cau a,b,c thay no co chung 1 dang do la

\(\sqrt[3]{a+m}+\sqrt[3]{a-m}\)

dang nay co 2 cach

C1: nhanh kho nhin de sai

VD: cau B

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

B^3=40+3(2)(B)

B^3=40+6B

B=4

C2: hoi dai nhung de nhin

dat \(a=\sqrt[3]{20+14\sqrt{2}};b=\sqrt[3]{20-14\sqrt{2}}\)

de thay B=a+b

            ab=2

            a^3+b^3=40

suy ra B^3=a^3+b^3+3ab(a+b)

B^3=40+6B

B=4

giai tuong tu

con co cach nay nhung it su dung vi kho tim

C3: dua ve tong lap phuong

VD:cau B

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

\(20-14\sqrt{2}=\left(2-\sqrt{2}\right)^3\)

de thay

B=4

cau d)

dung CT nay

\(\sqrt[m]{a}=\sqrt[m\cdot n]{\left(a\right)^n}\)

ap dung vao bai

\(\sqrt[3]{2\sqrt{3}-4\sqrt{2}}=\sqrt[6]{\left(2\sqrt{3}-4\sqrt{2}\right)^2}=\sqrt[6]{44-16\sqrt{6}}\)

nhanh vao

\(\sqrt[6]{\left(44-16\sqrt{6}\right)\left(44+16\sqrt{6}\right)}=\sqrt[6]{400}=\sqrt[3]{20}\)

(14,78-a)/(2,87+a)=4/1

14,78+2,87=17,65

Tổng số phần bằng nhau là 4+1=5

Mỗi phần có giá trị bằng 17,65/5=3,53

=>2,87+a=3,53

=>a=0,66.

Bài 1: 

Ta có:

\(\left(a-b+c\right)^3=a^3-b^3+c^3-3a^2b+3a^2c+3ab^2+3b^2c+3ac^2-3bc^2-6abc\)

\(\Rightarrow\left(\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\right)^3=\frac{1}{9}-\frac{2}{9}+\frac{4}{9}-\frac{1}{3}.\sqrt[3]{2}+\frac{1}{3}.\sqrt[3]{4}+\frac{1}{3}.\sqrt[3]{4}+\frac{2}{3}.\sqrt[3]{2}\)

\(+\frac{2}{3}.\sqrt[3]{2}-\frac{2}{3}.\sqrt[3]{4}-\frac{4}{3}=\sqrt[3]{2}-1\)

\(\Rightarrow\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\)

Hằng đẳng thức ???

Áp dụng BĐT \(x^2+y^2\ge2xy\) ta có:

\(\frac{x^4+y^4}{2}\ge\frac{\left(x^2\right)^2+\left(y^2\right)^2}{2}\ge\frac{2x^2y^2}{2}=x^2y^2\)

Tương tự cho 2 BĐT còn lại cũng có;

\(\frac{y^4+z^4}{2}\ge y^2z^2;\frac{z^4+x^4}{2}\ge x^2z^2\)

Cộng theo vế 3 BĐT trên ta có:

\(VT=\frac{x^4+y^4}{2}+\frac{y^4+z^4}{2}+\frac{z^4+x^4}{2}\ge x^2y^2+y^2z^2+z^2x^2=VP\)

Khi \(x=y=z\)

Áp dụng bđt Cô si cho 2 số không âm, ta có:

\(\hept{\begin{cases}\frac{x^4+y^4}{2}\ge\sqrt{x^4y^4}=x^2y^2\\\frac{y^4+z^4}{2}\ge\sqrt{y^4z^4}=y^2z^2\\\frac{z^4+x^4}{2}\ge\sqrt{z^4x^4}=z^2x^2\end{cases}}\)

\(\Rightarrow\frac{x^4+y^4}{2}+\frac{y^4+z^4}{2}+\frac{z^4+x^4}{2}\ge x^2y^2+y^2z^2+z^2x^2\)

Chuẩn hóa \(a+b+c=3\) thì cần c/m

\(\sqrt{\frac{a}{3-a}}+\sqrt{\frac{b}{3-b}}+\sqrt{\frac{c}{3-c}}\ge\frac{3\sqrt{2}}{2}\)

Ta có BĐT phụ \(\sqrt{\frac{a}{3-a}}\ge\frac{3\sqrt{2}}{8}a+\frac{\sqrt{2}}{8}\)

\(\Leftrightarrow\frac{\frac{3\left(a-1\right)^2\left(3a-1\right)}{32\left(3-a\right)}}{\sqrt{\frac{a}{3-a}}+\frac{3\sqrt{2}}{8}a+\frac{\sqrt{2}}{8}}\ge0\forall0< a< 3\)

Tương tự cho 2 BĐT còn lại cũng có:

\(\sqrt{\frac{b}{3-b}}\ge\frac{3\sqrt{2}}{8}b+\frac{\sqrt{2}}{8};\sqrt{\frac{c}{3-c}}\ge\frac{3\sqrt{2}}{8}c+\frac{\sqrt{2}}{8}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge\frac{3\sqrt{2}}{8}\left(a+b+c\right)+\frac{\sqrt{2}}{8}\cdot3=\frac{3\sqrt{2}}{2}\)

Vì sao a + b + c = 3 vậy bạn?