Nháp:

\(9\pm\sqrt{80}=9\pm4\sqrt{5}=\dfrac{72\pm32\sqrt{5}}{8}=\left(\dfrac{3\pm\sqrt{5}}{2}\right)^3\)

\(\Rightarrow \sqrt[3]{9+\sqrt{80}}=\dfrac{3+\sqrt{5}}{2}\); \(\Rightarrow \sqrt[3]{9-\sqrt{80}}=\dfrac{3-\sqrt{5}}{2}\)

\(S=\sqrt[3]{26+15\sqrt{3}}\left(2-\sqrt{3}\right)+\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\\ S=\sqrt[3]{\left(2+\sqrt{3}\right)^3}+\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\\ S=2+\sqrt{3}+\dfrac{3+\sqrt{5}}{2}+\dfrac{3-\sqrt{5}}{2}\\ S=2+\sqrt{3}+3\\ S=5+\sqrt{3}\)

3x-7^2=80

3x=80/49

x=80/147

 k nhé

\(x=\dfrac{\sqrt[3]{\left(2+\sqrt{3}\right)^3}\left(2-\sqrt{3}\right)}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}=\dfrac{1}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}\)

Đặt \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)\(\Leftrightarrow A^3=18+3\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow A^3=18+3A\sqrt[3]{1}\\ \Leftrightarrow A^3-3A-18=0\\ \Leftrightarrow A=3\\ \Leftrightarrow X=\dfrac{1}{3}\\ \Leftrightarrow Q=\left[3\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^2-1\right]^{2021}=\left(\dfrac{1}{9}-\dfrac{1}{9}-1\right)^{2021}=\left(-1\right)^{2021}=-1\)

Đặt A = \(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\)=> \(A^3=18+3A\Leftrightarrow A^3-3A-18=0\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\Leftrightarrow A-3=0\Leftrightarrow A=3\)

\(\dfrac{\sqrt[3]{26+15\sqrt{3}}\left(2-\sqrt{3}\right)}{\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}}=\dfrac{\sqrt[3]{\left(2+\sqrt{3}\right)^3}\left(2-\sqrt{3}\right)}{3}=\dfrac{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{3}=\dfrac{1}{3}\)

huhu help mik

ko có phân số nào

a) (1,1x-9).2/5=80%=4/5

1,1x-9=4/5:2/5=4/5*5/2=2

1,1x=2+9

1,1x=11

x=11/1,1

x=10

 (1,1x-9).2/5=80%=4/5

1,1x-9=4/5:2/5=4/5*5/2=2

1,1x=2+9

1,1x=11

x=11/1,1

x=10

Ta có \(\sqrt[3]{26+15\sqrt{3}}=\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}\)

\(=\sqrt[3]{2^3+3.2^2\sqrt{3}+3.2.\left(\sqrt{3}\right)^2+\left(\sqrt{3}\right)^3}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}\)

\(=2+\sqrt{3}\)

Đặt \(x=\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\)

Ta có \(x^3=\left(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\right)^3\)

\(=9+\sqrt{80}+9-\sqrt{80}+3.\left(\sqrt[3]{9+\sqrt{80}}\right)^2\left(\sqrt[3]{9-\sqrt{80}}\right)+3.\left(\sqrt[3]{9-\sqrt{80}}\right)^2\left(\sqrt[3]{9+\sqrt{80}}\right)\)

\(=18+3\sqrt[3]{9+\sqrt{80}}.\sqrt[3]{9-\sqrt{80}}\left(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\right)\)

\(=18+3\sqrt[3]{9^2-80}.x\)

\(=18+3x\)

Vậy \(x^3=18+3x\)

\(\Leftrightarrow x^3-3x-18=0\)

Vậy x = 3

Do đó \(M=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+3=2^2-3+3=4\)

Vậy M = 4.

`#Neoo`

`A=9 : 0,25 - 6,4 : 2/5 + 80`

`A=(9:0,25)-(6,4:2/5)+80`

`A=36-16+80`

`A=20+80`

`A=100`

A = 9 : 0,25 - 6,4 : 2/5 + 80

= ( 9 : 0,25 ) - ( 6,4 : 2/5 ) + 80

= 36 - 16 + 80

= 20 + 80

= 100.

A=36-16+80

=20+80=100