`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`

a) (-14) + 5 + 17 + 14 = [(-14) + 14] + 5 + 17 = 22.

b) 40 +13 + (-25) + (-13) - 40 + (-25) + [l3 + (-13)] = 15.

c) (-5) + (-146) + (-15) +146 - (-5) + (-15)+[(-146) +146] = -20

d) (-2) + (-5) + 20 + (-13) = [(-2) + (-5) + (-13)] + 20 - 0

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

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

<=> \(B=2\sqrt{3}\)

a) 35-{12-[-14+(-2)]}.

= 35 - 12 + 14 + 2

= 35 + 14 - (12 - 2)

= 49 - 10

= 39

b) 160:{17+[9×5-(14+2048:256)]}.

= 160: {17 + [45 - (14 + 8)]}

= 160: {17 + [45 - 22)]}

= 160:{17 + 23}

= 160: 40 = 4

c) 2021⁰-{225:[20×15-8×25]-25}.

= 1 - {225:[300 - 200] - 25}

= 1 - {225 : 100 - 25}

= 1 - {2,25 - 25}

= 1 + 25 - 2,25

= 23,75

\(A=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{2^3+3.2^2.\sqrt{2}+3.2.\left(\sqrt{2}\right)^2+\left(\sqrt{2}\right)^3}+\sqrt[3]{2^3-3.2^2.\sqrt{2}+3.2.\left(\sqrt{2}\right)^2-\left(\sqrt{2}\right)^3}\)\(=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}=2+\sqrt{2}+2-\sqrt{2}=4.\)

\(A=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}=2+\sqrt{2}+2-\sqrt{2}=4\)