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\(\sqrt{2,5}.\sqrt{360}\)
\(=\sqrt{25}.\sqrt{36}\)
\(=5.6\)
\(=30\)
\(\sqrt{\frac{-49}{-121}}\)
\(=\sqrt{\frac{49}{121}}\)
\(=\frac{\sqrt{49}}{\sqrt{121}}=\frac{7}{11}\)
Lời giải:
Coi yêu cầu đề là rút gọn. Lần sau bạn chú ý viết đầy đủ đề.
ĐK: $x>0; x\neq 1$
Gọi biểu thức đã cho là $P$. Ta có:
\(P=\frac{x-2+\sqrt{x}}{\sqrt{x}(\sqrt{x}+2)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{(\sqrt{x}-1)(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+2)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
ĐKXĐ: \(x\ge0\)
Từ biểu thức đầu suy ra: \(7\left(5\sqrt{x}-2\right)=2\left(8\sqrt{x}+2,5\right)\)
⇒ \(35\sqrt{x}-14=16\sqrt{x}+5\)
⇒ \(19\sqrt{x}=19\Rightarrow x=1\)(Thỏa mãn ĐKXĐ)
\(49+20\sqrt{6}=25+2.5.2\sqrt{6}+24=\left(5+2\sqrt{6}\right)^2=\left(3+2.\sqrt{3}\sqrt{2}+2\right)^2=\left(\sqrt{3}+\sqrt{2}\right)^4\)
\(\Leftrightarrow\sqrt[4]{49+20\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
tuiwng tự \(\Leftrightarrow\sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}\)
=> Cộng lại = > dpcm
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
Với \(k\in N;k\ne0\) ta có :
\(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{\left(k+1\right)}}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}\)
\(=\frac{\sqrt{k+1}+\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\)
\(=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Áp dụng ta có :
\(M=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}=1-\frac{1}{11}=\frac{10}{11}\)
a) \(\sqrt{36}.\sqrt{121}+\sqrt[3]{-64}-\sqrt[3]{125}\)
\(=6.11+\left(-4\right)-5=66-9=57\)
b) \(\sqrt{75}+\sqrt{\left(\sqrt{3}-2\right)^2}-30\sqrt{\frac{3}{25}}\)
\(=\sqrt{25.3}+\left|\sqrt{3}-2\right|-30.\frac{\sqrt{3}}{\sqrt{25}}\)
\(=5\sqrt{3}+2-\sqrt{3}-30.\frac{\sqrt{3}}{5}\)
\(=5\sqrt{3}+2-\sqrt{3}-6\sqrt{3}=2-2\sqrt{3}\)
c) \(\sqrt{11-4\sqrt{7}}-\frac{12}{1+\sqrt{7}}=\sqrt{7-4\sqrt{7}+4}-\frac{12}{1+\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\frac{12}{1+\sqrt{7}}=\left|\sqrt{7}-2\right|-\frac{12}{1+\sqrt{7}}\)
\(=\left(\sqrt{7}-2\right)-\frac{12}{\sqrt{7}+1}=\frac{\left(\sqrt{7}-2\right)\left(\sqrt{7}+1\right)}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}\)
\(=\frac{5-\sqrt{7}}{\sqrt{7}+1}-\frac{12}{\sqrt{7}+1}=\frac{-7-\sqrt{7}}{\sqrt{7}+1}\)
\(=\frac{-\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}+1}=-\sqrt{7}\)
\(\sqrt{8,1}.\sqrt{250}\)
\(=\sqrt{81}.\sqrt{25}\)
\(=9.5\)
\(=45\)
\(\sqrt{2,5}.\sqrt{360}\)
\(=\sqrt{25}.\sqrt{36}\)
\(=5.6\)
\(=30\)
\(\sqrt{\frac{-49}{-121}}=\sqrt{\frac{49}{121}}\)
\(=\frac{\sqrt{49}}{\sqrt{121}}\)
\(=\frac{7}{11}\)
\(\sqrt{\frac{-36}{-169}}=\sqrt{\frac{36}{169}}\)
\(=\frac{\sqrt{36}}{\sqrt{169}}=\frac{6}{13}\)