Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)
\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=32+8\sqrt{15}-8\sqrt{15}-30\)
=2
1. \(\sqrt{\left(x+3\right)\left(x+7\right)}=3\sqrt{x+3}+2\sqrt{x+7}-6\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x+7\right)}-3\sqrt{x+3}-2\sqrt{x+7}+6=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+7}-3\right)-2\left(\sqrt{x+7}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+7}-3\right)\left(\sqrt{x+3}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+7}-3=0\\\sqrt{x+3}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+7}=3\\\sqrt{x+3}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy...
2. \(2x^2+2x+1=\sqrt{4x+1}\)
\(\Leftrightarrow2x^2+2x+1-\sqrt{4x+1}=0\)
\(\Leftrightarrow4x^2+4x+2-2\sqrt{4x+1}=0\)
\(\Leftrightarrow4x+1-2\sqrt{4x+1}+1+4x^2=0\)
\(\Leftrightarrow\left(\sqrt{4x+1}-1\right)^2+4x^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4x+1}=1\\2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}4x+1=1\\x=0\end{matrix}\right.\)\(\Leftrightarrow x=0\)
Vậy...
3. \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}=\frac{x+3}{2}\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1=\frac{x+3}{2}\)
Đặt \(\sqrt{x-1}=a\)
\(\Leftrightarrow x-1=a^2\Leftrightarrow x+3=a^2+4\)
\(pt\Leftrightarrow\left|a-1\right|+a+1=\frac{a^2+4}{2}\)
+) Xét \(a\le1\Leftrightarrow a-1\le0\Leftrightarrow1\le x\le2\)
\(pt\Leftrightarrow1-a+a+1=\frac{a^2+4}{2}\)
\(\Leftrightarrow2=\frac{a^2+4}{2}\)
\(\Leftrightarrow a^2+4=4\)
\(\Leftrightarrow a=0\)
\(\Leftrightarrow\sqrt{x-1}=0\)
\(\Leftrightarrow x=1\) ( thỏa )
+) Xét \(a\ge1\Leftrightarrow a-1\ge0\Leftrightarrow x>2\)
\(pt\Leftrightarrow a-1+a+1=\frac{a^2+3}{2}\)
\(\Leftrightarrow2a=\frac{a^2+3}{2}\)
\(\Leftrightarrow a^2+3=4a\)
\(\Leftrightarrow a^2-4a+3=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(loai\right)\\x=10\left(thoa\right)\end{matrix}\right.\)
Vậy...
a)\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{10\left(4-\sqrt{15}\right)}+\sqrt{6\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
b) \(2\left(\sqrt{10}-\sqrt{2}\right)\left(4+\sqrt{6-2\sqrt{5}}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{\left(1-\sqrt{5}\right)^2}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{5}-1\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=6\sqrt{10}+2\sqrt{50}-6\sqrt{2}-2\sqrt{10}\)
\(=6\sqrt{10}+10\sqrt{2}-6\sqrt{2}-2\sqrt{10}\)
\(=4\sqrt{10}+4\sqrt{2}\)
c) \(\left(\sqrt{7}+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=7\sqrt{7}-7\sqrt{2}+\sqrt{98}-\sqrt{28}\)
\(=7\sqrt{7}-7\sqrt{2}+7\sqrt{2}-2\sqrt{7}\)
\(=5\sqrt{7}\)
d) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)
\(=\sqrt{\dfrac{289+42\sqrt{2}}{16}}\)
\(=\dfrac{\sqrt{289+42\sqrt{2}}}{\sqrt{4^2}}\)
\(=\dfrac{\sqrt{\left(1+12\sqrt{2}\right)^2}}{4}\)
\(=\dfrac{1+12\sqrt{2}}{4}\)
e) \(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{147}-\sqrt{63}+7-\sqrt{21}\)
\(=7\sqrt{3}-\sqrt{63}+7-\sqrt{21}\)
f) bạn xem đề lại nhé
\(=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
a) \(9+4\sqrt{5}=\left(\sqrt{5}\right)^2+2.\sqrt{5}.2+2^2=\left(\sqrt{5}+2\right)^2\)
b) \(23-8\sqrt{7}=4^2-2.4.\sqrt{7}+\left(\sqrt{7}\right)^2=\left(4-\sqrt{7}\right)^2\)
c) \(4-2\sqrt{3}=\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2=\left(\sqrt{3}-1\right)^2\)
d) \(11+6\sqrt{2}=3^2+2.3.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(3+\sqrt{2}\right)^2\)
a) \(9+4\sqrt{5}=\left(\sqrt{5}+2\right)^2\)
b) \(23-8\sqrt{7}=\left(4-\sqrt{7}\right)^2\)
c) \(4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
d) \(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)
10+2 căn 5
= 10+2 căn 10 . căn 2 trên 2+2 trên 4
= (căn 10+ căn 2 trên 2) 2
mik ko biết viết căn nhé, bạn tự dịch, còn kqua sai thì thôi nhé
\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
\(\sqrt{8-2\sqrt{12}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}=\left|\sqrt{6}-\sqrt{2}\right|=\sqrt{6}-\sqrt{2}\)
\(\sqrt{21+6\sqrt{6}}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\left|3\sqrt{2}-\sqrt{3}\right|=3\sqrt{2}-\sqrt{3}\)
\(\sqrt{15-6\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|=3-\sqrt{6}\)
\(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)
\(\sqrt{41+12\sqrt{5}}=\sqrt{\left(6+\sqrt{5}\right)^2}=6+\sqrt{5}\)
a, 4-2\(\sqrt{3}\)
=3-\(2\sqrt{1}.\sqrt{3}\)+1
=(\(\sqrt{3}\))2-\(2\sqrt{3}.\sqrt{1}+\left(\sqrt{1}\right)^2\)
=\(\left(\sqrt{3}-\sqrt{1}\right)^2\)
b,3+\(2\sqrt{2}\)
=\(2+2\sqrt{2}.\sqrt{1}+1\)
=\(\left(\sqrt{2}\right)^2+2.\sqrt{1}.\sqrt{2}+\left(\sqrt{1}\right)^2\)
=\(\left(\sqrt{2}+\sqrt{1}\right)^2\)
c, 11-2\(\sqrt{30}\)
=6-\(2\sqrt{5}.\sqrt{6}+5\)
=\(\left(\sqrt{6}\right)^2-2\sqrt{5}.\sqrt{6}+\left(\sqrt{5}\right)^2\)
=\(\left(\sqrt{6}-\sqrt{5}\right)^2\)
a/ \(4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
b/ \(3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)
c/ \(11-2\sqrt{30}=6-2\sqrt{30}+5=\left(\sqrt{6}-\sqrt{5}\right)^2\)