Cho x2-y2-z2=0. Chứng minh rằng:
(5x-3y+4z) (5x-3y-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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cho x^2-y^2-z^2=0 chứng minh rằng: (5x-3y+4z)*(5x-3y-4z)=(3x-5y)^2
nếu x^2=y^2+x^2
chứng minh rằng ( 5x-3y+4z)(5x-3y-4z)=(3x-5y)^2
Sửa đề: x2 = y2 + z2
=> z2 = x2 - y2
Ta có:
\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)\)
\(=\left(5x-3y\right)^2-\left(4z\right)^2\)
\(=25x^2-30xy+9y^2-16z^2\)
\(=25x^2-30xy+9y^2-16\left(x^2-y^2\right)\)
\(=\left(3x-5y\right)^2\)
=> ĐPCM
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nếu x^2=y^2+z^2
chứng minh rằng
(5x-3y+4z)(5x-3y-4z)=(3x-5y)^2
Ta có
\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-16z^2\)
\(=25x^2-30xy+9y^2-16z^2\left(!\right)\)
Thay \(x^2=y^2+z^2\) vào ! thì
\(25x^2-30xy+9y^2-16\left(x^2-y^2\right)\)
\(=\left(3x-5y\right)^2\)
Cho \(x^2-y^2-z^2=0\)
Chứng minh rằng: \(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)^2-\left(4z\right)^2=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)-16z^2-\left(3x-5y\right)^2=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-\left(9x^2-30xy+25y^2\right)=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-9x^2+30xy-25y^2=0\)
\(\Rightarrow25\left(x^2-y^2\right)+9\left(x^2-y^2\right)-16z^2=0\)
\(\Rightarrow34\left(x^2-y^2\right)-16z^2=0\)
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\(Đây\)\(mới\)\(là\)\(câu\)\(trả\)\(lời\)\(đúng\)
\(ta\)\(có\)\(16\left(x^2-y^2-z^2\right)=16\left(x^2-y^2\right)-16z^2=8\left(x-y\right)2\left(x-y\right)-\left(4z\right)^2=\left(8x-8y\right)\left(2x+2y\right)-\left(4z\right)^2=\left(5x-3y+3x-5y\right)\left(5x-3y-3x+5y\right)-\left(4z\right)^2\)
\(=\left(5x-3y\right)^2-\left(3x-5y\right)^2-16z^2\)
\(\Leftrightarrow\left(5x-3y\right)^2-\left(4z\right)^2=\left(3x-5y\right)^2\)
\(\Leftrightarrow\left(5x-3y-4z\right)\left(5x-3y+4z\right)=\left(3x-5y\right)^2\)
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Nếu x2=y2+z2. chứng minh rằng (5x-3y+4z)(5x-3y-4z)=(3x-5y)2
ta có
(5x - 3y + 4z)(5x - 3y - 4z) = (5x - 3y)² - 16z²
= 25x² - 30xy + 9y² - 16x² + 16y²
= 25y² - 30xy + 9x² = (5y - 3x)² = (3x - 5y)²
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lần sau suy nghĩ kĩ rồi hãy post lên nhé ^^
ta có
(5x - 3y + 4z)(5x - 3y - 4z) = (5x - 3y)² - 16z²
= 25x² - 30xy + 9y² - 16x² + 16y²
= 25y² - 30xy + 9x² = (5y - 3x)² = (3x - 5y)²
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chứng minh rằng : (5x-3y+4z)(5x- 3y-4z)=(3x-5y)2 nếu x2=y2+z2
Ta có:
(5x – 3y + 4z)( 5x –3y –4z) = (5x – 3y )2 –16z2= 25x2 –30xy + 9y2 –16 z2
Vì x2=y2 + z2 nên 25x2 –30xy + 9y2 –16 (x2 –y2) = (3x –5y)2
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a) cho x^2 = y^2+z^2. chứng minh: (5x-3y+4z)(5x-3y-4z)=(3x-5y)^2
b) cho 10x^2=10y^2+z^2. chứng minh: (7x-3y+2z)(7x-3y-2z)=(3x-7y)^2
Mọi người làm dùm mình với!!!
Cho x^2 - y^2 - z^2=0
Chứng minh (5x-3y+4z)(5x-2y-4z)=(3x-5y)^2
theo giả thiết x^2-y^2-z^2=0
<=> x^2-y^2=z^2
Ta có (5x-3y+4z)(5x-3y+4z) = (5x-3y)^2-(4z)^2
=25.x^2-30xy+9y^2 -16z^2
=25.x^2-30xy+9y^2 -16(x^2-y^2) ( vì x^2-y^2=z^2)
=25.x^2-30xy+9.y^2-16.x^2+16.y^2
=9.x^2-30xy+25.y^2
=(3x-5y)^2
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