A = \(10-\left(\sqrt{32}-\sqrt{8}-\sqrt{27}\right)\left(\sqrt{8}-\sqrt{32}-\sqrt{27}\right)\)
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NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 9 2019
ĐKXĐ: \(x\ge0;x\ne4\)
\(B=\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2+\sqrt{x}-2-2\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{1}{x-4}\)
\(A=10-\left(2\sqrt{2}-3\sqrt{3}\right)\left(-2\sqrt{2}-3\sqrt{3}\right)\)
\(=10+\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)\)
\(=10+\left(8-27\right)=-9\)
\(AB=-1\Leftrightarrow\frac{-9}{x-4}=-1\Rightarrow x-4=9\Rightarrow x=13\)
HS
11 tháng 9 2018
\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\\ =\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\\ =\left(-\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+\sqrt{2}\right)\\ =\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)\\ =\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=9.3-2=25\)
S
11 tháng 9 2018
\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=\left(3\sqrt{3}\right)^2-\sqrt{2}^2\)
\(=9.3-2=27-2=25\)
7 tháng 8 2018
\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)\)
\(=27-2\)
\(=25\)
7 tháng 8 2018
\(\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(\Leftrightarrow\left(4\sqrt{2}+3\sqrt{3}-5\sqrt{2}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)\)
\(\Leftrightarrow\left(4\sqrt{2}+3\sqrt{2}-5\sqrt{2}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(\Leftrightarrow\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(\Leftrightarrow\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+2\right)\)
\(\Leftrightarrow\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)
\(\Rightarrow25\)
Vậy: BT = 25
P/s: từ dòng thứ 2 trở xuống bạn tự phân ... Vấn đề là ở bạn thôi :)))
30 tháng 7 2021
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 10 2019
\(A=\left(\sqrt{5}+3\right)\left(5-\sqrt{15}\right)=5\sqrt{5}-5\sqrt{3}+15-3\sqrt{15}\)
Bạn ghi nhầm đề thì phải, ngoặc đầu là \(\sqrt{5}+\sqrt{3}\) mới rút gọn được theo HĐT số 3
\(B=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)
\(C=1-\left(3\sqrt{5}-2\sqrt{5}-\sqrt{3}\right)\left(2\sqrt{5}-3\sqrt{5}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{5}-\sqrt{3}\right)\left(-\sqrt{5}-\sqrt{3}\right)=1+\left(5-3\right)=3\)
\(D=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}}\right).\sqrt{6}=\frac{\left(3-2\right)}{\sqrt{6}}.\sqrt{6}=1\)
4 tháng 7 2018
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
\(A=10-\left(\sqrt{32}-\sqrt{8}-\sqrt{27}\right)\left(\sqrt{8}-\sqrt{32}-\sqrt{27}\right)\)
\(A=10-\left[-\sqrt{27}+\left(\sqrt{32}-\sqrt{8}\right)\right]\left[-\sqrt{27}-\left(\sqrt{32}-\sqrt{8}\right)\right]\)
\(A=10-\left[\left(-\sqrt{27}\right)^2-\left(\sqrt{32}-\sqrt{8}\right)^2\right]\)
\(A=10-\left(27-8\right)\)
\(A=-9\)