Cho x2=y2+4z2. Chứng minh: ( 5x-3y+8z)(5x-3y-8z)=(3x-5y)2
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10 tháng 9 2021
\(x^2-y^2=4z^2\\ \Leftrightarrow64z^2=16x^2-16y^2\)
\(\left(5x-3y+8z\right)\left(5x-3y-8z\right)\\ =\left(5x-3y\right)^2-64z^2\\ =25x^2-30xy+9y^2-64z^2\\ =25x^2-16x^2+9y^2+16y^2-30xy\\ =9x^2-30xy+25y^2=\left(3x-5y\right)^2\)
TN
1
A
22 tháng 9 2019
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
TN
2
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 9 2019
\(x^2=y^2+4z^2\Rightarrow x^2-y^2=4z^2\)
\(A=\left(5x-3y+8x\right)\left(5x-3y-8z\right)+1\)
\(=\left(5x-3y\right)^2-64z^2+1\)
\(=\left(5x-3y\right)^2-16\left(x^2-y^2\right)+1\)
\(=25x^2+9y^2-30xy-16x^2+16y^2+1\)
\(=9x^2-30xy+25y^2+1\)
\(=\left(3x-5y\right)^2+1>0\) \(\forall x;y\)
TN
1
VM
22 tháng 9 2019
Áp dụng hằng đẳng thức \(\left(a-b\right).\left(a+b\right)=a^2-b^2\) vào ta được:
\(\left(5x-3y+8z\right).\left(5x-3y-8z\right)=\left(5x-3y\right)^2-\left(8z\right)^2\)
\(=25x^2-30xy+9y^2-64z^2.\)
Ta dùng tính chất:
\(x^2=y^2+4z^2\Rightarrow x^2-y^2=4z^2.\)
\(\Leftrightarrow25x^2-30xy+9y^2-16.4z^2\)
\(=25x^2-30xy+9y^2-16.\left(x^2-y^2\right)\)
\(=25x^2+9y^2-30xy-16x^2+16y^2\)
\(=9x^2-30xy+25y^2\)
\(=\left(3x-5y\right)^2.\)
Ta có: \(\left(3x-5y\right)^2+1\ge0\) \(\forall x,y.\)
\(\Rightarrow\left(3x-5y\right)^2\) luôn dương.
\(\Rightarrow\left(5x-3y+8z\right).\left(5x-3y-8z\right)+1\) luông dương \(\forall x,y\left(đpcm\right).\)
Chúc bạn học tốt!
8 tháng 12 2023
1: \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
2: \(x^2-y^2+x-y\)
\(=\left(x^2-y^2\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+1\right)\)
3: \(3x-3y+x^2-y^2\)
\(=\left(3x-3y\right)+\left(x^2-y^2\right)\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y+3\right)\)
4: \(5x-5y+x^2-y^2\)
\(=\left(5x-5y\right)+\left(x^2-y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(5+x+y\right)\)
5: \(x^2-5x-y^2-5y\)
\(=\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
6: \(x^2-y^2+2x-2y\)
\(=\left(x^2-y^2\right)+\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+2\right)\)
7: \(x^2-4y^2+x+2y\)
\(=\left(x^2-4y^2\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+1\right)\)
8: \(x^2-y^2-2x-2y\)
\(=\left(x^2-y^2\right)-\left(2x+2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
9: \(x^2-4y^2+2x+4y\)
\(=\left(x^2-4y^2\right)+\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+2\right)\)
NH
0
MD
1
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
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