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Sửa lại đề nha: x+y+z=0
a)
Xét x+y+z=0
(x+y+z)2=02
x2+y2+z2+2xy+2yz+2zx=0
=> x2+y2+z2=-2xy-2yz-2zx
Xét \(\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
= \(\dfrac{x^2+y^2+z^2}{\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)}\)
=\(\dfrac{x^2+y^2+z^2}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2}\)
=\(\dfrac{x^2+y^2+z^2}{2x^2+2y^2+2z^2-2xy-2yz-2zx}\)(1)
Thay x2+y2+z2=-2xy-2yz-2zx vào (1)
=>\(\dfrac{x^2+y^2+z^2}{2x^2+2y^2+2z^2+x^2+y^2+z^2}\\=\dfrac{x^2+y^2+z^2}{3x^2+3y^2+3z^2}\\ =\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}\\ =\dfrac{1}{3}\)
b)
Xét x+y+z=0 ba lần:
- Lần 1:x+y+z=0
<=> x+y=0-z
<=>(x+y)2=(0-z)2
<=>x2+2xy+y2=z2
<=>x2+y2-z2=-2xy(1)
-Lần 2: x+y+z=0
<=> y+z=0-x
<=>(y+z)2=(0-x)2
<=>y2+2yz+z2=x2
<=>y2+z2-x2=-2yz(2)
-Lần 3: x+y+z=0
<=>z+x=0-y
<=>(z+x)2=(0-y)2
<=>z2+2zx+x2=y2
<=> z2+x2-y2=-2zx(3)
Thay (1),(2),(3) vào Q, ta có:
=>\(\dfrac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}=\dfrac{\left(-2xy\right)\left(-2yz\right)\left(-2zx\right)}{16xyz}\\=\dfrac{\left(-2yz\right)\left(-2zx\right)}{-8z}\\ =\dfrac{y\left(-2zx\right)}{4}\\ =\dfrac{-2xyz}{4}\\ =-\dfrac{xyz}{2}\)
a. Ta có:
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c+a-b\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
và \(ab^2-ac^2-b^3+bc^2=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)=\left(a-b\right)\left(b-c\right)\left(b+c\right)\)
Vậy, \(A=\frac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(b+c\right)}=\frac{c-a}{-c-b}=\frac{a-c}{c+b}\)
\(M=\frac{z^5.\left(x+y^2\right).\left(x^2-y^3\right).\left(x^2-y\right)}{x^2+y^2+z^2+1}=\frac{\left(-5\right)^5.\left(-4+16^2\right).\left[\left(-4\right)^2-16^3\right].\left[\left(-4\right)^2-16\right]}{\left(-4\right)^2+16^2+\left(-5\right)^2+1}\)
\(=\frac{\left(-5\right)^5.\left(-4+16^2\right).\left[\left(-4\right)^2-16^3\right].0}{\left(-4\right)^2+16^2+\left(-5\right)^2+1}=0\)
\(M=\left(\dfrac{x}{y}+1-5\right)^3=\left(\dfrac{x}{y}-4\right)^3\)
\(=\left(\dfrac{12}{2}-4\right)^3=\left(6-4\right)^3=2^3=8\)
a) Đặt \(x^2+3x+1=y\) khi đó ta có:
\(y\left(y-4\right)-5\)
\(=y^2-4y-5\)
\(=y\left(y-5\right)+\left(y-5\right)\)
\(=\left(y+1\right)\left(y-5\right)\)
Thay \(y=x^2+3x+1\):
\(\left(x^2+3x+1+1\right)\left(x^2+3x+1-5\right)\)
\(=\left(x^2+3x+2\right)\left(x^2+3x-4\right)\)
\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x-1\right)+4\left(x-1\right)\right]\)
\(=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x+4\right)\)
b) Biến đổi 3 số sau có chứa x2 + 2x rồi đặt ẩn.
c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+7=y'\)
Khi đó ta đc:
\(y'\left(y'+8\right)+15\)
\(=\left(y'\right)^2+8y'+15\)
\(=y'\left(y'+3\right)+5\left(y'+3\right)\)
\(=\left(y'+5\right)\left(y'+3\right)\)
....
d) \(x^2-2xy+y^2-7x+7y+12\)
Biến đổi chứa x - y rồi đặt ẩn.
Đỗ thị như quỳnh: làm tương tự thôi mà, nếu bạn ko hiểu chỗ nào thì nói đi :)