cho x^2 =y^2+4z^2 chứng minh (5x-3y+8z )(5x-3y-8z)+1 luôn duong
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cho x^2 =y^2+4z^2 chứng minh (5x-3y+8z )(5x-3y-8z)+1 luôn duong
\(x^2=y^2+4z^2\Rightarrow x^2-y^2=4z^2\)
\(A=\left(5x-3y+8x\right)\left(5x-3y-8z\right)+1\)
\(=\left(5x-3y\right)^2-64z^2+1\)
\(=\left(5x-3y\right)^2-16\left(x^2-y^2\right)+1\)
\(=25x^2+9y^2-30xy-16x^2+16y^2+1\)
\(=9x^2-30xy+25y^2+1\)
\(=\left(3x-5y\right)^2+1>0\) \(\forall x;y\)
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cho x^2 =y^2+4z^2 chứng minh (5x-3y+8z )(5x-3y-8z)+1 luôn Dương
có \(x^2=y^2+4x^2\)
\(x^2-y^2=4z^2\)
Tiếp tục với \(\left(5x-3y+8z\right)\left(5x-3y-8z\right)+1\)
\(=\left(5x-3y\right)^2-\left(8x\right)^2+1\)
\(=25x^2-30xy+9y^2-64x^2+1\)
\(=25x^2-30xy+9y^2-16\cdot4x^2+1\)
Thay \(x^2-y^2=4z^2\)
\(\Rightarrow25x^2-30xy+9y^2-16\cdot4x^2+1\)
\(=25x^2-30xy+9y^2-16\cdot\left(x^2-y^2\right)+1\)
\(=25x^2-30xy+9y^2-16x^2+16y^2+1\)
\(=9x^2-30xy+25y^2+1\)
\(=\left(9x^2-30xy+25y^2\right)+1\)
\(=\left(3x-5y\right)^2+1\)
ta có \(\left(3x-5y\right)^2\ge0\)
\(\Rightarrow\left(3x-5y\right)^2+1>0\)
\(\Rightarrow\left(5x-3x+8z\right)\left(5x-3y-8z\right)+1\)luôn dương với mọi x;y
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Cho x2=y2+4z2. Chứng minh: ( 5x-3y+8z)(5x-3y-8z)=(3x-5y)2
Áp dụng hằng đẳng thức ( a - b ) ( a + b ) = a2 - b2 ta đc:
\(\left(5x-3y+8z\right)\left(5x-3y-8z\right)=\left(5x-3y\right)^2-\left(8z\right)^2\)
\(=25x^2-30xy+9y^2-64z^2\)
Đề có sai ko vậy bn
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mk lấy kq của bạn Kia Cerato mk giải típ
tc \(x^2=y^2+4z^2\Leftrightarrow x^2-y^2=4z^2\)
\(\Leftrightarrow25x^2-30xy+9y^2-16.4z^2\)
\(=25x^2-30xy+9y^2-16\left(x^2-y^2\right)\)
\(=25x^2+9y^2-30xy-16x^2+16y^2\)
\(=9x^2-30xy+25y^2=\left(3x-5y\right)^2\)
ok
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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
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
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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Cho x2-y2-z2=0. Chứng minh rằng:
(5x-3y+4z) (5x-3y-4z) = (3x-5y)2
Ta có:
\(x^2-y^2-z^2=0\left(gt\right)\)
Nếu \(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)^2-16z^2=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)^2-\left(3x-5y\right)^2=16z^2\)
\(\Rightarrow\left(5x-3y-3x+5y\right)\left(5x-3y+3x-5y\right)=16z^2\)
\(\Rightarrow\left(2x+2y\right)\left(8x-8y\right)=16z^2\)
\(\Rightarrow2\left(x+y\right).8\left(x-y\right)=16z^2\)
\(\Rightarrow16\left(x^2-y^2\right)=16z^2\)
\(\Rightarrow x^2-y^2=z^2\)
\(\Rightarrow x^2-y^2-z^2=0\)
\(\Rightarrow\) Đúng với giả thuyết ban đầu
Vậy \(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\) với \(x^2-y^2-z^2=0\)
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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
\(25x^2-30xy+9y^2-16\left(x^2-y^2\right)\)
\(=\left(3x-5y\right)^2\)