\(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{a+b+c}\) thì \(\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}=\sqrt[n]{a+b+c}\)với n lẻ
Chứng minh nếu \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{a+b+c}\) thì với mọi số nguyên dương lẻ n có \(\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}=\sqrt[n]{a+b+c}\)
Cho \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{a+b+c}\)và cả ba số a,b,c đều lẻ thì \(\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}=\sqrt[n]{a+b+c}\)
Giúp mình với!!!!!!!!!
Cho \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{a+b+c}\)
CMR với mọi n lẻ thì \(\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}=\sqrt[n]{a+b+c}\)
Nếu \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{a+b+c}\)
Chứng minh rằng : Với mọi số nguyên dương lẻ n , ta đều có : \(\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}=\sqrt[n]{a+b+c}\)
\(\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)^3=a+b+c\)
\(\Leftrightarrow a+b+c+3\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\left(\sqrt[3]{b}+\sqrt[3]{c}\right)\left(\sqrt[3]{a}+\sqrt[3]{c}\right)=a+b+c\)
\(\Leftrightarrow3\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\left(\sqrt[3]{b}+\sqrt[3]{c}\right)\left(\sqrt[3]{a}+\sqrt[3]{c}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\b+c=0\\c+a=0\end{matrix}\right.\)
+Neu a+b =0 => \(\sqrt[n]{a}+\sqrt[n]{b}=0\)( n : le)=> \(VT=VP=\sqrt[n]{c}\)(dpcm)
Tuong tu cac TH
=> KL
Chứng minh rằng nếu \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{a+b+c}\) thì \(\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}=\sqrt[n]{a+b+c}\) với n là số nguyên dương lẻ.
đặt \(\sqrt[3]{a}=x;\sqrt[3]{b}=y;\sqrt[3]{c}=z\)
\(\rightarrow x+y+z=\sqrt[3]{x^3+y^3+z^3}\)
\(\left(x+y+z\right)^3=x^3+y^3+z^3\)
\(\left(x+y\right)\left(z+y\right)\left(x+z\right)=0\)
luôn tồn tại 2 số đối nhau => a,b,c luôn có 2 số đối nhau
mặt khác do n là số lẻ nên \(\sqrt[n]{}\) của 2 số cũng đối nhau
nên \(\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}=\sqrt[n]{a+b+c}\)
CMR nếu \(\sqrt[3]{a}\)+ \(\sqrt[3]{b}\)+\(\sqrt[3]{c}\)= \(\sqrt[3]{a+b+c}\)thì với mọi n lẻ ta có : \(\sqrt[n]{a}\)+\(\sqrt[n]{b}\)+\(\sqrt[n]{c}\)
\(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{a+b+c}\)\(\Leftrightarrow\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)^3=a+b+c\Leftrightarrow a+b+c+3.\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\left(\sqrt[3]{b}+\sqrt[3]{c}\right)\left(\sqrt[3]{c}+\sqrt[3]{a}\right)=a+b+c\)
\(\Rightarrow\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\left(\sqrt[3]{b}+\sqrt[3]{c}\right)\left(\sqrt[3]{c}+\sqrt[3]{a}\right)=0\)
CMR nếu \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{a+b+c}\)
thì mọi số nguyên dương lẻ n ta có: \(\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}=\sqrt[n]{a+b+c}\)
1. Rút gọn biểu thức
\(\frac{a^{-n}+b^{-n}}{a^{-n}-b^{-n}}-\frac{a^{-n}-b^{-n}}{a^{-n}+b^{-n}}\)
2. Tính các biểu thức
a. \(\sqrt{a\sqrt{a\sqrt{a\sqrt{a}}}}:a^{\frac{11}{16}}\)
b. \(\sqrt[3]{a\sqrt{a^3.\sqrt{a}}}:a^{\frac{1}{2}}\)
c. \(\sqrt[5]{\frac{b}{a}.\sqrt[3]{\frac{a}{b}}}\)
d.\(\frac{6^{3+\sqrt{5}}}{2^{2+\sqrt{5}}.3^{1+\sqrt{5}}}\)
Help me pleases ,thanks before
1) c/m \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\) với mọi số nguyên dương n
2)cho A=\(\frac{\sqrt{2}-\sqrt{1}}{1+2}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+\frac{\sqrt{4}-\sqrt{3}}{3+4}+....+\frac{\sqrt{25}-\sqrt{24}}{24+25}\)
C/m \(A< \frac{2}{5}\)
3)Cho 3 số a,b,c dương,c/m
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2\)
Bài 1:
Có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Có: \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}\)
xong bn áp dụng lên trên lm tiếp
Bài 3:
theo bđt cô si ta có:
\(\sqrt{\frac{b+c}{a}\cdot1}\le\left(\frac{b+c}{a}+1\right):2=\frac{b+c+a}{2a}\)
=> \(\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\) (1)
Tương tự ta có :
\(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\) (2)
\(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\) (3)
Cộng vế vs vế (1)(2)(3) ta có:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2a+2b+2c}{a+b+c}=2\)
Bài 2:
Ta có:
\(\frac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Nên:
\(A< \frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\right)=\frac{1}{2}\left(1-\frac{1}{5}\right)=\frac{2}{5}\)