Ta có: \(2014a=2016-3a-a^3\)

\(S=\sqrt[3]{3a^2+2014-2015}+\sqrt[3]{3a^2-2014a+2017}\)

\(=\sqrt[3]{3a^2+2016-3a-a^3-2015}+\sqrt[3]{3a^2-2016+3a+a^3+2017}\)

\(=\sqrt[3]{-a^3+3a^2-3a+1}+\sqrt[3]{a^3+3a^2+3a+1}\)

\(=\sqrt[3]{\left(1-a\right)^3}+\sqrt[3]{\left(1+a\right)^3}=1-a+1+a=2\)

sáng em đăng rồi mới giải xong hihi nhưng e vẫn sẽ k

Dat bieu thuc tren la A

ta co  \(\frac{1}{\sqrt{n+2}+\sqrt{n}}=\frac{\sqrt{n+2}-\sqrt{n}}{2}\)

ap dung dang thuc tren ta co\(\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}\)

                        tuong tu ta co \(\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{2}\)

                                              .........

                                            \(\frac{1}{\sqrt{2017}+\sqrt{2015}}=\frac{\sqrt{2017}-\sqrt{2015}}{2}\)

ta co

\(A=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+.....+\sqrt{2017}-\sqrt{2015}\right)=\frac{\sqrt{2017}-1}{2}\)

\(x=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\Rightarrow x^3=5+2\sqrt{13}+5-2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}.x\)

          \(=10+3x\sqrt[3]{25-52}\)

          \(=10+3x\sqrt[3]{-27}\)

           \(=10-9x\)

\(\Rightarrow x^3+9x-10=0\)

\(\Leftrightarrow x^3-x+10x-10=0\)

\(\Leftrightarrow x\left(x^2-1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+10\right)=0\)

Vì \(x^2+x+10=\left(x+\frac{1}{2}\right)^2+\frac{39}{4}>0\forall x\)

=> x - 1 = 0

=> x = 1

Thay vào A = 1²⁰¹⁵ - 1²⁰¹⁶ = 0

Vậy A = 0

Từ Từ đã nha!!

\(\text{Ta có: }x=\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}=\frac{3-\sqrt{5}}{\sqrt{9-5}}=\frac{3-\sqrt{5}}{2}.\)

\(A=x^5-6x^4+12x^3-4x^2-13x+2020\)

\(=\left(x^5-3x^4+x^3\right)-\left(3x^4-9x^3+3x^2\right)+\left(2x^3-6x^2+2x\right)+\left(5x^2-15x+5\right)+2015\)

\(=x^3\left(x^2-3x+1\right)-3x^2\left(x^2-3x+1\right)+2x\left(x^2-3x+1\right)+5\left(x^2-3x+1\right)+2015\)

\(=\left(x^2-3x+1\right)\left(x^3-3x^2+2x+5\right)+2015\)

Thay x vào A ta có: 

\(A=\left[\left(\frac{3-\sqrt{5}}{2}\right)^2-3.\frac{3-\sqrt{5}}{2}+1\right]\left(.....\right)+2015\)

\(=\left(\frac{14-6\sqrt{5}}{4}-\frac{9-3\sqrt{5}}{2}+1\right)\left(....\right)+2015\)

\(=0\cdot\left(......\right)+2015=2015\)

Vậy.....

x+√(x^2+3)=3/(y+√(y^3))=3(y-√(y^2+3)/-a(trục căn thức)

x+√(x^2+3)=-y+√(y^2+3) suy ra x+y=√(y^2+3)-√(x^2+3)(1)

Tương tự,x+y=√(x^2+3)-√(y^2+3)(2)

Cộng (1),(2) theo vế suy ra 2(x+y)=0 suy ra x+y=0

hay E=0.

Vậy E=0

nhân \(-x+\sqrt{x^2+3}\)  vào 2 vế ta đc : \(\left(-x^2+x^2+3\right)\left(y+\sqrt{y^2+3}\right)=\)\(3\left(-x+\sqrt{x^2+3}\right)\)
                         <=>  \(y+\sqrt{y^2+3}=-x+\sqrt{x^2+3}\)<=> \(y+\sqrt{y^2+3}+x-\sqrt{x^2+3}=0\)__(1)___
làm tương tự ta đc \(\left(-y+\sqrt{y^2+3}\right)\left(x+\sqrt{x^2+3}\right)\)\(=3\left(-y+\sqrt{y^2+3}\right)\)
                          <=> \(x+\sqrt{x^2+3}=-y+\sqrt{y^2+3}\)<=> \(x+\sqrt{x^2+3}+y-\sqrt{y^2+3}=0\)__(2)__
       lấy (1) + (2) => 2(x+y) =0 => x+y=0        
   lấy 

1)

dat \(a=\sqrt[3]{x+1};b=\sqrt[3]{7-x}\)

ta co b=2-a

a^3+b^3=x+1+7-x=8 

a^3+b^3=a^3+b^3+3ab(a+b)

ab(a+b)=0

suy ra a=0 hoac b=0 hoac a=-b

<=> x=-1; x=7 

a=-b

a^3=-b^3

x+1=x+7 (vo li nen vo nghiem)

cau B tuong tu

2)

tat ca cac bai tap deu chung 1 dang do la

\(\sqrt[3]{a+m}+\sqrt[3]{b-m}\)voi m la tham so

dang nay co 2 cach 

C1 lap phuong VD: \(B^3=10+3\sqrt[3]{< 5+2\sqrt{13}>< 5-2\sqrt{13}>}\left(B\right)\)

B^3=10-9B

B=1 cach nay nhanh nhung kho nhin

C2 dat an

\(a=\sqrt[3]{5+2\sqrt{13}};b=\sqrt[3]{5-2\sqrt{13}}\)

de thay B=a+b

a^3+b^3=10

ab=-3

B^3=10-9B

suy ra B=1

tuong tu giai cac cau con lai.

Bài 1:

a. Đặt \(a=\sqrt[3]{x+1}\); \(b=\sqrt[3]{7-x}\). Ta có:

\(\hept{\begin{cases}a+b=2\\a^3+b^3=8\end{cases}\Leftrightarrow a^3+\left(2-a\right)^3=8\Leftrightarrow...\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}a=0\\b=2\end{cases}}\)hoặc \(\hept{\begin{cases}a=2\\b=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt[3]{x+1}=0\\\sqrt[3]{7-x}=2\end{cases}}\)hoặc \(\hept{\begin{cases}\sqrt[3]{x+1}=2\\\sqrt[3]{7-x}=0\end{cases}}\)

\(\Leftrightarrow x=-1\)hoặc \(x=7\)

ai nay dung kinh nghiem la chinh

cau a)

ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)

\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)

khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)

\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)

\(x=\frac{3-1}{1}=2\)

suy ra 

x^3-4x+1=1

A=1^2018

A=1

b)

ta thay

\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)

khi do 

\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)

\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)

x=2

thay vao

x^3+3x-14=0

B=0^2018

B=0