cho x2-y2-z2=0.CMR
(5x-3y+4z).(5x-37-4z)=(3x-5y)2
Những câu hỏi liên quan
1)Cmr nếu a-b=1 thì (a+b)(a2+b2)(a4+b4)...(a32+b32) =a64-b64
2) Cho x2=y2+z2. CM (5x-3y+4z)(5x-3y-4z)=(3x-5y)2
1) Ta có: \(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\cdot...\cdot\left(a^{32}+b^{32}\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\cdot...\cdot\left(a^{32}+b^{32}\right)\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\cdot...\cdot\left(a^{32}+b^{32}\right)\)
\(=\left(a^4-b^4\right)\left(a^4+b^4\right)\cdot...\cdot\left(a^{32}+b^{32}\right)\)
\(=\left(a^8-b^8\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\left(a^{32}+b^{32}\right)\)
\(=\left(a^{16}-b^{16}\right)\left(a^{16}+b^{16}\right)\left(a^{32}+b^{32}\right)\)
\(=\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)\)
\(=a^{64}-b^{64}\)
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CMR : Nếu x^2 - y^2 - z^2 = 0 thì ( 5x-3y+4z ) . ( 5x-3y - 4z ) = ( 3x - 5y )^2
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)
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x^2-y^2-z^2=0.CMR
(5x-3y+4z).(5x-3y-4z)=(3x-5y)^2
CMR: (5x-3y+4z)(5x-3y-4z)=(3x-5y)^2
\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-\left(4z\right)^2\)
\(=\left(3x-5y\right)^2-16z^2\)
Đẳng thức chỉ đúng khi \(z=0\)
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Ta có:
\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)\)
\(=\left(5x-3y\right)^2-16z^2\)
\(=25x^2-30xy+9y^2-16z^2\left(#\right)\)
Vì \(x^2=y^2+z^2\Rightarrow\left(#\right)=25x^2-30xy+9y^2-16\left(x^2-y^2\right)=\left(3x-5y\right)^2\)
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Xem thêm câu trả lời
Cho \(x^2-y^2-z^2=0\)
CMR:(5x-3y+4z)(5x-3y-4z)=\(\left(3x-5y\right)^2\)
Ta có:
\(x^2-y^2-z^2=0\)
\(16x^2-16y^2-16z^2=0\)
\(25x^2-9x^2+9y^2-25y^2-16z^2+30xy-30xy=0\)
\(\left(5x-3y\right)^2-16z^2= \left(3x-5y\right)^2\)
\(\left(5x-3y-4z\right)\left(5x-3y+4z\right)=\left(3x-5y\right)^2\)
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Cho (5x-3y+4z).(5x-3y-4z)=(3x-5y)^2
CMR: x^2=y^2+z^2
Ta có \(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
\(\Leftrightarrow\left(5x-3y\right)^2-\left(4z\right)^2=\left(3x-5y\right)^2\)
\(\Leftrightarrow25x^2-30xy+9y^2-16z^2=9x^2-30xy+25y^2\)
\(\Leftrightarrow16x^2=16y^2+16z^2\Leftrightarrow x^2=y^2+z^2\)
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(5x - 3y + 4z) . (5x - 3y - 4z) = (3x - 5y)2
(5x - 3y)2 - 16z2 = (3x - 5y)2
25x2 - 2.5x.3y + 9y2 - 16z2 = 9x2 - 2.3x.5y + 25y2
16x2 + 9y2 - 16z2 - 25y2 = 0
16x2 - 16y2 - 16z2 = 0
x2 - y2 - z2 = 0
x2 = y2 + z2
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Cho x^2 -y ^2=4z^2 . CMR: (5x-3y+8z)(5x-3y-8z)=(3x-5y)^2
\(x^2-y^2=4z^2\\ \Leftrightarrow64z^2=16x^2-16y^2\)
\(\left(5x-3y+8z\right)\left(5x-3y-8z\right)\\ =\left(5x-3y\right)^2-64z^2\\ =25x^2-30xy+9y^2-64z^2\\ =25x^2-16x^2+9y^2+16y^2-30xy\\ =9x^2-30xy+25y^2=\left(3x-5y\right)^2\)
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cho x^2-y^2-z^2=0 chứng minh rằng: (5x-3y+4z)*(5x-3y-4z)=(3x-5y)^2
Cho \(^{x^2-y^2-z^2=0.CMR:\left(5X-3Y+4Z\right)\left(5Z-3Y-4Z\right)=\left(3X-5Y\right)^2}\)
Ta có \(x^2-y^2-z^2=0\Rightarrow z^2=x^2-y^2\)
Có \(VT=\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-\left(4z\right)^2\)\(=\left(5x-3y\right)^2-16z^2=\left(5x-3y\right)^2-16\left(x^2-y^2\right)\)
\(=25x^2-30xy+9y^2-16x^2+16y^2=9x^2-30xy+25y^2\)
\(=\left(3x\right)^2-2.3x.5y+\left(5y\right)^2=\left(3x-5y\right)^2=VP\left(đpcm\right)\)
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