Đặt \(am^3=bn^3=cp^3=k\)

Ta có \(\sqrt[3]{k}=\sqrt[3]{a}m=\sqrt[3]{b}n=\sqrt[3]{c}p=\frac{\sqrt[3]{a}}{\frac{1}{m}}=\frac{\sqrt[3]{b}}{\frac{1}{n}}=\frac{\sqrt[3]{c}}{\frac{1}{p}}\)

\(=\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\frac{1}{m}+\frac{1}{n}+\frac{1}{p}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\) \(\left(TCDTSBN\right)\)\(\left(1\right)\)

Ta cũng có \(k=\frac{am^2}{\frac{1}{m}}=\frac{bn^2}{\frac{1}{n}}=\frac{cp^2}{\frac{1}{p}}=\frac{am^2+bn^2+cp^2}{\frac{1}{m}+\frac{1}{n}+\frac{1}{p}}=am^2+bn^2+cp^2\)  \(\left(TCDTSBN\right)\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{am^2+bn^2+cp^2}=\sqrt[3]{k}\)

cách khác nhé: 

Đặt:   \(am^3=bn^3=cp^3=k^3\)

\(\Rightarrow\)\(a=\frac{k^3}{m^3};\)\(b=\frac{k^3}{n^3};\)\(c=\frac{k^3}{p^3}\)

Ta có:

\(VT=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)

\(=\sqrt[3]{\frac{k^3}{m^3}}+\sqrt[3]{\frac{k^3}{n^3}}+\sqrt[3]{\frac{k^3}{p^3}}\)

\(=\frac{k}{m}+\frac{k}{n}+\frac{k}{p}=k\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)=k\)    (do 1/m + 1/n + 1/p = 1)

\(VP=\sqrt[3]{am^2+bn^2+cp^2}\)

\(=\sqrt[3]{\frac{k^3}{m^3}.m^2+\frac{k^3}{n^3}.n^2+\frac{k^3}{p^3}.p^2}\)

\(=\sqrt[3]{k^3\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)}=\sqrt[3]{k^3}=k\)   (do 1/m + 1/n + 1/p = 1)

suy ra:   \(VT=VP=k\) (đpcm)

chtt đi. tớ làm bài tương tự r 

chtt là cái j v?

\(A=\sqrt[3]{am^2+bn^2+cp^2}=\sqrt[3]{\frac{am^3}{m}+\frac{bn^3}{n}+\frac{cp^3}{p}}=\sqrt[3]{am^3\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)}\)
\(\Rightarrow A=\sqrt[3]{am^3}=m\sqrt[3]{a}\Rightarrow\frac{A}{m}=\sqrt[3]{a}\)
Tương tự :.. \(\Rightarrow A\left(\frac{1}{m}+\frac{1}{p}+\frac{1}{n}\right)=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)

Đặt   \(am^3=bn^3=cp^3=k^3\)

\(\Rightarrow\)\(a=\frac{k^3}{m^3};\) \(b=\frac{k^3}{n^3};\) \(c=\frac{k^3}{p^3}\)

Ta có:  \(VT=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)

                  \(=\sqrt[3]{\frac{k^3}{m^3}}+\sqrt[3]{\frac{k^3}{n^3}}+\sqrt[3]{\frac{k^3}{p^3}}\)

                  \(=\frac{k}{m}+\frac{k}{n}+\frac{k}{p}=k\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)=k\)

          \(VP=\sqrt[3]{am^2+bn^2+cp^2}\)

                 \(=\sqrt[3]{\frac{k^3}{m}+\frac{k^3}{n}+\frac{k^3}{p}}\)

                 \(=\sqrt[3]{k^3\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)}\)

                \(=\sqrt[3]{k^3}=k\)

suy ra: đpcm

bài này ở trong Sách nâng cao và phát triển toán 9 tập 1 của ông Vũ Hữu Bình ý

Áp dụng BĐT Holder ta có:

\(\left(am^2+bn^2+cp^2\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\)

\(\ge\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)^3\)

Suy ra ĐPCM 

Dấu "=" xảy ra khi \(am^3=bn^3=cp^3\)

À mình viết lộn đề câu 1, co mình sửa lại nhá!

 1) Tìm số nguyên n thỏa:

   \(\sqrt[3]{n+\sqrt{n^2+27}}+\sqrt[3]{n-\sqrt{n^2+27}}=4\)

Khi đó nếu bỏ chữ số tận cùng thì số mới là abc

Ta có:

abc3 - abc = (1000a + 100b + 10c + 3) - (100a + 10b + c)

                 => 900a + 90b + 9c + 3=1992

                 => 900a + 90b + 9c=1989

                 => 9(100a + 10b + c)=1989

                 => 100a + 10b + c = 221

                 => abc = 221

                 => abc3 = 2213

              Vậy số cần tìm là 2213

Bạn lên google mà tìm

\(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}=1\Leftrightarrow\)\(\frac{mn+np+mp}{mnp}=1\Leftrightarrow mn+np+mp=mnp\)

Ta có: \(am^3=bn^3=cp^3\Leftrightarrow\)\(\sqrt[3]{am^3}=\sqrt[3]{bn^3}=\sqrt[3]{cp^3}\)\(\Leftrightarrow\sqrt[3]{a}m=\sqrt[3]{b}n=\sqrt[3]{c}p\)

\(\frac{\sqrt[3]{a}m}{mnp}=\frac{\sqrt[3]{b}n}{mnp}=\frac{\sqrt[3]{c}p}{mnp}\Leftrightarrow\)\(\frac{\sqrt[3]{a}}{np}=\frac{\sqrt[3]{b}}{mp}=\frac{\sqrt[3]{c}}{mn}\Leftrightarrow\)\(\frac{\sqrt[3]{a}}{np}=\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{mn+np+mp}\Leftrightarrow\)\(\frac{\sqrt[3]{a}}{np}=\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{mnp}\Leftrightarrow\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{a}m\)

Mặt khác: \(am^3=bn^3=cp^3\Leftrightarrow\)\(\frac{am^3}{mnp}=\frac{bn^3}{mnp}=\frac{cp^3}{mnp}\Leftrightarrow\)\(\frac{am^2}{np}=\frac{bn^2}{mp}=\frac{cp^2}{mn}\Leftrightarrow\)

\(\frac{am^2}{np}=\frac{am^2+bn^2+cp^2}{mn+np+mp}=\frac{am^2+bn^2+cp^2}{mnp}\)\(\Leftrightarrow am^2+bn^2+cp^2=am^3\Leftrightarrow\sqrt[3]{am^2+bn^2+cp^2}=\sqrt[3]{a}m\)

Vậy =>dpcm

a. Đặt \(S_{AOB}=c^2;S_{BOC}=a^2;S_{COA}=b^2\Rightarrow S_{ABC}=a^2+b^2+c^2\)

Ta có \(\frac{AM}{OM}=\frac{S_{ABC}}{S_{BOC}}=\frac{a^2+b^2+c^2}{a^2}=1+\frac{b^2+c^2}{a^2}\)

Vậy thì \(\frac{OA}{OM}=\frac{AM}{OM}-1=\frac{b^2+c^2}{a^2}\Rightarrow\sqrt{\frac{OA}{OM}}=\sqrt{\frac{b^2+c^2}{a^2}}\ge\frac{1}{\sqrt{2}}\left(\frac{b}{a}+\frac{a}{b}\right)\)

Tương tự, ta có: \(\sqrt{\frac{OA}{OM}}+\sqrt{\frac{OB}{ON}}+\sqrt{\frac{OC}{OP}}\ge\frac{1}{\sqrt{2}}\left(\frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}\right)\ge\frac{1}{\sqrt{2}}.6=3\sqrt{2}\)

đặt \(am^3=bn^3=cp^3=k^3\)

\(\Rightarrow a=\dfrac{k^3}{m^3};b=\dfrac{k^3}{n^3};c=\dfrac{k^3}{p^3}\)

VT=\(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\dfrac{k}{m}+\dfrac{k}{n}+\dfrac{k}{p}=k\)

VF=\(\sqrt[3]{\dfrac{k^3}{m}+\dfrac{k^3}{n}+\dfrac{k^3}{p}}=\sqrt[3]{k^3}=k\)

do đó VT=VF, đẳng thức được chứng minh