Biến đổi các biểu thức sau thành biểu thức 1 tổng và 1 hiệu : 23 + 8 .\(\sqrt{7}\)
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\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+\sqrt{4}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.3+3^2}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{5-\left(\sqrt{5}-1\right)}\)\(=\sqrt{6-\sqrt{5}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{20-2.3.2\sqrt{5}+9}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{5-\sqrt{5-2.1.\sqrt{5}+1}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{5-\sqrt{5}+1}\)
Nếu là căn 5 thì = 1
chẳng bít có đúng không
\(=\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}.\frac{\sqrt{a}+1}{\sqrt{b}-1}}=\sqrt{\frac{a-1}{b-1}}=\sqrt{\frac{7,25-1}{3,25-1}}=\sqrt{\frac{625}{225}}=\frac{5}{3}\)
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}-1}}:\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}=\frac{\sqrt{\sqrt{a}-1}}{\sqrt{\sqrt{b+1}}}\cdot\frac{\sqrt{\sqrt{a}+1}}{\sqrt{\sqrt{b}-1}}\)
\(=\frac{\sqrt{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}}{\sqrt{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}}=\frac{\sqrt{a-1}}{\sqrt{b-1}}\)
Thay a = 7,25 ; b = 3,25 ta có
\(\frac{\sqrt{7,25-1}}{\sqrt{3,25-1}}=\frac{\sqrt{6,25}}{\sqrt{2,25}}=\frac{2,5}{1,5}\)
Đúng nha
\(=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
\(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}}\)
1)
a)\(\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2=x-y\)
b)\(\left(\sqrt{x}-3\right).\left(\sqrt{x}+2\right)=\left(\sqrt{x}\right)^2+2\sqrt{x}-3\sqrt{x}-6=x-\sqrt{x}-6\)
c)\(\sqrt{\left(2-\sqrt{5}\right)^2.\left(2+\sqrt{5}\right)^2}=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-\left(\sqrt{5}\right)^2\)
=\(4-5=1\)
d)\(\sqrt{\left(-5\right)^2.3^2}=\left(5\right).3=15\)
e)\(\sqrt{\frac{5}{27}.\frac{8}{20}}=\sqrt{\frac{2}{27}}\)
ĐÂy này nhớ **** vài câu nha
a, \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{2}\sqrt{5-\sqrt{21}}\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{7-2.\sqrt{3}.\sqrt{7}+3}\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=7-3=4\)
= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) ( vì căn 16 = 4)
=\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) (vì căn 4 = 2 mà 2 + 2 = 4 nên tách luôn thành căn 4 )
= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
Đúng nha lần sau mình giải tiếp cho
a) đk: a>=0\(M=\frac{x^4-x}{x^2+x+1}-\frac{x^4+x}{x^2-x+1}=\frac{x\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}-\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}=x^2-x-x^2-x=-2x\)
\(\Leftrightarrow M=-2\sqrt{a}\)
b) \(P=\sqrt{-2\sqrt{a}+a+1-1}=\sqrt{a-2\sqrt{a}+1-1}=\sqrt{\left(\sqrt{a}-1\right)^2-1}=\sqrt{\left(a-1-1\right)\left(a-1+1\right)}=\sqrt{a\left(a-2\right)}\)