Ta có \(x^2-y^2-z^2=0\Rightarrow z^2=x^2-y^2\)
Có \(VT=\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-\left(4z\right)^2\)\(=\left(5x-3y\right)^2-16z^2=\left(5x-3y\right)^2-16\left(x^2-y^2\right)\)
\(=25x^2-30xy+9y^2-16x^2+16y^2=9x^2-30xy+25y^2\)
\(=\left(3x\right)^2-2.3x.5y+\left(5y\right)^2=\left(3x-5y\right)^2=VP\left(đpcm\right)\)
Cho \(x^2-y^2-z^2=0\)
Chứng minh rằng: \(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
a Rút gọn biểu thức \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)+...+\left(2^{256}+1\right)+1\)
b. Nếu \(x^2=y^2+z^2\). Cmr: \(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
CMR : Nếu x^2 - y^2 - z^2 = 0 thì ( 5x-3y+4z ) . ( 5x-3y - 4z ) = ( 3x - 5y )^2
cho x^2-y^2-z^2=0 chứng minh rằng: (5x-3y+4z)*(5x-3y-4z)=(3x-5y)^2
Cho x2-y2-z2=0
C/m (5x-3y+4z).(5x-3y-4z)=(3x-5y)2
cho x2-y2-z2=0
chúng minh : (5x-3y+4z)(5x-3y-4z)=(3x-5y)2
CMR vs mọi số nguyên x,y thì
a.\(\left(2x^2+3y\right)⋮17\Leftrightarrow\left(9x^2+5y\right)⋮17\)
b.\(\left(5x^2-4y\right)⋮23\Leftrightarrow\left(3x^2-7y\right)⋮23\)
Cho x^2 -y ^2=4z^2 . CMR: (5x-3y+8z)(5x-3y-8z)=(3x-5y)^2
giả sử x,y,z là các số thực dương thỏa mãn x+y+z=xyz. cmr \(\frac{x}{1+x^2}+\frac{18y}{1+y^2}+\frac{4z}{1+z^2}\)=\(\frac{xyz\left(22x+5y+19z\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
