Question

1 . rút gọn 3(2^2+1)(2^4+1)(2^8+1)(2^32+1)(2^64+1)(2^128+1)

2 . Cho a+b+c=0 và a^2+b^2+c^2=10. Tính 1/a^4+b^4+c^4

3 . giải phương trình x^2+1/x^2+y^2+1/y^2=4

Answer 1

1/ \(3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{128}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{128}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{128}+1\right)\)

..................................................................

\(=\left(2^{128}-1\right)\left(2^{128}+1\right)=2^{256}-1\)

Answer 2

2/ Ta có: \(a+b+c=0\Leftrightarrow a+b=-c\)

\(\Leftrightarrow a^2+2ab+b^2=c^2\Leftrightarrow a^2+b^2-c^2=-2ab\)

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2=4a^2b^2\)

\(\Leftrightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\)

Ta lại có: \(a^2+b^2+c^2=10\)

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=100\)

\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=100\Leftrightarrow a^4+b^4+c^4=50\)

\(\Leftrightarrow\frac{1}{a^4+b^4+c^4}=\frac{1}{50}\)

Answer 3

3/ \(x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}=4\)

\(\Leftrightarrow\left(x^2-2+\frac{1}{x^2}\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)

\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)

\(\Leftrightarrow\left\{\begin{matrix}x=\frac{1}{x}\\y=\frac{1}{y}\end{matrix}\right.\)

\(\Leftrightarrow\left(x,y\right)=\left(-1,-1;-1,1;1,-1;1,1\right)\)