right)}{\sqrt[]{1}+\sqrt{3}}+\dfrac{\left(8+\sqrt{15}\right)}{\sqrt{3}+\sqrt{5}}+...+\dfrac{2k+\sqrt{k^2-1}}{\sqrt{k-1}+\sqrt{k+1}}+...+\dfrac{240+\sqrt{14399}}{\sqrt{119}+\sqrt{121}}\)
F = \(\left(\dfrac{2a+1}{a\sqrt{a}-1}-\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\) với a >= 0 và a khác 1
Rút gọn các biểu thức : a) \(\sqrt{\left(3+\sqrt{5}\right)^2}\)
b) \(\sqrt{\left(5-\sqrt{5}\right)^2}\)
c) \(\sqrt{\left(4-\sqrt{11}\right)^2}+\sqrt{11}\)
d) \(\sqrt{\left(\sqrt{8}-7\right)^2}-\sqrt{8}\)
\(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2^6\right)}\)
rút gọn:giải chi tiết hộ mình nha
a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
1)chứng minh
a)\(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)
b)\(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}=6\)
2)chứng minh
a)\(8-2\sqrt{7}=\left(\sqrt{7}-1\right)^2\)
b)\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=2\)
a, phân tích vế trái ta được:
11+6\(\sqrt{2}\)=9+2.3.\(\sqrt{2}\)+2=(3+\(\sqrt{2}\))2\(\)=VP(dpcm)
b,phân tích vế trái ta được
\(\sqrt{11+6\sqrt{ }2}\)+\(\sqrt{11-6\sqrt{ }2}\)=|3+\(\sqrt{2}\)|+|3-\(\sqrt{2}\)|=6=VP(dpcm)
a,phân tích vế trái ta được
8-2\(\sqrt{7}\)=7-2\(\sqrt{7}\)+1=(\(\sqrt{7}\)-1)2
câu b sai đề nha
Ta có a) \(11+6\sqrt{2}=9+2\times3\times\sqrt{2}+2=\left(3+\sqrt{2}\right)^2\)
b) \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
1. \(\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}\) .
2. \(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}.}\)
3. \(\sqrt{11}-\sqrt{20-6\sqrt{11}}=3\)
4.\(\sqrt{41+12\sqrt{5}}-\sqrt{41-12\sqrt{5}}=2\sqrt{5.}\)
\(1.\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}=5+\sqrt{7}-\sqrt{7-2\sqrt{7}+1}=5+\sqrt{7}-\sqrt{7}+1=6\)
\(2.\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{3-2\sqrt{3}+1}=\sqrt{3}+1-\sqrt{3}+1=2\)
\(3.VT=\sqrt{11}-\sqrt{20-6\sqrt{11}}=\sqrt{11}-\sqrt{11-2.3\sqrt{11}+9}=\sqrt{11}-\sqrt{11}+3=3=VP\)
Vậy , đẳng thức được chứng minh .
\(4.VT=\sqrt{41+12\sqrt{5}}-\sqrt{41-12\sqrt{5}}=\sqrt{36+2.6\sqrt{5}+5}-\sqrt{36-2.6\sqrt{5}+5}=6+\sqrt{5}-6+\sqrt{5}=2\sqrt{5}=VP\)
Vậy , đẳng thức được chứng minh .