a) \(A=\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+\sqrt{4901}}\)

\(\Leftrightarrow A^3=70-\sqrt{4901}+70+\sqrt{4901}+3\sqrt[3]{\left(70-\sqrt{4901}\right)\left(70+\sqrt{4901}\right)}\cdot A\)

\(\Leftrightarrow A^3=140+3\sqrt[3]{4900-4901}\cdot A\)

\(\Leftrightarrow A^3=140-3A\)

\(\Leftrightarrow A^3+3A-140=0\)

\(\Leftrightarrow\left(A-5\right)\left(A^2+5A+28\right)=0\)

\(\Leftrightarrow A=5\) ( do \(A^2+5A+28>0\forall A\) )

Vậy \(A=5\)

\(B=\frac{1}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}+\sqrt[3]{2}\\ =\frac{1}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}+\frac{\sqrt[3]{2}\left(\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}\right)}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}\\ =\frac{1+\sqrt[3]{12}+2+\sqrt[3]{18}}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}=\frac{3+\sqrt[3]{12}+\sqrt[3]{18}}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}\\ =\frac{\sqrt[3]{3}\left(\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}\right)}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}=\sqrt[3]{3}\)

\(\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+\sqrt{4901}}\\ =\dfrac{1}{2}.2\left(\sqrt[3]{70-13\sqrt{29}}+\sqrt[3]{70+13\sqrt{29}}\right)\\ =\dfrac{1}{2}\left(\sqrt[3]{560-104\sqrt{29}}+\sqrt[3]{560+104\sqrt{29}}\right)\\ =\dfrac{1}{2}\left(\sqrt[3]{5^3-3.5^2.\sqrt{29}+3.5.29-29\sqrt{29}}+\sqrt[3]{5^3+3.5^2.\sqrt{29}+3.5.29+29\sqrt{29}}\right)\\ =\dfrac{1}{2}\left[\sqrt[3]{\left(5-\sqrt{29}\right)^3}+\sqrt[3]{\left(5+\sqrt{29}\right)^3}\right]\\ =\dfrac{1}{2}\left(5-\sqrt{29}+5+\sqrt{29}\right)\\ =\dfrac{1}{2}.10=5\)

Đặt \(A=\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+\sqrt{4901}}\)

\(\Rightarrow A^3=140+3.\left(-1\right)A\)

\(\Leftrightarrow\left(A^3-5A^2\right)+\left(5A^2-25A\right)+\left(28A-140\right)=0\)

\(\Leftrightarrow\left(A-5\right)\left(A^2+5A+28\right)=0\)

\(\Leftrightarrow A=5\)

\(D=\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+\sqrt{4901}}\) =>\(D^3=70-\sqrt{4901}+70+\sqrt{4901}+3\sqrt[3]{\left(70-\sqrt{4901}\right)\left(70+\sqrt{4901}\right)}.D\) \(=140-3D\) => \(D^3+3D-140=0\) <=>\(D^3-25D+28D-140=0\) <=>\(D\left(D^2-25\right)+28\left(D-5\right)=0\) <=>\(D\left(D+5\right)\left(D-5\right)+28\left(D-5\right)=0\) <=>\(\left(D^2+5D+28\right)\left(D-5\right)=0\) Vì \(D^2+5D+28>0\) =>D-5=0 =>D=5

Kết quả là 5 nha bạn mik ko coppy đáp án đâu nhé

\(D^3=70-\sqrt{4901}+70+\sqrt{4901}+3\cdot D\cdot\sqrt[3]{70^2-4901}\)

\(=140+3\cdot D\cdot\left(-1\right)\)

=>D^3+3D-140=0

=>D=5

\(a^3=140+3.a\)

Vậy a nghiệm của  phương trình.x^3-3x-140 =0

nhầm dấu

a^3=140-3a

đa thức cần tim là x^3+3x-140

nhắc lại HĐT: (a+b)^3=a^3+b^3+3ab(a+b) 

\(a=\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+\sqrt{4901}}\)

\(a^3=\left(\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+\sqrt{4901}}\right)^{^3}\)

\(a^3=\left(70-\sqrt{4901}\right)+\left(70+\sqrt{4901}\right)+3.\left(\sqrt[3]{70^2-4901}\right).a\)

\(a^3=70+70+3.\sqrt[3]{-1}.a=140-3a\)

2) Đặt VT là A: Áp dụng công thức:\(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

\(A^3=70-\sqrt{4901}+70+\sqrt{4901}+3.\sqrt[3]{70-\sqrt{4901}}.\sqrt[3]{70+\sqrt{4901}}\left(\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+\sqrt{4901}}\right)\)

\(A^3=140-3A\)

P/S:\(\left(3\sqrt[3]{70+\sqrt{4901}}.\sqrt[3]{70-\sqrt{4901}}\right)=3\sqrt[3]{\left(70+\sqrt{4901}\right)\left(70-\sqrt{4901}\right)}=3\sqrt[3]{70^2-4901}=-3\)

\(A^3+3A-140=0\)

\(\left(A+5\right)\left(A^2-5A+28\right)=0\)

\(A^2-5A+28=\left(A-\dfrac{5}{2}\right)^2+\dfrac{87}{4}>0\)

Vậy: A=5

1) T nghĩ là tìm Max

Hãy tìm Min của mẫu, lấy 1 chia ra là Max

mình ghi lộn bài 1 phải là: Tìm GTLN mới đúng

Câu 1: 

a: \(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(=2\sqrt{x}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=2\sqrt{x}+2\)

b: Để A=4 thì \(2\sqrt{x}=2\)

=>x=1(loại)