a ) \(\frac{1}{\left(x-y\right)\left(y-z\right)}+\frac{1}{\left(y-z\right)\left(z-x\right)}+\frac{1}{\left(z-x\right)\left(x-y\right)}\)

     = \(\frac{z-x}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{x-y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

    = \(\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)

b ) \(\frac{4}{\left(y-x\right)\left(z-x\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)

 = \(\frac{-4}{\left(y-x\right)\left(x-z\right)}+\frac{3}{\left(y-x\right)\left(y-z\right)}+\frac{3}{\left(y-z\right)\left(x-z\right)}\)

= \(\frac{-4\left(y-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(x-z\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}+\frac{3\left(y-x\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)

= \(\frac{-4y+4z+3x-3z+3y-3x}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}=\frac{z-y}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}\)

= \(\frac{-\left(y-x\right)}{\left(x-z\right)\left(y-z\right)\left(y-x\right)}=\frac{-1}{\left(x-z\right)\left(y-z\right)}=\frac{1}{\left(x-z\right)\left(x-y\right)}\)

Chúc bạn học tốt !!!

\(\frac{1}{\left(x-y\right)\left(y-z\right)}+\frac{1}{\left(y-z\right)\left(z-x\right)}+\frac{1}{\left(z-x\right)\left(x-y\right)}\)

\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

\(=0\)

Giúp mình với các bạn

a: \(2x^2+3xy-14y^2\)

\(=2x^2+7xy-4xy-14y^2\)

\(=\left(2x^2+7xy\right)-\left(4xy+14y^2\right)\)

\(=x\left(2x+7y\right)-2y\left(2x+7y\right)\)

\(=\left(2x+7y\right)\left(x-2y\right)\)

b: \(\left(x-7\right)\left(x-5\right)\left(x-3\right)\left(x-1\right)+7\)

\(=\left(x-7\right)\left(x-1\right)\left(x-5\right)\left(x-3\right)+7\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)+7\)

\(=\left(x^2-8x\right)^2+15\left(x^2-8x\right)+7\left(x^2-8x\right)+105+7\)

\(=\left(x^2-8x\right)^2+22\left(x^2-8x\right)+112\)

\(=\left(x^2-8x\right)^2+8\left(x^2-8x\right)+14\left(x^2-8x\right)+112\)

\(=\left(x^2-8x\right)\left(x^2-8x+8\right)+14\left(x^2-8x+8\right)\)

\(=\left(x^2-8x+8\right)\left(x^2-8x+14\right)\)

c: \(\left(x-3\right)^2+\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)

\(=\left(x-3\right)^2+2\left(x-3\right)\left(3x-1\right)-\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)

\(=\left(x-3\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]-\left(3x-1\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]\)

\(=\left(x-3+6x-2\right)\left(x-3-3x+1\right)\)

\(=\left(7x-5\right)\left(-2x-2\right)\)

\(=-2\left(x+1\right)\left(7x-5\right)\)

d: \(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)

\(=x^2y-xy^2+y^2z-yz^2+zx\left(z-x\right)\)

\(=\left(x^2y-yz^2\right)-\left(xy^2-y^2z\right)+xz\left(z-x\right)\)

\(=y\left(x^2-z^2\right)-y^2\left(x-z\right)-xz\left(x-z\right)\)

\(=y\cdot\left(x-z\right)\left(x+z\right)-\left(x-z\right)\left(y^2+xz\right)\)

\(=\left(x-z\right)\left(xy+zy-y^2-xz\right)\)

\(=\left(x-z\right)\left[\left(xy-y^2\right)+\left(zy-zx\right)\right]\)

\(=\left(x-z\right)\left[y\cdot\left(x-y\right)-z\left(x-y\right)\right]\)

\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)

Đặt \(A = \frac{y^2 + z^2 - x^2}{2yz}, \quad B = \frac{z^2 + x^2 - y^2}{2xz}, \quad C = \frac{x^2 + y^2 - z^2}{2xy}\)

\(1+A=\frac{y^2+z^2-x^2}{2yz}+1=\frac{y^2+2yz+z^2-x^2}{2yz}=\frac{\left(y+z\right)^2-x^2}{2yz}=\frac{\left(y+z-x\right)\left(y+z+x\right)}{2yz}\)

\(1 + B = \frac{(z + x - y)(x + y + z)}{2xz}\)

\(1 + C = \frac{(x + y - z)(x + y + z)}{2xy}\)

\(1-A=1-\frac{y^2+z^2-x^2}{2yz}=\frac{x^2-\left(y^2-2yz+z^2\right)}{2yz}\)

\(=\frac{x^2-\left(y-z\right)^2}{2yz}=\frac{\left(x-y+z\right)\left(x+y-z\right)}{2yz}\)

\(1 - B = \frac{(y - z + x)(y + z - x)}{2xz}\)

\(1 - C = \frac{(z - x + y)(z + x - y)}{2xy}\)

(1-A)(1-B)(1-C)

\(=\frac{(x + y - z)(y + z - x)(z + x - y) \cdot(x - y + z)(y - z + x)(z - x + y)}{8x^2 y^2 z^2}\)

\(= \frac{(x + y - z)^2 (y + z - x)^2 (z + x - y)^2}{8x^2 y^2 z^2}\)

(1+A)(1+B)(1+C)

\(=\frac{\left(x+y+z\right)\cdot\left(x+y+z\right)\left(x+y+z\right)(y+z-x)\cdot\left.(z+x-y\right)(x+y-z)}{8x^2y^2z^2}\)

\(=\frac{(x + y + z)^3 (y + z - x)(z + x - y)(x + y - z)}{8x^2 y^2 z^2}\)

Theo đề, ta có: A+B+C=1

=>\(x(y^2 + z^2 - x^2) + y(z^2 + x^2 - y^2) + z(x^2 + y^2 - z^2) = 2xyz\)

=>\(xy^2 + xz^2 - x^3 + yz^2 + yx^2 - y^3 + zx^2 + zy^2 - z^3 = 2xyz\)

=>\(-(x^3 + y^3 + z^3) + (xy^2 + x^2y) + (yz^2 + y^2z) + (zx^2 + z^2x) - 2xyz = 0\)

=>\((x + y - z)(y + z - x)(z + x - y) = 0\)

TH1: x+y-z=0

=>z=x+y

\(A = \frac{y^2 + (x+y)^2 - x^2}{2y(x+y)} = \frac{y^2 + x^2 + 2xy + y^2 - x^2}{2y(x+y)} = \frac{2y^2 + 2xy}{2y(x+y)} = \frac{2y(x+y)}{2y(x+y)} = 1\)

\(B = \frac{(x+y)^2 + x^2 - y^2}{2x(x+y)} = \frac{x^2 + 2xy + y^2 + x^2 - y^2}{2x(x+y)} = \frac{2x^2 + 2xy}{2x(x+y)} = \frac{2x(x+y)}{2x(x+y)} = 1\)

\(C = \frac{x^2 + y^2 - (x+y)^2}{2xy} = \frac{x^2 + y^2 - (x^2 + 2xy + y^2)}{2xy} = \frac{-2xy}{2xy} = -1\)

=>A=B=1; C=-1(1)

TH2: y+z-x=0

=>x=y+z

\(A = \frac{y^2 + z^2 - (y+z)^2}{2yz} = \frac{-2yz}{2yz} = -1\)

\(B = \frac{z^2 + (y+z)^2 - y^2}{2z(y+z)} = \frac{2z(y+z)}{2z(y+z)} = 1\)

\(C = \frac{(y+z)^2 + y^2 - z^2}{2y(y+z)} = \frac{2y(y+z)}{2y(y+z)} = 1\)

Do đó: B=C=1; A=-1(2)

TH3: z+x-y=0

=>y=x+z

\(A = \frac{y^2 + z^2 - x^2}{2yz} = \frac{(x+z)^2 + z^2 - x^2}{2(x+z)z}\)

\(= \frac{(x^2 + 2xz + z^2) + z^2 - x^2}{2z(x+z)} = \frac{2xz + 2z^2}{2z(x+z)} = \frac{2z(x+z)}{2z(x+z)} = 1\)

\(B = \frac{z^2 + x^2 - y^2}{2xz} = \frac{z^2 + x^2 - (x+z)^2}{2xz}\)

\(= \frac{z^2 + x^2 - (x^2 + 2xz + z^2)}{2xz} = \frac{-2xz}{2xz} = -1\)

\(C = \frac{x^2 + y^2 - z^2}{2xy} = \frac{x^2 + (x+z)^2 - z^2}{2x(x+z)}\)

\(= \frac{x^2 + (x^2 + 2xz + z^2) - z^2}{2x(x+z)} = \frac{2x^2 + 2xz}{2x(x+z)} = \frac{2x(x+z)}{2x(x+z)} = 1\)

Do đó: A=C=1; B=-1(3)

Từ (1),(2),(3) suy ra trong 3 phân thức A,B,C; sẽ có hai phân thức bằng 1 và phân thức còn lại bằng -1

Đặt \(A = \frac{y^2 + z^2 - x^2}{2yz}, \quad B = \frac{z^2 + x^2 - y^2}{2xz}, \quad C = \frac{x^2 + y^2 - z^2}{2xy}\)

\(1+A=\frac{y^2+z^2-x^2}{2yz}+1=\frac{y^2+2yz+z^2-x^2}{2yz}=\frac{\left(y+z\right)^2-x^2}{2yz}=\frac{\left(y+z-x\right)\left(y+z+x\right)}{2yz}\)

\(1 + B = \frac{(z + x - y)(x + y + z)}{2xz}\)

\(1 + C = \frac{(x + y - z)(x + y + z)}{2xy}\)

\(1-A=1-\frac{y^2+z^2-x^2}{2yz}=\frac{x^2-\left(y^2-2yz+z^2\right)}{2yz}\)

\(=\frac{x^2-\left(y-z\right)^2}{2yz}=\frac{\left(x-y+z\right)\left(x+y-z\right)}{2yz}\)

\(1 - B = \frac{(y - z + x)(y + z - x)}{2xz}\)

\(1 - C = \frac{(z - x + y)(z + x - y)}{2xy}\)

(1-A)(1-B)(1-C)

\(=\frac{(x + y - z)(y + z - x)(z + x - y) \cdot(x - y + z)(y - z + x)(z - x + y)}{8x^2 y^2 z^2}\)

\(= \frac{(x + y - z)^2 (y + z - x)^2 (z + x - y)^2}{8x^2 y^2 z^2}\)

(1+A)(1+B)(1+C)

\(=\frac{\left(x+y+z\right)\cdot\left(x+y+z\right)\left(x+y+z\right)(y+z-x)\cdot\left.(z+x-y\right)(x+y-z)}{8x^2y^2z^2}\)

\(=\frac{(x + y + z)^3 (y + z - x)(z + x - y)(x + y - z)}{8x^2 y^2 z^2}\)

Theo đề, ta có: A+B+C=1

=>\(x(y^2 + z^2 - x^2) + y(z^2 + x^2 - y^2) + z(x^2 + y^2 - z^2) = 2xyz\)

=>\(xy^2 + xz^2 - x^3 + yz^2 + yx^2 - y^3 + zx^2 + zy^2 - z^3 = 2xyz\)

=>\(-(x^3 + y^3 + z^3) + (xy^2 + x^2y) + (yz^2 + y^2z) + (zx^2 + z^2x) - 2xyz = 0\)

=>\((x + y - z)(y + z - x)(z + x - y) = 0\)

TH1: x+y-z=0

=>z=x+y

\(A = \frac{y^2 + (x+y)^2 - x^2}{2y(x+y)} = \frac{y^2 + x^2 + 2xy + y^2 - x^2}{2y(x+y)} = \frac{2y^2 + 2xy}{2y(x+y)} = \frac{2y(x+y)}{2y(x+y)} = 1\)

\(B = \frac{(x+y)^2 + x^2 - y^2}{2x(x+y)} = \frac{x^2 + 2xy + y^2 + x^2 - y^2}{2x(x+y)} = \frac{2x^2 + 2xy}{2x(x+y)} = \frac{2x(x+y)}{2x(x+y)} = 1\)

\(C = \frac{x^2 + y^2 - (x+y)^2}{2xy} = \frac{x^2 + y^2 - (x^2 + 2xy + y^2)}{2xy} = \frac{-2xy}{2xy} = -1\)

=>A=B=1; C=-1(1)

TH2: y+z-x=0

=>x=y+z

\(A = \frac{y^2 + z^2 - (y+z)^2}{2yz} = \frac{-2yz}{2yz} = -1\)

\(B = \frac{z^2 + (y+z)^2 - y^2}{2z(y+z)} = \frac{2z(y+z)}{2z(y+z)} = 1\)

\(C = \frac{(y+z)^2 + y^2 - z^2}{2y(y+z)} = \frac{2y(y+z)}{2y(y+z)} = 1\)

Do đó: B=C=1; A=-1(2)

TH3: z+x-y=0

=>y=x+z

\(A = \frac{y^2 + z^2 - x^2}{2yz} = \frac{(x+z)^2 + z^2 - x^2}{2(x+z)z}\)

\(= \frac{(x^2 + 2xz + z^2) + z^2 - x^2}{2z(x+z)} = \frac{2xz + 2z^2}{2z(x+z)} = \frac{2z(x+z)}{2z(x+z)} = 1\)

\(B = \frac{z^2 + x^2 - y^2}{2xz} = \frac{z^2 + x^2 - (x+z)^2}{2xz}\)

\(= \frac{z^2 + x^2 - (x^2 + 2xz + z^2)}{2xz} = \frac{-2xz}{2xz} = -1\)

\(C = \frac{x^2 + y^2 - z^2}{2xy} = \frac{x^2 + (x+z)^2 - z^2}{2x(x+z)}\)

\(= \frac{x^2 + (x^2 + 2xz + z^2) - z^2}{2x(x+z)} = \frac{2x^2 + 2xz}{2x(x+z)} = \frac{2x(x+z)}{2x(x+z)} = 1\)

Do đó: A=C=1; B=-1(3)

Từ (1),(2),(3) suy ra trong 3 phân thức A,B,C; sẽ có hai phân thức bằng 1 và phân thức còn lại bằng -1