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: \(=\dfrac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)

b: \(=\dfrac{\sqrt{10}\left(\sqrt{11}+\sqrt{7}\right)}{\sqrt{2}\left(\sqrt{11}+\sqrt{7}\right)}=\sqrt{\dfrac{10}{2}}=\sqrt{5}\)

c: \(=\dfrac{\sqrt{6}\left(\sqrt{7}-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{7}-\sqrt{6}\right)}=\sqrt{\dfrac{6}{3}}=\sqrt{2}\)

1) \(\dfrac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{9\sqrt{5}+3\sqrt{9\cdot3}}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{9\sqrt{5}+3\cdot3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{9\cdot\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{9}{1}=9\)

2) \(\dfrac{\sqrt{110}+\sqrt{70}}{\sqrt{22}+\sqrt{14}}\)

\(=\dfrac{\sqrt{10}\cdot\sqrt{11}+\sqrt{10}\cdot\sqrt{7}}{\sqrt{2}\cdot\sqrt{11}+\sqrt{2}\cdot\sqrt{7}}\)

\(=\dfrac{\sqrt{10}\cdot\left(\sqrt{11}+\sqrt{7}\right)}{\sqrt{2}\cdot\left(\sqrt{11}+\sqrt{7}\right)}\)

\(=\dfrac{\sqrt{10}}{\sqrt{2}}=\sqrt{\dfrac{10}{2}}\)

\(=\sqrt{5}\)

3) \(\dfrac{\sqrt{42}-6}{\sqrt{21}-\sqrt{18}}\)

\(=\dfrac{\sqrt{6}\cdot\sqrt{7}-\sqrt{6}\cdot\sqrt{6}}{\sqrt{3}\cdot\sqrt{7}-\sqrt{3}\cdot\sqrt{6}}\)

\(=\dfrac{\sqrt{6}\cdot\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{3}\cdot\left(\sqrt{7}-\sqrt{3}\right)}\)

\(=\dfrac{\sqrt{6}}{\sqrt{3}}=\sqrt{\dfrac{6}{3}}\)

\(=\sqrt{2}\)

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

Đặt \(A=\sqrt[3]{99-70\sqrt{2}}+\sqrt[3]{99+70\sqrt{2}}\)

Ta có: \(A^3=\left(\sqrt[3]{99-70\sqrt{2}}+\sqrt[3]{99+70\sqrt{2}}\right)^3\)

\(=99-70\sqrt{2}+99+70\sqrt{2}+3\cdot\sqrt[3]{\left(99-70\sqrt{2}\right)\left(99+70\sqrt{2}\right)}\cdot A\)

\(=198+3A\)

\(\Leftrightarrow A^3-198-3A=0\)

\(\Leftrightarrow A^3-3A-198=0\)

\(\Leftrightarrow A^3-6A^2+6A^2-36A+33A-198=0\)

\(\Leftrightarrow A^2\left(A-6\right)+6A\left(A-6\right)+33\left(A-6\right)=0\)

\(\Leftrightarrow\left(A-6\right)\left(A^2+6A+33\right)=0\)

mà \(A^2+6A+33>0\forall A\)

nên A-6=0

hay A=6

Vậy: \(\sqrt[3]{99-70\sqrt{2}}+\sqrt[3]{99+70\sqrt{2}}=6\)

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 \(A=5\)

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

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

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