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Ta có: \(\left\{{}\begin{matrix}x+y+z=0\\xy+yz+zx=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\2\left(xy+yz+zx\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2xy+2yz+2xz=0\\2xy+2yz+2xz=0\end{matrix}\right.\)
\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz-2xy-2yz-2xz=0\)
\(\Rightarrow x^2+y^2+z^2=0\Rightarrow\left\{{}\begin{matrix}x^2\ge0\forall x\\y^2\ge0\forall y\\z^2\ge0\forall z\end{matrix}\right.\Rightarrow x^2+y^2+z^2\ge0\)
\("="\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\\z^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)
\(\Rightarrow x=y=z=0\Rightarrow dpcm\)
\(x+y+z=0\Leftrightarrow\left(x+y+z\right)^2=0\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=x^2+y^z+z^2+0=0\)
\(\Leftrightarrow x^2+y^2+z^2=0\Leftrightarrow x=y=z=0\)
b) Bằng chứ ^^
\(\left(x+y\right)^2=x^2+2xy+y^2=4xy\)
\(\Leftrightarrow x^2-2xy+y^2=0\Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\)
A ) xy(z+y)+yz(y+z)+zx(z+x)
=y.[x(z+y)+z(y+z)]+zx(z+x)
=y.(xz+xy+zy+z2)+zx(z+x)
=y.(xz+z2+xy+zy)+zx(z+x)
=y.[z.(z+x)+y.(z+x)]+zx(z+x)
=y.(z+x)(z+y)+zx(z+x)
=(z+x)[y(z+y)+zx]
=(z+x)(yz+y2+zx)
B )xy(x+y)-yz(y+z)-zx(z-x)
=y.[x(x+y)-z(y+z)]-zx(z-x)
=y.(x2+xy-zy-z2)-zx(z-x)
=y.(x2-z2+xy-zy)-zx(z-x)
=y.[(x+z)(x-z)+y.(x-z)]-zx(z-x)
=y.(x-z)(x+z+y)+zx(x-z)
=(x-z)[y(x+z+y)+zx]
=(x-z)(yx+yz+y2+zx)
=(x-z)(yx+zx+yz+y2)
=(x-z)[x.(y+z)+y.(y+z)]
=(x-z)(y+z)(x+y)
b. \(\text{ xy(x+y)-yz(y+z)-xz(z-x) =xy(x+y+z-z)+yz(y+z)+xz(x-z) =xy(x-z)+xy(y+z)+yz(y+z)+xz(x-z) =(x+y)(y+z)(x-z) }\)
a/ \(A=xy-4y-5x+20\)
\(=x\left(y-5\right)-4\left(y-5\right)\)
\(=\left(x-4\right)\left(y-5\right)\)
Thay \(x=14;y=5,5\) vào biểu thức A ta có :
\(A=\left(14-4\right)\left(5,5-5\right)\)
\(=10.0,5=5\)
Vậy...
b/ \(B=xyz-\left(xy+yz+zx\right)+x+y+z-1\)
\(=xyz-xy-yz-zx+x+y+z-1\)
\(=\left(xyz-xy\right)-\left(yz-y\right)-\left(zx-x\right)+\left(z-1\right)\)
\(=xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)\)
\(=\left(z-1\right)\left(xy-y-x+1\right)\)
\(=\left(z-1\right)\left[y\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)
Thay \(x=9,y=10,z=11\) vào biểu thức B ta có :
\(B=\left(9-1\right)\left(10-1\right)\left(11-1\right)\)
\(=720\)
Vậy....
c/ \(C=x^3-x^2y-xy^2+y^3\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)^2\left(x+y\right)\)
Thay \(x=5,75,y=4,25\) vào biểu thức C ta có :
\(C=\left(5,75-5,25\right)^2\left(5,75+5,25\right)=11,25\)
Vậy..
Bài 3:
Gọi bốn số nguyên dương liên tiếp là x,x+1,x+2,x+3
Theo đề, ta có: \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=120\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=120\)
\(\Leftrightarrow\left(x^2+3x\right)^2+2\left(x^2+3x\right)-120=0\)
\(\Leftrightarrow\left(x^2+3x\right)^2+12\left(x^2+3x\right)-10\left(x^2+3x\right)-120=0\)
\(\Leftrightarrow\left(x^2+3x+12\right)\left(x^2+3x-10\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
mà x là số nguyên dương
nên x=2
Vậy: Bốn số cần tìm là 2;3;4;5
a,\(\dfrac{x-y}{xy}+\dfrac{y-z}{yz}+\dfrac{z-x}{zx}\)
=\(\dfrac{\left(x-y\right).z}{xyz}+\dfrac{\left(y-z\right).x}{xyz}+\dfrac{\left(z-x\right).y}{xyz}\)
=\(\dfrac{xz-yz}{xyz}+\dfrac{xy-xz}{xyz}+\dfrac{yz-xy}{xyz}\)
=\(\dfrac{xz-yz+xy-xz+yz-xy}{xyz}\)
=\(\dfrac{0}{xyz}\)=0
Vậy biểu thức trên ko phụ thuộc vào x,y,z
b,\(\dfrac{1}{\left(x-y\right).\left(y-z\right)}-\dfrac{1}{\left(x-z\right).\left(y-z\right)}-\dfrac{1}{\left(x-y\right).\left(x-z\right)}\)
=\(\dfrac{1.\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}-\dfrac{\left(x-y\right).1}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}-\dfrac{1\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
=\(\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)=\(\dfrac{0}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)=0
Vậy biểu thức trên ko phụ thuộc vào x,y,z