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a)
\((\sqrt{3}-2\sqrt{12}+2\sqrt{4})(\sqrt{27}+\sqrt{144}-2\sqrt{16})\)
\(=(\sqrt{3}-4\sqrt{3}+4)(3\sqrt{3}+12-8)\)
\(=(-3\sqrt{3}+4)(3\sqrt{3}+4)=4^2-(3\sqrt{3})^2=16-27=-11\)
b)
\((2\sqrt{5}+2\sqrt{3})^2-4\sqrt{60}\)
\(=(2\sqrt{5})^2+2.2\sqrt{5}.2\sqrt{3}+(2\sqrt{3})^2-8\sqrt{15}\)
\(=32+8\sqrt{15}-8\sqrt{15}=32\)
c)
\(\sqrt{6}(3\sqrt{12}-4\sqrt{3}+\sqrt{48}-5\sqrt{6})\)
\(=3\sqrt{72}-4\sqrt{18}+\sqrt{6.48}-5.\sqrt{36}\)
\(=18\sqrt{2}-12\sqrt{2}+12\sqrt{2}-30=18\sqrt{2}-30\)
d)
\((\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})\)
\(=(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})(\sqrt{6}+\sqrt{2})\)
\(=(2-3)(\sqrt{6}+\sqrt{2})=-(\sqrt{6}+\sqrt{2})\)
e) Biểu thức bên trong căn lớn âm nên biểu căn bậc 2 không có nghĩa
f)
\((\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}).\frac{1}{\sqrt{3}+5}\)
\(=(\frac{2\sqrt{3}+15}{3-\sqrt{3}}+\frac{3}{\sqrt{3}-2}).\frac{1}{\sqrt{3}+5}\)
\(=\frac{2\sqrt{3}+15)(\sqrt{3}-2)+3(3-\sqrt{3})}{(3-\sqrt{3})(\sqrt{3}-2)}.\frac{1}{\sqrt{3}+5}\)
\(=\frac{-15+8\sqrt{3}}{(-9+5\sqrt{3})(\sqrt{3}+5)}=\frac{-15+8\sqrt{3}}{-30+16\sqrt{3}}=\frac{-15+8\sqrt{3}}{2(-15+8\sqrt{3})}=\frac{1}{2}\)
5.
ĐKXĐ: \(-\frac{1}{2}\le x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{2}+x+2\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=1\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
6.
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x^2-1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\left(vn\right)\end{matrix}\right.\)
2.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)
\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-x+1\\x+1=4x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-3=0\\4x^2-5x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)
Bài 1
a) \(A=\left(4-\sqrt{15}\right)\left(\sqrt{10}+\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\sqrt{\left(4-\sqrt{15}\right)\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}.\left(\sqrt{5}+\sqrt{3}\right).\sqrt{2}=\sqrt{\left(4-\sqrt{15}\right).\left(16-15\right).2}.\left(\sqrt{5}+\sqrt{3}\right)=\sqrt{8-2\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)=\sqrt{5-2\sqrt{5}.\sqrt{3}+3}.\left(\sqrt{5}+\sqrt{3}\right)=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}.\left(\sqrt{5}+\sqrt{3}\right)=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2\)
Ta có công thức tổng quát\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
Vậy \(B=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{16}-\sqrt{15}=\sqrt{16}-\sqrt{1}=4-1=3\)
b) \(6x^4-7x^2-3=0\Leftrightarrow6x^4-9x^2+2x^2-3=0\Leftrightarrow3x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\Leftrightarrow\left(2x^2-3\right)\left(3x^2+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}2x^2-3=0\\3x^2+1=0\left(ktm\right)\end{matrix}\right.\)\(\Leftrightarrow\)\(2x^2-3=0\Leftrightarrow2x^2=3\Leftrightarrow x^2=\frac{3}{2}\Leftrightarrow x=\frac{\pm\sqrt{6}}{2}\)
Vậy S={\(\frac{-\sqrt{6}}{2};\frac{\sqrt{6}}{2}\)}
\(B=\frac{5\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\frac{\sqrt{5}}{2}\right)^2+\left(\sqrt{4-2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\frac{\sqrt{3}}{2}\right)^2}{2}\)
\(=\frac{5\left(\sqrt{3}+\frac{\sqrt{5}}{2}\right)^2+\left(\sqrt{5}+\frac{\sqrt{3}}{2}-2\right)^2}{2}\)
\(=\frac{\frac{85}{4}+5\sqrt{15}+\frac{39}{4}+\sqrt{15}-2\sqrt{3}-4\sqrt{5}}{2}=\frac{31+3\sqrt{15}-4\sqrt{5}-2\sqrt{3}}{2}\)
\(5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}-\sqrt{3}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{3}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{3}+1+\sqrt{5}-1-\sqrt{5}\right)^2+\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}+1-\sqrt{3}\right)^2\)
\(=\frac{5}{2}\left(\sqrt{3}\right)^2+\frac{1}{2}\left(\sqrt{5}\right)^2=\frac{15}{2}+\frac{5}{2}=\frac{20}{2}=10\)