Cho x2y - y2z + x2z - z2x + y2z + z2y = 2xyz
CMR: Trong 3 số x,y,z ít nhất có hai số bằng nhau hoặc đối nhau
phân tích đa thức thành nhân tử
a)70a+84b-20ab-24b2
b) x2y+xy2+x2z+xz2+y2z+yz2+3xyz
c) x2y+xy2+x2z+xz2+y2z+yz2+2xyz
a: \(70a+84b-20ab-24b^2\)
\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)
\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)
\(=\left(5a+6b\right)\left(14-4b\right)\)
\(=2\left(7-2b\right)\left(5a+6b\right)\)
b: \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+x^2z\right)+\left(xy^2+xz^2\right)+\left(y^2z+yz^2\right)+3xyz\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2\right)+yz\left(y+z\right)+3xyz\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2\right)+yz\left(y+z\right)+2xyz+xyz\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2+2yz\right)+yz\left(y+z+x\right)\)
\(=x^2\left(y+z\right)+x\left(y+z\right)^2+yz\left(y+z+x\right)\)
\(=\left(y+z\right)\cdot x\left(x+y+z\right)+yz\left(y+z+x\right)\)
\(=\left(y+z+x\right)\cdot\left(xy+xz+yz\right)\)
c: \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=\left(x^2y+x^2z\right)+\left(xy^2+xz^2+2xyz\right)+\left(y^2z+yz^2\right)\)
\(=x^2\left(y+z\right)+x\left(y^2+z^2+2xz\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(x^2+yz\right)+x\left(y+z\right)^2\)
\(=\left(y+z\right)\left(x^2+yz+xy+xz\right)\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
Cho 3 số x,y,z biết 0≤x,y,z≤1
CMR: x2+y2+z2≤1+x2y+y2z+z2x
Lời giải:
Vì $0\leq x,y,z\leq 1$ nên:
$x(x-1)(y-1)\geq 0$
$\Leftrightarrow x^2y\geq x^2+xy-x$
Tương tự và cộng theo vế:
$x^2y+y^2z^2+z^2x+1\geq x^2+y^2+z^2+(xy+yz+xz)-(x+y+z)+1(*)$
Lại có:
$(x-1)(y-1)(z-1)\leq 0$
$\Leftrightarrow xyz-(xy+yz+xz)+(x+y+z)-1\leq 0$
$\Leftrightarrow xy+yz+xz-(x+y+z)\geq xyz-1\geq -1$ do $xyz\geq 0(**)$
Từ $(*); (**)\Rightarrow x^2y+y^2z+z^2x+1\geq x^2+y^2+z^2$
Ta có đpcm
Dấu "=" xảy ra khi $(x,y,z)=(0,1,1); (0,0,1)$ và hoán vị.
phân tích đa thức thành nhân tử
a)70a+84b-20ab-24b2
b) x2y+xy2+x2z+xz2+y2z+yz2+3xyz
c) x2y+xy2+x2z+xz2+y2z+yz2+2xyz
a) \(70a+84b-20ab-24b^2\)
\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)
\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)
\(=\left(5a+6b\right)\left(14-4b\right)\)
\(=2\left(5a+6b\right)\left(7-2b\right)\)
b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xyz+xz^2\right)+\left(xyz+y^2z+yz^2\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+y+z\right)+yz\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(xy+yz+xz\right)\)
c) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=\left(x^2y+xy^2\right)+\left(xz^2+yz^2\right)+\left(x^2z+2xyz+y^2z\right)\)
\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x^2+2xy+y^2\right)\)
\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x+y\right)^2\)
\(=\left(x+y\right)\left[xy+z^2+z\left(x+y\right)\right]\)
\(=\left(x+y\right)\left(xy+z^2+xz+yz\right)\)
\(=\left(x+y\right)\left[\left(xy+yz\right)+\left(xz+z^2\right)\right]\)
\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
a, 70a + 84b - 20ab - 24b2
= 14.(5a + 6b) - 4b(5a + 6b)
= (5a + 6b).(14 - 4b)
a, 70a + 84b - 20ab - 24b2
= (70a + 84b) - (20ab + 24b2)
= 14.(5a + 6b) - 4b.(5a + 6b)
= (5a + 6b).(14 - 4b)
Cho \(x^2y-xy^2+x^2z-xz^2+y^2z+yz^2=2xyz\). CMR: trong 3 số \(x,y,z\) có ít nhất hai số bằng nhau hoặc đối nhau.
\(x^2y-xy^2+x^2z-xz^2+y^2z+yz^2=2xyz\)
\(\Leftrightarrow\left(x^2y-xy^2\right)+\left(x^2z-xyz\right)-\left(xz^2-yz^2\right)-\left(xyz-y^2z\right)=0\)
\(\Leftrightarrow xy\left(x-y\right)+xz\left(x-y\right)-z^2\left(x-y\right)-yz\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(xy+xz-z^2-yz\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[x\left(y+z\right)-z\left(y+z\right)\right]=0\)
\(\Leftrightarrow\left(x-y\right)\left(x-z\right)\left(y+z\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=z\\y=-z\end{matrix}\right.\)\(\left(đpcm\right)\)
phân tích đa thức thành nhân tử bằng cách nhóm hạng tử
3) x2 (x+2y) - x - 2y
4) x3 - 4x2 - 9x + 36
5) x2y + xy2 + x2z + y2z + 2xyz
3) \(x^2\left(x+2y\right)-x-2y\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x^2-1\right)\left(x+2y\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(x+2y\right)\)
4) \(x^3-4x^2-9x+36\)
\(=\left(x^3-4x^2\right)-\left(9x-36\right)\)
\(=x^2\cdot\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x^2-9\right)\)
\(=\left(x-4\right)\left(x+3\right)\left(x-3\right)\)
\(x^2\left(x+2y\right)-x-2y\\ =x^2\left(x+2y\right)-\left(x+2y\right)\\ =\left(x^2-1\right)\left(x+2y\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+2y\right)\\ ---\\ x^3-4x^2-9x+36\\ =x^2\left(x-4\right)-9\left(x-4\right)\\ =\left(x^2-9\right)\left(x-4\right)\\ =\left(x-3\right)\left(x+3\right)\left(x-4\right)\)
Cho x^2y-y^2x+x^2z-z^2x+y^2z+z^2y=2xyz. Chứng minh x,y,z ít nhất cũng có hai số bằng nhau hoặc đối nhau.
(x2 y - y2 x) + (x2 z - xyz) + (z2 y - z2 x) + (y2 z - xyz) = (x-y)(xy+zx-z2 -yz)=(x-y)(x-z)(y+z)=0
Giải giùm rồi đấy bạn
Cho \(x^2y-y^2x+x^2z-z^2x+y^2z+z^2y=2xyz\)
Chứng minh rằng trong 3 số \(x;y;z\)ít nhất cũng có 2 số bằng nhau hoặc đối nhau.
Cho x^2 y - y^2 x + x^2 z - z^2 x + y^2 z + z^2 y = 2xyz
Cmr trong 3 số x,y,z ít nhất có 2 số = nhau hoặc đối nhau ?
\(x^2y-y^2x+x^2z-z^2x+y^2z+z^2y=2xyz\)\(\Leftrightarrow\left(x^2y-xy^2\right)+\left(x^2z-xyz\right)+\left(z^2y-z^2x\right)+\left(y^2z-xyz\right)=0\)\(\Leftrightarrow xy\left(x-y\right)+xz\left(x-y\right)-z^2\left(x-y\right)-yz\left(x-y\right)=0\)\(\Leftrightarrow\left(xy+xz-z^2-yz\right)\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[x\left(y+z\right)-z\left(y+z\right)\right]=0\)
\(\Leftrightarrow\left(x-y\right)\left(x-z\right)\left(y+z\right)=0\Rightarrow\left[{}\begin{matrix}x-y=0\\x-z=0\\y+z=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x=z\\y=-z\end{matrix}\right.\Rightarrowđpcm\)
\(x^2y-y^2x+x^2z-z^2x+y^2z+z^2y=2xyz\)
\(x^2y-y^2x+x^2z-z^2x+y^2z+z^2y-2xyz=0\)
\(\left(x^2y-y^2x\right)+\left(x^2z-xyz\right)+\left(z^2y-z^2x\right)=\left(y^2z-xyz\right)+\left(y^2z-xyz\right)=0\)
\(\left[\left(x-y\right)\left(xy\right)\right]+\left[\left(x-y\right)\left(zx\right)\right]+\left[\left(x-y\right)\left(-z^2\right)\right]+\left[\left(x-y\right)\left(-yz\right)\right]\)
\(\left(x-y\right)\left(xy+zx-z^2-yz\right)=\left(x-y\right)\left(x-z\right)\left(y+z\right)\)
đpcm
x2y - y2x+x2z - z2x +y2z +z2y - 2xyz = 0
chứng minh rằng trong ba số x, y, z ít nhất cũng có hai số bằng nhau hoặc đối nhau
x2y - y2x+x2z - z2x +y2z +z2y - 2xyz = 0
=> xy.(x - y) + xz. (x - z) + zy.(y + z) - xyz - xyz = 0
=> [xy.(x - y) - xyz] + [xz.(x - z) - xyz] + zy,(y +z) = 0
=> xy.(x - y - z) + xz.(x - z - y) + zy.(y +z) = 0
<=> (x-y-z). (y+z).x + zy.(y +z) = 0
<=> (y +z). [x(x - y - z) + zy] = 0
<=> y + z = 0 hoặc x(x - y - z) + zy = 0
+) y + z = 0 => y;z đối nhau
+) x(x- y - z) + zy = 0 => x (x - y) - z.(x - y) = 0 => (x - z)(x - y) = 0 => x = z hoặc x = y
Vậy ....