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VP=\(A^2X^2+B^2Y^2+C^2Z^2+A^2Y^2+B^2X^2+A^2Z^2+C^2X^2+B^2Z^2+C^2Y^2\)
=\(A^2\left(X^2+Y^2+Z^2\right)+B^2\left(X^2+Y^2+Z^2\right)+C^2\left(X^2+Y^2+Z^2\right)\)
=\(\left(X^2+Y^2+Z^2\right)\left(A^2+B^2+C^2\right)\)
\(\frac{ay-bx}{c}=\frac{cx-az}{b}=\frac{bz-cy}{a}\)
\(\Rightarrow\frac{acy-bcx}{c^2}=\frac{bcx-abz}{b^2}=\frac{abz-acy}{a^2}=\frac{0}{a^2+b^2+c^2}=0\)
\(\Rightarrow\hept{\begin{cases}ay-bx=0\\cx-az=0\\bz-cy=0\end{cases}}\)
\(\Rightarrow\left(ay-bx\right)^2+\left(cx-az\right)^2+\left(bz-ay\right)^2=0\)
\(\Rightarrow a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2bycz\)
\(+c^2y^2=0\)
\(\Rightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)
\(\Rightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)
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Akai Haruma Võ Đông Anh Tuấn
mấy anh chị cm cho e thêm cái : \(\dfrac{ay+bx}{c}=\dfrac{bz+cy}{a}=\dfrac{cx+az}{b}\)
\(\left(ax+by\right)^2-\left(ay+bx\right)^2\)
\(=\left(ax+by+ay+bx\right)\left(ax+by-ay-bx\right)\)
\(=\left[a\left(x+y\right)+b\left(x+y\right)\right]\left[a\left(x-y\right)-b\left(x-y\right)\right]\)
\(=\left(a+b\right)\left(a-b\right)\left(x+y\right)\left(x-y\right)\)
\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)
\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)
\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)
\(=\left(a+b-1\right)\left(a+b+1\right)\left(a-b-3\right)\left(a-b+3\right)\)
a)
(ax+by)2 - (ay+bx)2
=(ax+by-ay-bx)(ax+by+ay+bx)
=[ a(x-y) -b(x-y)][ a(x+y) + b(x+y)]
=(a-b)(x-y)(a+b)(x+y)
b)(a2+b2-5)2 - 4(ab+2)2
=(a2+b2-5-2ab-4)(a2+b2-5+2ab+4)
=[ (a-b)2 -9][ (a+b)2 -1]
=(a-b-3)(a-b+3)(a+b-1)(a+b+1)
a) Sửa đề: \(\left(ax+by+cx\right)^2+\left(bx-ay\right)^2+\left(cy-bz\right)^2+\left(az-cx\right)^2\)
= a2x2 + b2y2 + c2x2 + 2axby + 2bycz + 2axcz + b2x2 - 2bxay + a2y2 + c2y2 - 2cybz + b2z2 + a2z2 - 2azcx + c2x2
= a2x2 + b2y2 + c2x2 + b2x2 + a2y2 + c2y2 + b2z2 + a2z2 + c2x2
= a2(x2+y2+z2) + b2(x2+y2+z2) + c2(x2+y2+z2)
= (a2+b2+c2)(x2+y2+z2) (đpcm)
b) Đặt x = b; y = c; z = a, ta có:
\(\left(ay+bz+cx\right)^2+\left(az-by\right)^2+\left(bx-cz\right)^2+\left(cy-ax\right)^2\)
= a2y2 + b2z2 + c2x2 + 2aybz + 2bzcx + 2aycx + a2z2 - 2azby + b2y2 + b2x2 - 2bxcz + c2z2 + c2y2 - 2cyax + a2x2
= a2y2 + b2z2 + c2x2 + a2z2 + b2y2 + b2x2 + c2z2 + c2y2 + a2x2
= (a2+b2+c2)(x2+y2+z2)
Thay b = x, c = y, a = z, ta có:
(a2+b2+c2)(x2+y2+z2) = (a2+b2+c2)2 (đpcm)
g ) \(4x^2\left(x-2y\right)-\left(4x+1\right)\left(2y-x\right)\)
\(=4x^2\left(x-2y\right)+\left(4x+1\right)\left(x-2y\right)\)
\(=\left(4x^2+4x+1\right)\left(x-2y\right)\)
\(=\left(2x+1\right)^2\left(x-2y\right)\)
h ) \(x^2-ax^2-y+ay+cx^2-cy\)
\(=x^2\left(1-a+c\right)-y\left(1-a+c\right)\)
\(=\left(x^2-y\right)\left(1-a+c\right)\)
Ta có:
\(X-A\)\(=\)\(by+cz-cy-bz=\left(b-c\right)y+\left(c-b\right)z=\left(b-c\right)\left(y-z\right)\)
\(X-B\)\(=\)\(ax+by-bx-ay=\left(a-b\right)x+\left(b-a\right)y=\left(a-b\right)\left(x-y\right)\)
\(X-C\)\(=\)\(ax+cz-cx-az=\left(a-c\right)x+\left(c-a\right)z=\left(a-c\right)\left(x-z\right)\)
\(Y-A\)\(=\)\(cx+ay-ax-cy=\left(c-a\right)x+\left(a-c\right)y=\left(c-a\right)\left(x-y\right)\)
\(Y-B\)\(=\)\(cx+bz-bx-cz=\left(c-b\right)x+\left(b-c\right)z=\left(c-a\right)\left(x-z\right)\)
\(Y-C\)\(=\)\(zy+bz-by-az=\left(a-b\right)y+\left(b-a\right)z=\left(a-b\right)\left(y-z\right)\)
\(Z-A\)\(=\)\(bx+az-ax-bz=\left(b-a\right)x+\left(a-b\right)z=\left(b-a\right)\left(x-z\right)\)
\(Z-B\)\(=\)\(cy+az-ay-cz=\left(c-a\right)y+\left(a-c\right)z=\left(c-a\right)\left(y-z\right)\)
\(Z-C\)\(=\)\(bx+cy-cx-by=\left(b-c\right)x+\left(c-b\right)y=\left(b-c\right)\left(x-y\right)\)
Từ đó có:
\(\left(X-A\right)\left(X-B\right)\left(X-C\right)=\left(b-c\right)\left(a-b\right)\left(a-c\right)\left(y-z\right)\left(x-y\right)\left(x-z\right)\)
\(\left(Y-A\right)\left(Y-B\right)\left(Y-C\right)=\left(c-a\right)\left(c-b\right)\left(a-b\right)\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
\(\left(Z-A\right)\left(Z-B\right)\left(Z-C\right)=\left(b-a\right)\left(c-a\right)\left(b-c\right)\left(x-z\right)\left(y-z\right)\left(x-z\right)\)
Ta thấy , vế phải của ba đẳng thức trên là tích của sáu thừa số . Các thừa số đều có mặt trong các tích nếu ta áp dụng quy tắc đổi dấu