Cho \(^{x^2-y^2-z^2=0.CMR:\left(5X-3Y+4Z\right)\left(5Z-3Y-4Z\right)=\left(3X-5Y\right)^2}\)
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\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)^2-\left(4z\right)^2=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)-16z^2-\left(3x-5y\right)^2=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-\left(9x^2-30xy+25y^2\right)=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-9x^2+30xy-25y^2=0\)
\(\Rightarrow25\left(x^2-y^2\right)+9\left(x^2-y^2\right)-16z^2=0\)
\(\Rightarrow34\left(x^2-y^2\right)-16z^2=0\)
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a) Đề sai nha bạn :) mấy dấu cộng bạn phỉa chuyển thành dấu nhân nhé
\(A=\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)
\(A=2^{512}-1+1\)
\(A=2^{512}\)
b . ( 5x - 3y + 4z )( 5x - 3y - 4z ) = ( 5x - 3y )^2 - ( 4z )^2 = 25x^2 - 30xy + 9y^2 - 16z^2 = 25( y^2 + z^2 ) - 30xy + 9y^2 - 16z^2 = 9z^2 + 34y^2 - 30xy ( 1 )
( 3x - 5y )^2 = 9x^2 - 30xy + 25y^2 = 9( y^2 + z^2 ) - 30xy + 25y^2 = 34y^2 + 9z^2 - 30xy ( 2 )
Tu ( 1 ) va ( 2 ) => dpcm
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4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)
=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2
=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc
Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0
=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm
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\(\hept{\begin{cases}\left|x^2+y^2+z^2-1\right|=0\\\left(3y-4z\right)^4\ge0\\\left(3x-2y\right)^2\ge0\end{cases}}\Rightarrow\left|x^2+y^2+z^2-1\right|+\left(3y-4z\right)^4+\left(3x-2y\right)^2\ge0\)
dấu = xảy ra khi \(\hept{\begin{cases}\left|x^2+y^2+z^2-1\right|=0\\\left(3y-4z\right)^4=0\\\left(3x-2y\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x^2+y^2+z^2=1\\3y=4z\\3x-2y=0\end{cases}}\Rightarrow\hept{\begin{cases}x^2+y^2+z^2=1\\y=\frac{4z}{3}\\x=\frac{2y}{3}\end{cases}}\)
Vậy ...
p/s bài này chắc chỉ có dạng chung thôi bn :)
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Vì \(x^2-y^2-z^2=0\Rightarrow x^2-y^2=z^2\)
Biến đổi vế trái ta có :
\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-16z^2\)
\(=25x^2-30xy+9y^2-16\left(x^2-y^2\right)\)
\(=25x^2-30xy+9y^2-16x^2+16y^2\)
\(=9x^2-30xy+25y^2\)
\(=\left(3x-5y\right)^2\) ( ĐPCM)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(x^2-y^2-z^2=0\Rightarrow z^2=x^2-y^2\)
Có \(VT=\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-\left(4z\right)^2\)\(=\left(5x-3y\right)^2-16z^2=\left(5x-3y\right)^2-16\left(x^2-y^2\right)\)
\(=25x^2-30xy+9y^2-16x^2+16y^2=9x^2-30xy+25y^2\)
\(=\left(3x\right)^2-2.3x.5y+\left(5y\right)^2=\left(3x-5y\right)^2=VP\left(đpcm\right)\)