Rut gon bieu thuc
(\(\sqrt[3]{49}\) +\(\sqrt[3]{35}\)+\(\sqrt[3]{25}\))×(\(\sqrt[3]{7}\) - \(\sqrt[3]{5}\) )
1/ Rut gon bieu thuc sau:
a) \(\sqrt{12-2\sqrt{35}}+\sqrt{7-2\sqrt{10}}-\sqrt{\sqrt{49}}\)
b) \(\frac{\sqrt{7}-5}{2}-\frac{6}{\sqrt{7}-2}+\frac{1}{3+\sqrt{7}}+\frac{3}{5+2\sqrt{7}}\)
\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
Rut gon bieu thuc
Điều kiện : x>=0
\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{1}-x}{\sqrt{1}+\sqrt{x}}=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\sqrt{x}+1-\sqrt{x}=1\)
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) \(\frac{\sqrt{7-4\sqrt{3}}}{1-\sqrt{3}}=\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{1-\sqrt{3}}=\frac{2-\sqrt{3}}{1-\sqrt{3}}\)
Rut Gon bieu thuc
\(\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}\)
\(\sqrt{3-2\sqrt{2}}=\sqrt{1-2\sqrt{2}+2}=\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|\)
\(\sqrt{5-2\sqrt{6}}=\sqrt{2-2\sqrt{6}+3}=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}=\left|\sqrt{2}-\sqrt{3}\right|\)
Mà\(1< \sqrt{2};\sqrt{2}< \sqrt{3}\)
\(\Rightarrow\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}\)
\(=\sqrt{3}-1\)
ta có: \(\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}.\)
\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}=\sqrt{3}-1\)
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)
\(B^2=24+2\sqrt{12^2-4.11}=24+2\sqrt{100}=24+20=44\)
\(B=\sqrt{44}=2\sqrt{11}\)
\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\)\(\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Rut gon bieu thuc
tum x de bieu thuc = \(\frac{-1}{7}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(A=\)\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}.\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\)\(\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(\Rightarrow\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=-\frac{1}{7}\Rightarrow-7\left(-5\sqrt{x}+2\right)=\sqrt{x}+3\)
\(\Rightarrow35\sqrt{x}-14=\sqrt{x}+3\)
\(\Rightarrow34\sqrt{x}=17\)
\(\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\left(tm\right)\)
Vậy với \(x=\frac{1}{4}\)thì \(A=-\frac{1}{7}\)
rut gon bieu thuc
\(\left(3\sqrt{2}+\sqrt{6}\right)\times\sqrt{6-3\sqrt{3}}\)
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: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) rut gon bieu thuc gium em a thanks
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)