Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
+) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
\(=3\sqrt{4.5}-2\sqrt{9.5}+4\sqrt{5}\)
\(=6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
\(=4\sqrt{5}\)
+) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=\left(2\sqrt{7}-\sqrt{28}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=\left(2\sqrt{7}-2\sqrt{7}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=7+7\sqrt{8}\)
\(A=\sqrt{24+8\sqrt{5}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{5+2.4\sqrt{5}+16}+\sqrt{4-2.2\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{5}+4\right)}^2+\sqrt{\left(2-\sqrt{3}\right)}^2\)
\(=|\sqrt{5}+4|+|2-\sqrt{3}|\)
\(=\sqrt{5}+4+4-\sqrt{3}\)
\(=\sqrt{5}-\sqrt{3}+8\)
Ko biết đề sai ko?
Chờ từ trưa không idol nào đụng thì thôi em xin vậy :))
BT1:
Ta có: \(A\cdot B=\sqrt{4+\sqrt{10+2\sqrt{5}}}\cdot\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(=\sqrt{16-10-2\sqrt{5}}\)
\(=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
Từ đó thay vào: \(\left(A-B\right)^2\)
\(=A^2-2AB+B^2\)
\(=4+\sqrt{10+2\sqrt{5}}-2\left(\sqrt{5}-1\right)+4-\sqrt{10+2\sqrt{5}}\)
\(=10-2\sqrt{5}\)
\(\Rightarrow A-B=\sqrt{10-2\sqrt{5}}\)
BT2:
Đặt \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(\Leftrightarrow B^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)
\(=8-2\sqrt{16-7}=8-2\cdot3=2\)
\(\Rightarrow B=\sqrt{2}\)
\(\Rightarrow A=B-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
BT3:
đk: \(\orbr{\begin{cases}x\ge2\\x< -2\end{cases}}\)
\(C=\frac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\frac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
\(C=\frac{\left(x+2+\sqrt{x^2-4}\right)^2}{\left(x+2\right)^2-\left(x^2-4\right)}+\frac{\left(x+2-\sqrt{x^2-4}\right)^2}{\left(x+2\right)^2-\left(x^2-4\right)}\)
\(C=\frac{\left(x+2\right)^2+2\left(x+2\right)\sqrt{x^2-4}+x^2-4+\left(x+2\right)^2-2\left(x+2\right)\sqrt{x^2-4}+x^2-4}{x^2+4x+4-x^2+4}\)
\(C=\frac{2x^2+8x+8+2x^2-8}{4x+8}\)
\(C=\frac{4x^2+8x}{4x+8}=x\)
Vậy C = x
\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\) ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))
\(=\sqrt{\left(2+\sqrt{7}+2-\sqrt{7}\right)\left(2+\sqrt{7}-2+\sqrt{7}\right)}\)
\(=\sqrt{4\cdot\sqrt{7}}\)
\(2,\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}-\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}=\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2=\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2-\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(=\left(3\sqrt{5}-5\sqrt{2}+5\sqrt{2}+3\sqrt{5}\right)\left(3\sqrt{5}-5\sqrt{2}-5\sqrt{2}-3\sqrt{5}\right)\)
\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)
\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}\)
cuối lười tính nên thôi nhá :>
Trả lời:
\(\frac{4}{\sqrt{7}-\sqrt{3}}+\frac{6}{3+\sqrt{3}}+\frac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\frac{4.\left(\sqrt{7}+\sqrt{3}\right)}{7-3}+\frac{6.\left(3-\sqrt{3}\right)}{9-3}-\frac{7-\sqrt{7}}{\sqrt{7}-1}\)
\(=\frac{4.\left(\sqrt{7}+\sqrt{3}\right)}{4}+\frac{6.\left(3-\sqrt{3}\right)}{6}-\frac{\sqrt{7}.\left(\sqrt{7}-1\right)}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
\(=3\)
Học tốt
a) \(\sqrt{\frac{165^2-124^2}{164}}=\sqrt{\frac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\frac{41.289}{164}}\)
\(=\sqrt{\frac{11849}{164}}=\sqrt{72,25}=8,5\)
b)\(\sqrt{\frac{149^2-76^2}{457^2-384^2}}=\sqrt{\frac{\left(149-76\right)\left(149+76\right)}{\left(457-384\right)\left(457+384\right)}}\) \(=\sqrt{\frac{73.225}{73.841}}=\sqrt{\frac{225}{841}}=\sqrt{\frac{15^2}{29^2}}=\frac{15}{29}\)
c)\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\) \(=\sqrt{2^2+3+2.2.\sqrt{3}}-\sqrt{2^2+3-2.2.\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\sqrt{3}^2}-\sqrt{2^2-2.2.\sqrt{3}+\sqrt{3}^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}=\left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right)\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
a) \(=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2.2}=\sqrt{4}=2\)
b) \(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{8+2\sqrt{15}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{15}-4\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\left(\sqrt{5}+\sqrt{3}\right)^2\left(\sqrt{15}-4\right)=2\left(4+\sqrt{15}\right)\left(\sqrt{15}-4\right)\)
\(=-2\)
\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(\Leftrightarrow C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(\Leftrightarrow C=\left|\sqrt{3}-1\right|-\left|2+\sqrt{3}\right|\)
\(\Leftrightarrow C=\sqrt{3}-1-2-\sqrt{3}\)
\(\Leftrightarrow C=-3\)
\(=\sqrt{8-2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(\sqrt{2}.\sqrt{4-\sqrt{7}}\\ =\dfrac{1}{2}.\sqrt{2}.\sqrt{8-2\sqrt{7}}\\ =\dfrac{1}{2}.\sqrt{2}.\sqrt{7-2\sqrt{7}.1+1}\\ =\dfrac{1}{2}.\sqrt{2}.\sqrt{\left(\sqrt{7}-1\right)^2}\\ =\dfrac{1}{2}.\sqrt{2}.\left(\sqrt{7}-1\right)\\ =\dfrac{1}{2}.\sqrt{2}.\sqrt{7}-\dfrac{1}{2}.\sqrt{2}\\ =\dfrac{1}{2}.\sqrt{14}-\dfrac{1}{2}.\sqrt{2}\\ =\dfrac{1}{2}.\left(\sqrt{14}-\sqrt{2}\right)\)