Rút gọn biểu thức.
a) \(\sqrt{\left(1-\sqrt{2}\right)^2}-\sqrt{3+2\sqrt{2}}+\sqrt{\left(-2\right)^6}\)
b) \(\sqrt{7+4\sqrt{3}}+\sqrt{13-4\sqrt{3}}\)
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Rút gọn biểu thức:
\(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)
\(B=3\sqrt{\left(1,5\right)^2}-4\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(A=\left|2-\sqrt{7}\right|+7-2\sqrt{7}+1\)
\(=\sqrt{7}-2+8-2\sqrt{7}\) \(=6-\sqrt{7}\)
\(B=3\cdot1,5-4\cdot\left|3-\sqrt{2}\right|\) \(=4,5-4\left(3-\sqrt{2}\right)\)
\(=4,5-12+4\sqrt{2}\) \(=4\sqrt{2}-7,5\)
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Ta có: \(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)
\(=\sqrt{7}-2+8-2\sqrt{7}\)
\(=6-\sqrt{7}\)
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rút gọn biểu thức a) left(sqrt{7}-sqrt{2}right).left(sqrt{9+2sqrt{14}}right)b) sqrt{sqrt{13}-sqrt{3-sqrt{13}}-4sqrt{3}}c) sqrt{80-sqrt{321-16sqrt{5}}-sqrt{226-80sqrt{5}-sqrt{89-25sqrt{5}}}}d) dfrac{1}{sqrt{8}+sqrt{7}}+sqrt{175}-dfrac{6sqrt{2}-4}{3-sqrt{2}}e) dfrac{sqrt{6-sqrt{11}}}{sqrt{22}-sqrt{2}}+dfrac{6}{sqrt{2}}-dfrac{3}{sqrt{2}+1}f) dfrac{sqrt{2}}{2sqrt{2}+sqrt{3}+sqrt{5}}+dfrac{sqrt{2}}{2sqrt{2}-sqrt{3}-sqrt{5}}g) dfrac{sqrt{2}+sqrt{3}+sqrt{6}+sqrt{8}+4}{sqrt{2}+sqrt{3}+sqrt{4}}
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rút gọn biểu thức
a) \(\left(\sqrt{7}-\sqrt{2}\right).\left(\sqrt{9+2\sqrt{14}}\right)\)
b) \(\sqrt{\sqrt{13}-\sqrt{3-\sqrt{13}}-4\sqrt{3}}\)
c) \(\sqrt{80-\sqrt{321-16\sqrt{5}}-\sqrt{226-80\sqrt{5}-\sqrt{89-25\sqrt{5}}}}\)
d) \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
e) \(\dfrac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}+\dfrac{6}{\sqrt{2}}-\dfrac{3}{\sqrt{2}+1}\)
f) \(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3}+\sqrt{5}}+\dfrac{\sqrt{2}}{2\sqrt{2}-\sqrt{3}-\sqrt{5}}\)
g) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)
\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)
=7-2
=5
d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)
\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)
\(=4\sqrt{7}\)
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rút gọn biểu thứcGdfrac{sqrt[3]{a}.a^{dfrac{2}{3}}}{left(a^{4-2sqrt{3}}right)^{4+2sqrt{3}}}Gdfrac{a^{sqrt{7}+1}.a^{2-sqrt{7}}}{left(a^{sqrt{2}-2}right)^{sqrt{2}+2}}Hdfrac{a^2.left(a^{-2}.b^3right).b^{-1}}{left(a^{-1}.bright)^3.a^{-5}.b^{-2}}Hdfrac{b^3.a^{-4}.left(ab^2right)^3}{left(a^2right)^{-2}.left(ab^3right)^2.b^2} Hdfrac{b^3.a^{-4}.left(ab^2right)^3}{left(a^2right)^{-2}.left(ab^3right)^2.b^2} Hdfrac{b^3.a^{-4}.left(ab^2right)^3}{left(a^2right)^{-2}.left(ab^3right)^2.b^2}
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rút gọn biểu thức
\(G=\dfrac{\sqrt[3]{a}.a^{\dfrac{2}{3}}}{\left(a^{4-2\sqrt{3}}\right)^{4+2\sqrt{3}}}\)
\(G=\dfrac{a^{\sqrt{7}+1}.a^{2-\sqrt{7}}}{\left(a^{\sqrt{2}-2}\right)^{\sqrt{2}+2}}\)
\(H=\dfrac{a^2.\left(a^{-2}.b^3\right).b^{-1}}{\left(a^{-1}.b\right)^3.a^{-5}.b^{-2}}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
Rút gọn các biểu thức saua) 2sqrt{32} + 3sqrt{72}-7sqrt{50}+sqrt{2} b)sqrt{left(3-sqrt{3}right)^2}+sqrt{left(2-sqrt{3}right)^2} c) sqrt{11+6sqrt{2}}-3+sqrt{2}d) x-4+sqrt{16-8x+x^2}left(x4right) e) dfrac{1}{a-b}sqrt{a^4left(a-bright)^2}vớia b
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Rút gọn các biểu thức sau
a) 2\(\sqrt{32}\) + 3\(\sqrt{72}-7\sqrt{50}+\sqrt{2}\) b)\(\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\) c) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)
d) \(x-4+\sqrt{16-8x+x^2}\left(x>4\right)\) e) \(\dfrac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}vớia< b\)
a) \(2\sqrt{32}+3\sqrt{72}-7\sqrt{50}+\sqrt{2}\)
\(=2\cdot4\sqrt{2}+3\cdot6\sqrt{2}-7\cdot5\sqrt{2}+\sqrt{2}\)
\(=8\sqrt{2}+18\sqrt{2}-35\sqrt{2}+\sqrt{2}\)
\(=-8\sqrt{2}\)
b) \(\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|3-\sqrt{3}\right|+\left|2-\sqrt{3}\right|\)
\(=3-\sqrt{3}+\sqrt{3}-2\)
\(=1\)
c) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)
\(=\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-3+\sqrt{2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
d) \(x-4+\sqrt{16-8x+x^2}\left(x>4\right)\)
\(=x-4+\sqrt{x^2-8x+16}\)
\(=x-4+\sqrt{\left(x-4\right)^2}\)
\(=x-4+\left|x-4\right|\)
\(=x-4+x-4\)
\(=2x-8\)
e) \(\dfrac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}\left(a< b\right)\)
\(=\dfrac{1}{a-b}\sqrt{\left[a^2\left(a-b\right)\right]^2}\)
\(=\dfrac{1}{a-b}\left|a^2\left(a-b\right)\right|\)
\(=\dfrac{-a^2\left(a-b\right)}{a-b}\)
\(=-a^2\)
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Rút gọn các biểu thức:
\(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{x-4}{3\sqrt{x}}\)
\(B=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}+\dfrac{6-7\sqrt{a}}{a-4}\right).\left(\sqrt{a}+2\right)\)
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
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rút gọn biểu thức
\(D=\left(\sqrt{a}\right)^7.\left(\sqrt[3]{a}\right).\left(\sqrt[4]{a}\right)^7\) (a>0)
\(D=a^{\sqrt{2}-1}.\left(a^2\right)^{\sqrt{2}}.\left(a^3\right)^{1-\sqrt{2}}\)
\(D=a^{\dfrac{7}{2}}.a^{\dfrac{1}{3}}.a^{\dfrac{7}{4}}=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}=\sqrt[12]{a^{67}}\)
\(D=a^{\sqrt{2}-1}.a^{2\sqrt{2}}.a^{3-3\sqrt{2}}=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{3}}=a^2\)
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\(D=\left(\sqrt{a}\right)^7\cdot\left(\sqrt[3]{a}\right)\left(\sqrt[4]{a}\right)^7\)
\(=a^{\dfrac{1}{2}\cdot7}\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{4}\cdot7}\)
\(=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}\)
b: \(D=a^{\sqrt{2}-1}\cdot\left(a^2\right)^{\sqrt{2}}\cdot\left(a^3\right)^{1-\sqrt{2}}\)
\(=a^{\sqrt{2}-1}\cdot a^{2\sqrt{2}}\cdot a^{3-3\sqrt{2}}\)
\(=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{2}}=a^2\)
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Rút gọn các biểu thức sau:j) left(dfrac{1}{sqrt{7-2sqrt{10}}}-dfrac{sqrt{2}}{sqrt{10}+2}+1right):left(sqrt{2}+1right)^2k) sqrt{5}left(sqrt{6}+1right):dfrac{sqrt{2sqrt{3}+sqrt{2}}}{sqrt{2sqrt{3}}-sqrt{2}}o) left(4+sqrt{15}right)left(sqrt{10}-sqrt{6}right)sqrt{4-sqrt{15}}p) left(sqrt{5}+3right)left(sqrt{10}-sqrt{2}right)sqrt{3-sqrt{5}}
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Rút gọn các biểu thức sau:
j) \(\left(\dfrac{1}{\sqrt{7-2\sqrt{10}}}-\dfrac{\sqrt{2}}{\sqrt{10}+2}+1\right):\left(\sqrt{2}+1\right)^2\)
k) \(\sqrt{5}\left(\sqrt{6}+1\right):\dfrac{\sqrt{2\sqrt{3}+\sqrt{2}}}{\sqrt{2\sqrt{3}}-\sqrt{2}}\)
o) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
p) \(\left(\sqrt{5}+3\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
j.
\(J=\left[\frac{1}{\sqrt{(\sqrt{5}-\sqrt{2})^2}}-\frac{\sqrt{2}}{\sqrt{2}(\sqrt{5}+\sqrt{2})}+1\right].\frac{1}{(\sqrt{2}+1)^2}\)
\(=\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right).\frac{1}{(\sqrt{2}+1)^2}\)
\(=[\frac{\sqrt{5}+\sqrt{2}-(\sqrt{5}-\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}+1].\frac{1}{(\sqrt{2}+1)^2}=(\frac{2\sqrt{2}}{3}+1).\frac{1}{(\sqrt{2}+1)^2}=\frac{3+2\sqrt{2}}{3}.\frac{1}{3+2\sqrt{2}}=\frac{1}{3}\)
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k. Đề sai sai, bạn xem lại
o.
\(O=(4+\sqrt{15})(\sqrt{5}-\sqrt{3}).\sqrt{2}.\sqrt{4-\sqrt{15}}\)
\(=(4+\sqrt{15}(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})=(4+\sqrt{15})(8-2\sqrt{15})\)
\(=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)
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p: Ta có: \(\left(3+\sqrt{5}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\cdot\sqrt{3-\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\cdot\left(6-2\sqrt{5}\right)\)
\(=18-6\sqrt{5}+6\sqrt{5}-20\)
=-2
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Rút gọn các biểu thức sau:Aleft(dfrac{1}{sqrt{x}-3}+dfrac{1}{sqrt{x}+3}right)left(1-dfrac{3}{sqrt{x}}right)Bleft(dfrac{sqrt{x}}{sqrt{x}-2}+dfrac{1}{sqrt{x}+2}+dfrac{6-7sqrt{x}}{x-4}right)left(sqrt{x}+2right)Cleft(dfrac{sqrt{a}}{sqrt{a}-1}-dfrac{sqrt{a}}{a-sqrt{1}}right):dfrac{sqrt{a}+1}{a-1}Dleft(dfrac{x-2}{x+2sqrt{x}}+dfrac{1}{sqrt{x}+2}right).dfrac{sqrt{x}+1}{sqrt{x}-1}Eleft(1+dfrac{x+sqrt{x}}{sqrt{x}+1}right)left(1+dfrac{x-sqrt{x}}{1-sqrt{x}}right)giúp mình với ạ!mình đang cần gấp
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Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}+\dfrac{6-7\sqrt{x}}{x-4}\right)\left(\sqrt{x}+2\right)\)
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(D=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(E=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
giúp mình với ạ!mình đang cần gấp
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
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2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
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3. ĐKXĐ: $a\geq 0; a\neq 1$
\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)
\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)
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Xem thêm câu trả lời
Bài 1: Rút gọn biểu thức:Afrac{a^3-3a+left(a^2-1right)sqrt{a^2-4}-2}{a^3-3a+left(a^2-1right)sqrt{a^2-4}+2}left(a2right)Bsqrt{frac{1}{a^2+b^2}+frac{1}{left(a+bright)^2}+sqrt{frac{1}{a^4}+frac{1}{b^4}+frac{1}{left(a^2+b^2right)^2}}}left(abne0right)Bài 2: Tính giá trị của biểu thức:Efrac{1}{1sqrt{2}+2sqrt{1}}+frac{1}{2sqrt{3}+3sqrt{2}}+frac{1}{3sqrt{4}+4sqrt{3}}+...+frac{1}{2017sqrt{2018}+2018sqrt{2017}}Bài 3: Chứng minh rằng các biểu thức sau có gúa trị là số nguyênAleft(sqrt{57}+3sqrt{6}+sqrt{38}...
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Bài 1: Rút gọn biểu thức:
\(A=\frac{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}-2}{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}+2}\left(a>2\right)\)
\(B=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}}\left(ab\ne0\right)\)
Bài 2: Tính giá trị của biểu thức:
\(E=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}+2018\sqrt{2017}}\)
Bài 3: Chứng minh rằng các biểu thức sau có gúa trị là số nguyên
\(A=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
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}{4-3sqrt{2}} - dfrac{1}{4+3sqrt{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}}
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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: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)
\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)
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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) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)
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