Rut Gon bieu thuc
\(\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}\)
Những câu hỏi liên quan
Cho bieu thuc:
P=\(\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}\)
a. Rut gon bieu thuc P
b.Tim GTLN cua P sau khi rut gon
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
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Rut gon bieu thuc
1)\(\frac{\sqrt{6-2\sqrt{5}}}{2-2\sqrt{5}}\)
2)\(\frac{\sqrt{7-4\sqrt{3}}}{1-\sqrt{3}}\)
1) \(\frac{\sqrt{6-2\sqrt{5}}}{2-2\sqrt{5}}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{2\left(1-\sqrt{5}\right)}=\frac{\sqrt{5}-1}{2\left(1-\sqrt{5}\right)}=-\frac{1}{2}\)
2) \(\frac{\sqrt{7-4\sqrt{3}}}{1-\sqrt{3}}=\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{1-\sqrt{3}}=\frac{2-\sqrt{3}}{1-\sqrt{3}}\)
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\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
Rut gon bieu thuc
Điều kiện : x>=0
\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{1}-x}{\sqrt{1}+\sqrt{x}}=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\sqrt{x}+1-\sqrt{x}=1\)
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\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) rut gon bieu thuc gium em a thanks
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
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rut gon bieu thuc
\(\left(3\sqrt{2}+\sqrt{6}\right)\times\sqrt{6-3\sqrt{3}}\)
Giup minh nhe!!
1. Rut gon bieu thuc:
a) Asqrt{11-6sqrt{2}}+3+sqrt{2}
b) Bsqrt{12+2sqrt{11}}+sqrt{12-2sqrt{11}}
2. Rut gon bieu thuc:
a) Asqrt{x-2+2sqrt{x-3}}+sqrt{x+6+6sqrt{x-3}},voix3
3. Tim GTNN cua bieu thuc:
a) Asqrt{4x^2-12x+9}+sqrt{x^2-10x+25}+sqrt{9x^2-6x+1}+sqrt{16x^2-72x+81}
b) Bdfrac{1}{2}sqrt{x^2}+sqrt{x^2-2x+1}
Đọc tiếp
Giup minh nhe!!
1. Rut gon bieu thuc:
a) \(A=\sqrt{11-6\sqrt{2}}+3+\sqrt{2}\)
b) \(B=\sqrt{12+2\sqrt{11}}+\sqrt{12-2\sqrt{11}}\)
2. Rut gon bieu thuc:
a) \(A=\sqrt{x-2+2\sqrt{x-3}}+\sqrt{x+6+6\sqrt{x-3}},voix>=3\)
3. Tim GTNN cua bieu thuc:
a) \(A=\sqrt{4x^2-12x+9}+\sqrt{x^2-10x+25}+\sqrt{9x^2-6x+1}+\sqrt{16x^2-72x+81}\)
b) \(B=\dfrac{1}{2}\sqrt{x^2}+\sqrt{x^2-2x+1}\)
1)
a)
\(\sqrt{11-6\sqrt{2}}=\sqrt{2-2.3.\sqrt{2}+9}=\left|\sqrt{2}-3\right|=3-\sqrt{2}\)
\(A=3-\sqrt{2}+3+\sqrt{2}=6\)
b)
\(B^2=24+2\sqrt{12^2-4.11}=24+2\sqrt{100}=24+20=44\)
\(B=\sqrt{44}=2\sqrt{11}\)
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rut gon bieu thuc \(y=3\sqrt{5}-2\sqrt{5}+4\sqrt{5}\)
1/ Rut gon bieu thuc sau:
a) \(\sqrt{12-2\sqrt{35}}+\sqrt{7-2\sqrt{10}}-\sqrt{\sqrt{49}}\)
b) \(\frac{\sqrt{7}-5}{2}-\frac{6}{\sqrt{7}-2}+\frac{1}{3+\sqrt{7}}+\frac{3}{5+2\sqrt{7}}\)
rut gon bieu thuc a) \(\sqrt{15+2\sqrt{5}-\sqrt[]{21-4\sqrt{5}}}\)
b)\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
a) \(\sqrt{15+2\sqrt{5}-\sqrt{21-4\sqrt{5}}}\)
\(=\sqrt{15+2\sqrt{5}-\sqrt{\left(1-2\sqrt{5}\right)^2}}\)
\(=\sqrt{15+2\sqrt{5}-\left(2\sqrt{5}-1\right)}\)
\(=\sqrt{15+2\sqrt{5}-\left(2\sqrt{5}-1\right)}\)
\(=\sqrt{15+2\sqrt{5}-2\sqrt{5}+1}\)
\(=\sqrt{16}\)
\(=4\)
b) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt[4]{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt[4]{5-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)
\(=\sqrt[4]{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt[4]{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt[4]{5-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt[4]{5-\sqrt{\left(1-\sqrt{5}\right)^2}}\)
\(=\sqrt[4]{5-\left(\sqrt{5}-1\right)}\)
\(=\sqrt[4]{5-\sqrt{5}+1}\)
\(=\sqrt[4]{6-\sqrt{5}}\)
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