.

a) Ta có: \(VT=\left(x-y-z\right)^2\)

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

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

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

=VP(đpcm)

b) Ta có: \(VT=\left(x+y-z\right)^2\)

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

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

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

=VP(đpcm)

c) Sửa đề: Chứng minh \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)=x^4-y^4\)

Ta có: \(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)

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

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

=VP(đpcm)

d) Ta có: \(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)

\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)

\(=x^5+y^5\)

=VP(đpcm)

a, b, nhân vào là ra à

c, nghe cứ là lạ

d, cũng nhân là ra hà

\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5\)

Đáp án D

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

a) Ta có: \(VP=x^2+y^2+z^2-2xy+2yz-2zx\)

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

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

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

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

\(=\left(x-y-z\right)^2=VT\)(đpcm)

b) Ta có: \(VP=x^2+y^2+z^2+2xy-2yz-2zx\)

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

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

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

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

\(=\left(x+y-z\right)^2=VT\)(đpcm)

c) Ta có: \(VP=x^4-y^4\)

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

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

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

d) Ta có: \(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)

\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)

\(=x^5+y^5=VP\)(đpcm)

a)biến đổi vế trái ta đc:x(y+z)-y(x-z)=xy+xz-xy+yz

                                                        =(xz+yz)+(xy-xy)

                                                        =z(x+y)=vế phải(đpcm)

b)biến đổi vế trái ta đc:x(y-z)-x(y+a)=xy-xz-xy-xa

                                                         =(xy-xy)-(xz+xa)

                                                         =-(xz+xa)

                                                         =-x(z+a)=vế phải(đpcm)  

a;\(x\left(y+z\right)-y\left(x-z\right)=\left(x+y\right)z\)

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

\(xz+yz=\left(x+y\right)z\)

\(\left(x+y\right)z=\left(x+y\right)z\left(ĐPCM\right)\)

b;\(x\left(y-z\right)-x\left(y+a\right)=-x\left(z+a\right)\)

\(xy-xz-xy-xa=-x\left(z+a\right)\)

\(-xz-xa=-x\left(z+a\right)\)

\(-x\left(z+a\right)=-x\left(z+a\right)\left(ĐPCM\right)\)

P/S: sai thì thôi nha 

\(\Rightarrow\left(x+y+z\right)^2\ge\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\ge3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{3\left(x+y+z\right)}{xyz}\Rightarrow x+y+z\ge\dfrac{3}{xyz}\)

\(x+y+z=\dfrac{x+y+z}{3}+\dfrac{2\left(x+y+z\right)}{3}\ge\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{2}{3}.\dfrac{3}{xyz}\ge\dfrac{1}{3}\left(\dfrac{9}{x+y+z}\right)+\dfrac{2}{xyz}=\dfrac{3}{x+y+z}+\dfrac{2}{xyz}\left(đpcm\right)\)

\(dấu"="xảy\) \(ra\Leftrightarrow x=y=z=1\)