Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Cho biểu thức ban đầu là A
Đặt 3 = a ; \(\sqrt{9+\dfrac{125}{27}}\)= b
⇔A = \(\sqrt[3]{a+b} . \sqrt[3]{b-a}\)
⇔A= \(\sqrt[3]{(a+b)(b-a)}\)
⇔A= \(\sqrt[3]{b^2-a^2}\)
⇔A= \(\sqrt[3]{9+\dfrac{125}{27}-9}\)
⇔A= \(\sqrt[3]{\dfrac{125}{27}}\)
⇔A = \(\dfrac{5}{3}\) ( ĐPCM)
\(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)
=\(\sqrt[3]{-\left(3+\sqrt{9+\dfrac{125}{27}}\right)\left(3-\sqrt{9+\dfrac{125}{27}}\right)}\)
=\(\sqrt[3]{-\left[9-\left(9+\dfrac{125}{27}\right)\right]}\)
=\(\sqrt[3]{\dfrac{125}{27}}\)
=5/3
a) \(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
=\(\sqrt[3]{16+8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}\)
=\(\sqrt[3]{\left(1+\sqrt{5}\right)^3}+\sqrt[3]{\left(1-\sqrt{5}\right)^3}\)
=\(1+\sqrt{5}+1-\sqrt{5}=2\)
b) \(\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}\)
=\(\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}\)
=\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)
c) xem lại đề
Bạn Thái làm sai rồi
a)do ban đầu cậu nhân 2 cho hai vế nhưng bạn chưa chia lại.mik bổ sung ý tiếp cho bạn là
2A=2=>A=1.
mik lam tiep cau b la
B=\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
=4-3
=1.
còn câu c mik pó tay :))
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
9.
\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\)
\(=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=-\sqrt{5}\)
10.
\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\)
\(=5\sqrt{3}-\sqrt{5+\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2+\dfrac{2}{3}}+6\sqrt{3}\)
\(=11\sqrt{3}-\sqrt{\dfrac{16}{3}}+\dfrac{9}{2}\sqrt{\dfrac{8}{3}}\)
\(=11\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}\)
\(=\dfrac{29\sqrt{3}}{3}+3\sqrt{6}\)
đâu cần lập đặt 2 ẩn a;b là 2 cái căn 3 đó xong đưa về hệ phương trình là được mà đăng lên hỏi chơi thôi
\(A=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)
\(\Leftrightarrow A=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}+\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\)
\(\Leftrightarrow A^3=6+3A.\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}.\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\)
\(\Leftrightarrow A^3=6+3A.\left(-\dfrac{5}{3}\right)\)
\(\Leftrightarrow A^3+5A-6=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+6\right)=0\)
\(\Leftrightarrow A=1\)
chưa hiểu chỗ\(A^3\)