Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Xét: \(A=\sqrt{26+15\sqrt{3}}\) dễ thấy A > 0
\(\Leftrightarrow A^2=52-2\sqrt{26^2-15^2.3}=50\Leftrightarrow A=\sqrt{50}\)
Vậy: \(A=2+\sqrt{3}.\sqrt{26+15\sqrt{3}}-2\sqrt{3}.\sqrt{26-15\sqrt{3}}\)
\(=2+\sqrt{3}.A=2+\sqrt{3}.\sqrt{50}=5\sqrt{6}+10\sqrt{2}\)
Đặt bt là A
\(A\sqrt{2}=\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\sqrt{\left(3\sqrt{3}-5\right)^2}\)\(=\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)\)
khai triển hết ra ta đc \(A\sqrt{2}=2\Rightarrow A=\sqrt{2}\)
\(B=\left(\sqrt{x}+\frac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\right)\frac{\left(\sqrt{x}-\sqrt{5}\right)^2}{\sqrt{x}-\sqrt{5}}=\left(\sqrt{x}+\sqrt{5}\right)\left(\sqrt{x}-\sqrt{5}\right)=x-5\)
\(a=\sqrt[3]{15\sqrt{3}+26}+\sqrt[3]{15\sqrt{3}-26}\)
\(a^3=30\sqrt{3}+3a.\sqrt[3]{15^2.3-26^2}=30\sqrt{3}-3a\)
\(\Leftrightarrow a^3+3a-30\sqrt{3}=0\)
\(\Leftrightarrow\left(a-2\sqrt{3}\right)\left(a^2+2\sqrt{3}a+15\right)=0\)
\(\Rightarrow a=2\sqrt{3}\)
mơn đại ca
Sửa đề
\(A=\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{26-15\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\left(2+\sqrt{3}\right)\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=0\)
Thu gọn:
\(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
Ta có: \(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
\(=\frac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\left(2-\sqrt{3}\right)\cdot\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)
\(=\frac{\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\cdot\sqrt{\left(3\sqrt{3}-5\right)^2}}{\sqrt{2}}\)
\(=\frac{\left(2-\sqrt{3}\right)\left|3\sqrt{3}+5\right|-\left(2+\sqrt{3}\right)\left|3\sqrt{3}-5\right|}{\sqrt{2}}\)
\(=\frac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)(Vì \(3\sqrt{3}>5>0\))
\(=\frac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+1-\left(\sqrt{3}-1\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Lời giải:
Gọi biểu thức trên là $A$
Đặt \(\sqrt[3]{15\sqrt{3}-26}=a; \sqrt[3]{15\sqrt{3}+26}=b\). Ta có:
\(a^3-b^3=-52\)
\(ab=-1\)
\(A^3=(a-b)^3=a^3-3ab(a-b)-b^3=-52+3A\)
\(\Leftrightarrow A^3-3A+52=0\)
\(\Leftrightarrow A^2(A+4)-4A(A+4)+13(A+4)=0\)
\(\Leftrightarrow (A+4)(A^2-4A+13)=0\)
Dễ thấy $A^2-4A+13>0$ nên $A+4=0$
$\Leftrightarrow A=-4$