Thu gọn biểu thức
\(A=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)
Các bạn giúp mình với
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 4 2020
\(A=43+24\sqrt{3}-8\sqrt{20+2\sqrt{\left(3\sqrt{3}+4\right)^2}}\)
\(=43+24\sqrt{3}-8\sqrt{20+2\left(3\sqrt{3}+4\right)}\)
\(=43+24\sqrt{3}-8\sqrt{28+6\sqrt{3}}\)
\(=43+24\sqrt{3}-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)
\(=43+24\sqrt{3}-8\left(3\sqrt{3}+1\right)\)
\(=43-8=35\)
L
2 tháng 12 2020
\(B=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)
\(=\left(2\sqrt{3}-1\right)^2\left(2+\sqrt{3}\right)^2-8\sqrt{20+2\sqrt{\left(4+3\sqrt{3}\right)^2}}\)
\(=\left(3\sqrt{3}+4\right)^2-8\sqrt{20+2\left(4+3\sqrt{3}\right)}\)
\(=\left(3\sqrt{3}+4\right)^2-8\sqrt{28+6\sqrt{3}}\)
\(=\left(3\sqrt{3}+4\right)^2-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)
\(=43+24\sqrt{3}-8\left(3\sqrt{3}+1\right)=35\)
26 tháng 7 2019
\(D=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)
\(=\left(2\sqrt{3}-1\right)^2\left(\sqrt{3}+2\right)^2-8\sqrt{20+2\left(3\sqrt{3}+4\right)}\)
\(=\left(4+3\sqrt{3}\right)^2-8\sqrt{28+6\sqrt{3}}\)\(=\left(4+3\sqrt{3}\right)^2-8\left(3\sqrt{3}+1\right)\)
\(=43+24\sqrt{3}-24\sqrt{3}-8=35\)
26 tháng 5 2022
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
26 tháng 5 2022
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 4 2019
\(A=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-4}+\frac{\sqrt{x}-10}{x-4}\)
\(A=\frac{x+2\sqrt{x}+x-3\sqrt{x}+2+\sqrt{x}-10}{x-4}\)
\(A=\frac{2x-8}{x-4}=\frac{2\left(x-4\right)}{x-4}=2\)
\(B=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(3\sqrt{3}+4\right)^2}}\)
\(B=43+24\sqrt{3}-8\sqrt{20+6\sqrt{3}+8}\)
\(B=43+24\sqrt{3}-8\sqrt{28+6\sqrt{3}}\)
\(B=43+24\sqrt{3}-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)
\(B=43+24\sqrt{3}-24\sqrt{3}-8\)
\(B=35\)
19 tháng 8 2021
a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)
\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)
\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)
b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)
\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)
19 tháng 8 2021
b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)
\(A=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\) \(A=\left(6+7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(16+2.4.3\sqrt{3}+27\right)}}\)
\(A=6\left(7+4\sqrt{3}\right)+\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{\left(4+3\sqrt{3}\right)^2}}\)Trong căn là hằng đẳng thức (a+b)^2
\(A=42+24\sqrt{3}+7^2-\left(4\sqrt{3}\right)^2-8\sqrt{20+2\left(4+3\sqrt{3}\right)}\) sử dụng hằng đẳng thức a^2 -b^2\(A=43+24\sqrt{3}-8\sqrt{20+8+2.3\sqrt{3}}\)
\(A=43+24\sqrt{3}-8\sqrt{1+2.3\sqrt{3}+27}\)trong căn tiếp tục là hằng đẳng thức (a+b)^2\(A=43+24\sqrt{3}-8\sqrt{\left(1+3\sqrt{3}\right)^2}\)
\(A=43+24\sqrt{3}-8\left(1+3\sqrt{3}\right)\)
\(A=35\)
chúc bạn thành công nhé
cảm ơn bạn nhiều