ngu ngưoi viet cai de cung sai

Ta có: \(\frac{1}{x}+\frac{2}{y}=2\ge2\sqrt{\frac{2}{xy}}\Leftrightarrow\sqrt{\frac{2}{xy}}\le1\Leftrightarrow xy\ge2\)

\(5x^2+y-4xy+y^2=\left(2x-y\right)^2+x^2+y\)

\(\ge x^2+y=x^2+\frac{y}{2}+\frac{y}{2}\ge3\sqrt[3]{\frac{\left(xy\right)^2}{4}}\ge3\left(đpcm\right)\)

Dấu "="\(\Leftrightarrow x=1,y=2\)

Ta có \(\frac{n\left(2n-1\right)}{26}=k^2\Leftrightarrow2n^2-n-26k^2=0\)

\(\Delta=208k^2+1=t^2\)(vì n nguyên dương)

\(\Rightarrow\left(t+4\sqrt{13}k\right)\left(t-4\sqrt{13}k\right)=1\)

\(\Leftrightarrow\hept{\begin{cases}t+4\sqrt{13}k=1\\t-4\sqrt{13}k=1\end{cases}\Leftrightarrow\hept{\begin{cases}k=0\\t=1\end{cases}}}\)

Thế vào tìm được \(\orbr{\begin{cases}n=0\\n=\frac{1}{2}\end{cases}}\)

Vậy không có giá trị n nguyên dương nào thỏa mãn cái đó

\(\frac{n\left(2n-1\right)}{26}\text{ là SCP }\Leftrightarrow n\left(2n-1\right)=26k^2\)

\(\Delta_n=208k^2+1=y^2\Leftrightarrow y^2-208k^2=1\underrightarrow{\text{PELL}}\)

\(k=\pm\frac{\left(649-180\sqrt{13}\right)^m-\left(649+180\sqrt{13}\right)^m}{8\sqrt{13}}\)

\(n=\frac{1}{8}\left[-\left(649-180\sqrt{13}\right)^m-\left(649+180\sqrt{13}\right)^m+2\right]\left(m\inℤ,m\ge0\right)\)

\(3\left(x^2+y^2\right)-\left(x^3+y^3\right)+1\)

\(=3x^2+3y^2-\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=3x^2+3y^2-2\left(x^2-xy+y^2\right)\)

\(=3x^2+3y^2-2x^2+2xy-2y^2\)

\(=x^2+2xy+y^2=\left(x+y\right)^2=2^2=4\)

sai rồi bạn . Mình nhập vào nó hiện lên sai

a = 1
b = 2
=> a³ + b³ = 1³ + 2³ = 1 + 8 = 9
=> a + b = 1 + 2 = 3

a = 1

b = 2

\(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^2=0\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)

\(\Rightarrow1+2\left(ab+bc+ac\right)=0\)

\(\Rightarrow ab+bc+ac=-\frac{1}{2}\)

\(\Rightarrow\left(ab+bc+ac\right)^2=\frac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2=\frac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2ab\left(a+b+c\right)=\frac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2ab.0=\frac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=\frac{1}{4}\)

Có:  \(a^2+b^2+c^2=1\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=1\)

\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=1\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=1\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=1\)

\(\Rightarrow a^4+b^4+c^4+2.\frac{1}{4}=1\)

\(\Rightarrow a^4+b^4+c^4+\frac{1}{2}=1\)

\(\Rightarrow a^4+b^4+c^4=\frac{1}{2}\)

Mình làm kĩ nên hơi dài  :)