Đặt \(x=\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}\)

\(\Rightarrow x^3=40+3\sqrt[3]{\left(20+14\sqrt[]{2}\right)\left(20-14\sqrt[]{2}\right)}.\left(\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}\right)\)

\(\Rightarrow x^3=40+6x\)

\(\Rightarrow x^3-6x-40=0\)

\(\Rightarrow\left(x-4\right)\left(x^2+4x+10\right)=0\)

\(\Rightarrow x=4\)

Vậy \(\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}=4\)

em cảm ơn ạ

`x=root{3}{14sqrt2+20}+sqrt{-14sqrt2+20}`

`<=>x^3=14sqrt2+20-14sqrt2+20+3root{3}{(14sqrt2+20)(20-14sqrt2)}(root{3}{14sqrt2+20}+sqrt{-14sqrt2+20})`

`<=>x^3=40+3root{3}{400-392}.x`

`<=>x^3=40+6x`

`<=>x^3-6x=40`

\(x=\sqrt[3]{30+14\sqrt{2}}-\sqrt[3]{20+14\sqrt{2}}\)

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

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

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

\(=4\)

Vậy x = 4.

a)\(A=^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\)

=>  \(A^3=\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)^3\)

\(=20+14\sqrt{2}+20-14\sqrt{2}\)

\(+3\left(\text{​​}^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\right)\left(^3\sqrt{20+14\sqrt{2}}.^3\sqrt{20-14\sqrt{2}}\right)\)

\(=40+3A.^3\sqrt{\left(20+14\sqrt{2}\right)\left(20+14\sqrt{2}\right)}\)

\(\Rightarrow A^3=40+3.A.2\)

=> \(A^3-6A-40=0\)

<=> \(A^3-16A+10A-40=0\)

<=> \(A\left(A-4\right)\left(A+4\right)+10\left(A-4\right)=0\)

<=> \(\left(A-4\right)\left(A^2+4A+10\right)=0\)

<=> A = 4 ( vì \(A^2+4A+10=\left(A+2\right)^2+6>0\))

Vậy A = 4.

b/ \(B=^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\)

=> \(B^3=\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right)^3\)

\(=26+15\sqrt{3}-26+15\sqrt{3}\)

\(-3\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right).^3\sqrt{26+15\sqrt{3}}.^3\sqrt{26-15\sqrt{3}}\)

\(=30\sqrt{3}-3B.1\)

=> \(B^3+3B-30\sqrt{3}=0\)

<=> \(B^3-12B+15B-30\sqrt{3}=0\)

<=> \(B\left(B-2\sqrt{3}\right)\left(B+2\sqrt{3}\right)+15\left(B-2\sqrt{3}\right)=0\)

<=> \(\left(B-2\sqrt{3}\right)\left(B^2+2\sqrt{3}B+15\right)=0\)

<=> \(B-2\sqrt{3}=0\)( vì \(B^2+2\sqrt{3}B+15=\left(B+\sqrt{3}\right)^2+12>0\))

<=> \(B=2\sqrt{3}\)

Áp dụng:   \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

\(x=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)

=>  \(x^3=\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)^3\)

           \(=20+14\sqrt{2}+20-14\sqrt{2}+3\sqrt[3]{\left(20+14\sqrt{2}\right)\left(20-14\sqrt{2}\right)}\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)\)

\(=40+6x\)

=>  \(x^3-6x=40\)

ta có \(x^3=\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)^3\)\(=20+14\sqrt{2}+3\sqrt[3]{\left(20+14\sqrt{2}\right)^2}.\sqrt[3]{20-14\sqrt{2}}+20-14\sqrt{2}\)\(+3\sqrt[3]{20+14\sqrt{2}}.\sqrt[3]{\left(20-14\sqrt{2}\right)^2}=\)\(40+3\sqrt[3]{\left(20+14\sqrt{2}\right)\left(20-14\sqrt{2}\right)}\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)\)

\(=40+3\sqrt[3]{20^2-14\sqrt{2}^2}.x\)x này là đề bài cho nên thay vào nha bạn

\(=40+3.2.x\)\(hay\)\(x^3=6x+40\Leftrightarrow x^3-6x=40\)(đây là kết quả cần tìm)

a, c.Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath