\(a,VP=\dfrac{x^2+4x+3}{x^2+6x+9}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)^2}=\dfrac{x+1}{x+3}=VT\)

Vậy ta có đpcm 

b, \(VP=\dfrac{3x\left(x+y\right)^2}{9x^2\left(x+y\right)}=\dfrac{x+y}{3x}=VT\)

Vậy ta có đpcm 

a) Ta có: \(\dfrac{x^2+4x+3}{x^2+6x+9}\)

\(=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x+3\right)}\)

\(=\dfrac{x+1}{x+3}\)

b: Ta có: \(\dfrac{3x\left(x+y\right)^2}{9x^2\left(x+y\right)}\)

\(=\dfrac{3x\left(x+y\right)\left(x+y\right)}{3x\cdot3x\cdot\left(x+y\right)}\)

\(=\dfrac{x+y}{3x}\)

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

Bài 1: 

c: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)

1: \(\frac{1-x}{2-y}\)

\(=\frac{-\left(x-1\right)}{-\left(y-2\right)}\)

\(=\frac{x-1}{y-2}\)

2: \(\frac{2a}{-5b}=\frac{2a\cdot\left(-1\right)}{\left(-5b\right)\cdot\left(-1\right)}=\frac{-2a}{5b}\)

3: \(\frac{2^3-x^3}{x\left(x^2+2x+4\right)}\)

\(=\frac{\left(2-x\right)\left(2^2+2\cdot x+x^2\right)}{x\left(x^2+2x+4\right)}\)

\(=\frac{\left(2-x\right)\left(x^2+2x+4\right)}{x\left(x^2+2x+4\right)}=\frac{2-x}{x}=\frac{x-2}{-x}\)

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

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

\(=x^3-1=VP\)

b) \(VT=\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)\)

\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)

\(=x^4-y^4=VP\)

c) \(VT=\left(x+y+z\right)^2\)

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

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

\(=x^2+y^2+z^2+2xy+2yz+2zx=VP\)

Chúc bạn học tốt.