\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)

\(\Rightarrow\left(5x-3y\right)^2-\left(4z\right)^2=\left(3x-5y\right)^2\)

\(\Rightarrow\left(5x-3y\right)-16z^2-\left(3x-5y\right)^2=0\)

\(\Rightarrow25x^2-30xy+9y^2-16z^2-\left(9x^2-30xy+25y^2\right)=0\)

\(\Rightarrow25x^2-30xy+9y^2-16z^2-9x^2+30xy-25y^2=0\)

\(\Rightarrow25\left(x^2-y^2\right)+9\left(x^2-y^2\right)-16z^2=0\)

\(\Rightarrow34\left(x^2-y^2\right)-16z^2=0\)

câu o0o trả lời là sai

\(Đây\)\(mới\)\(là\)\(câu\)\(trả\)\(lời\)\(đúng\)

\(ta\)\(có\)\(16\left(x^2-y^2-z^2\right)=16\left(x^2-y^2\right)-16z^2=8\left(x-y\right)2\left(x-y\right)-\left(4z\right)^2=\left(8x-8y\right)\left(2x+2y\right)-\left(4z\right)^2=\left(5x-3y+3x-5y\right)\left(5x-3y-3x+5y\right)-\left(4z\right)^2\)

\(=\left(5x-3y\right)^2-\left(3x-5y\right)^2-16z^2\)

\(\Leftrightarrow\left(5x-3y\right)^2-\left(4z\right)^2=\left(3x-5y\right)^2\)

\(\Leftrightarrow\left(5x-3y-4z\right)\left(5x-3y+4z\right)=\left(3x-5y\right)^2\)

Ta có:

\(x^2-y^2-z^2=0\)

\(16x^2-16y^2-16z^2=0\)

\(25x^2-9x^2+9y^2-25y^2-16z^2+30xy-30xy=0\)

\(\left(5x-3y\right)^2-16z^2= \left(3x-5y\right)^2\)

\(\left(5x-3y-4z\right)\left(5x-3y+4z\right)=\left(3x-5y\right)^2\)

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

cho mình hỏi câu a bạn kia giải sao (2+1) tách ra (2-1)(2+1) được

\(\left(3x-2y\right)^2+\left(3y-4z\right)^4+\left(x^2+y^2+z^2-1\right)=0\)

Vì \(\left(3x-2y\right)^2\ge0;\left(3y-4x\right)^4\ge0\)

\(\Rightarrow VT=0\Leftrightarrow3x-2y=0;3y-4z=0;x^2+y^2+z^2-1=0\)

....... ( típ theo tự làm nhé eiu)

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

2) (a+b+c)²=3(ab+bc+ac) =>(a-b)²+(b-c)²+(c-a)²=0

=>a-b=b-c=c-a=0 =>a=b=c

\(\hept{\begin{cases}\left|x^2+y^2+z^2-1\right|=0\\\left(3y-4z\right)^4\ge0\\\left(3x-2y\right)^2\ge0\end{cases}}\Rightarrow\left|x^2+y^2+z^2-1\right|+\left(3y-4z\right)^4+\left(3x-2y\right)^2\ge0\)

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 :)

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)