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\(x=\frac{1}{2}\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}=\frac{1}{2}.\left(\sqrt{2}-1\right)\)
\(\Rightarrow2x=\sqrt{2}-1\Rightarrow2x+1=\sqrt{2}\)
\(\Rightarrow4x^2+4x+1=2\Rightarrow4x^2+4x-1=0\)
\(B=\left[x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1-1\right]^{2018}+2018\)
\(=\left(-1\right)^{2018}+2018=2019\)
Lời giải:
\(Q=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(1+\sqrt{2})(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
Ta có : \(\frac{bc}{\sqrt{3a+bc}}=\frac{bc}{\sqrt{\left(a+b+c\right)a+bc}}=\frac{bc}{\sqrt{a^2+ab+ac+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng bđt Cauchy , ta có : \(\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{bc}{2}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
Tương tự : \(\frac{ac}{\sqrt{3b+ac}}=\frac{ac}{\sqrt{\left(a+b\right)\left(b+c\right)}}\le\frac{ac}{2}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\); \(\frac{ab}{\sqrt{3c+ab}}=\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\frac{ab}{2}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)
\(\Rightarrow P=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{ac}{\sqrt{\left(b+a\right)\left(b+c\right)}}+\frac{ab}{\sqrt{\left(a+c\right)\left(c+b\right)}}\)
\(\le\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ac}{a+b}+\frac{ac}{b+c}\right)\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ac}{a+b}\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
Suy ra : Max P \(=\frac{3}{2}\Leftrightarrow a=b=c=1\)
đây nhé Câu hỏi của Steffy Han - Toán lớp 8 | Học trực tuyến
\(A=\sqrt{2}\left(\sqrt{2}-2\right)+\left(\sqrt{2}+1\right)^2\)
\(A=2-2\sqrt{2}+2+2\sqrt{2}+1\)
\(A=5\)
Lời giải:
\(N=\sqrt{4\sqrt{6}+8\sqrt{3}+4\sqrt{2}+18}\)
\(=\sqrt{2\sqrt{24}+4(2\sqrt{3}+\sqrt{2})+18}\)
\(=\sqrt{12+2\sqrt{24}+2+4(\sqrt{12}+\sqrt{2})+4}\)
\(=\sqrt{(\sqrt{12}+\sqrt{2})^2+4(\sqrt{12}+\sqrt{2})+4}\)
\(=\sqrt{(\sqrt{12}+\sqrt{2}+2)^2}=\sqrt{12}+\sqrt{2}+2=2\sqrt{3}+\sqrt{2}+2\)
a: \(=\dfrac{1}{\sqrt{6}-1+1}-\dfrac{1}{\sqrt{6}+1-1}\)
\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}\)
=0
b: \(=\dfrac{3+\sqrt{7}-3+\sqrt{7}}{2}=\dfrac{2\sqrt{7}}{2}=\sqrt{7}\)
c: \(=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\)
a/ \(A=\sqrt{6-2\sqrt{5}}-\sqrt{5}\)\(=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1^2}-\sqrt{5}\)\(=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}\)\(=\sqrt{5}-1-\sqrt{5}\)\(=-1.\)
Bạn kiểm tra lại câu b với c đi, hình như sai đề rồi.
\(A=\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)
=\(\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\)
\(=3-\sqrt{3}+2\sqrt{3}-3\)
=\(\sqrt{3}\)