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Bạn xem lại đề bài 1 và 2.b nhé !
2/ \(A=\sqrt{\left(3-5\sqrt{2}\right)^2}-\sqrt{51+10\sqrt{2}}\)
\(A=5\sqrt{2}-3-\sqrt{\left(5\sqrt{2}+1\right)^2}\)
\(A=5\sqrt{2}-3-5\sqrt{2}-1\)
\(A=-4\)
\(B=\frac{5\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\frac{\sqrt{5}}{2}\right)^2+\left(\sqrt{4-2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\frac{\sqrt{3}}{2}\right)^2}{2}\)
\(=\frac{5\left(\sqrt{3}+\frac{\sqrt{5}}{2}\right)^2+\left(\sqrt{5}+\frac{\sqrt{3}}{2}-2\right)^2}{2}\)
\(=\frac{\frac{85}{4}+5\sqrt{15}+\frac{39}{4}+\sqrt{15}-2\sqrt{3}-4\sqrt{5}}{2}=\frac{31+3\sqrt{15}-4\sqrt{5}-2\sqrt{3}}{2}\)
6.
Đặt \(\left\{{}\begin{matrix}\sqrt{5x^2+6x+5}=a\\4x=b\end{matrix}\right.\)
\(\Rightarrow a\left(a^2+1\right)=b\left(b^2+1\right)\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{5x^2+6x+5}=4x\left(x\ge0\right)\)
\(\Leftrightarrow5x^2+6x+5=16x^2\)
\(\Leftrightarrow11x^2-6x-5=0\)
\(\Rightarrow x=1\)
4. Bạn coi lại đề (chính xác là pt này ko có nghiệm thực)
5.
\(\Leftrightarrow x^2+x+6-\left(2x+1\right)\sqrt{x^2+x+6}+6x-6=0\)
Đặt \(\sqrt{x^2+x+6}=t>0\)
\(t^2-\left(2x+1\right)t+6x-6=0\)
\(\Delta=\left(2x+1\right)^2-4\left(6x-6\right)=\left(2x-5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{2x+1+2x-5}{2}=2x-2\\t=\frac{2x+1-2x+5}{2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+6}=2x-2\left(x\ge1\right)\\\sqrt{x^2+x+6}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=4x^2-8x+4\left(x\ge1\right)\\x^2+x+6=9\end{matrix}\right.\)
\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
Có \(x=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}=\frac{2}{\sqrt[3]{16}+\sqrt[3]{8}+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{4}\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right)}=\frac{\sqrt[3]{2}}{\sqrt[3]{4}+\sqrt[3]{2}+1}\)
\(y=\frac{2}{2\sqrt[3]{2}-2+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{16}-\sqrt[3]{8}+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{4}\left(\sqrt[3]{4}-\sqrt[3]{2}+1\right)}=\frac{\sqrt[3]{2}}{\sqrt[3]{4}-\sqrt[3]{2}+1}\)
Đặt \(\sqrt[3]{2}=a\)
=> \(x=\frac{a}{a^2+a+1}\) ,\(y=\frac{a}{a^2-a+1}\)
Có: \(x+y=\frac{a}{a^2+a+1}+\frac{a}{a^2-a+1}=\frac{a^3-a^2+a+a^3+a^2+a}{\left(a^2+a+1\right)\left(a^2-a+1\right)}=\frac{2a^3+2a}{a^4+a^2+1}\)
\(x-y=\frac{a}{a^2+a+1}-\frac{a}{a^2-a+1}=\frac{a^3-a^2+a-a^3-a^2-a}{\left(a^2+a+1\right)\left(a^2-a+1\right)}=\frac{-2a^2}{a^4+a^2+1}\)
Có x2-y2= (x-y)(x+y)=\(\frac{2a^3+2a}{a^4+a^2+1}.\frac{-2a^2}{a^4+a^2+1}=\frac{-2a^2.2a\left(a^2+1\right)}{\left(a^4+a^2+1\right)^2}=\frac{-4a^3\left(a^2+1\right)}{\left(a^4+a^2+1\right)^2}=\frac{-4.2\left(\sqrt[3]{4}+1\right)}{\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)^2}\)
=\(\frac{-8\left(\sqrt[3]{4}+1\right)}{\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)^2}\)