được nghỉ thêm 2 tuần vẫn có bài

\(\left(x-2y-z\right)+\left(-2x+y-z\right)-\)\(\left(-x-y-2z\right)\)

\(=x-2y-z-2x+y-z+x+y+2z\)

\(=0\)

đã tắt máy chưa để cho mình giải nha

Giúp mik nha mọi người :)

Chưa tắt máy, ai giúp mik giải với!!!

a) A = x - y + z + z + y + x - 2y

A = (x + x) + (-y + y) + (z + z) - 2y

A = 2x + 0 + 2z - 2y 

A = 2 .(x + z - y)

b) Thay x = 3 ; y = -1 ; z = 2 vào biểu thức A , ta được :

A = 2 .[3 + 2 - (-1)]

A = 12

Vậy A = 12

Chúc bạn học tốt !

\(B=\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{x^2y-x^2z+y^2z-y^3}\)

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

\(=\frac{\left(x^2y-z^2y\right)-\left(xy^2-zy^2\right)-\left(x^2z-z^2x\right)}{\left(x^2-y^2\right)\left(y-z\right)}\)

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

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

\(=\frac{\left[\left(xy-y^2\right)-\left(xz-zy\right)\right]\left(x-z\right)}{\left(x-y\right)\left(x+y\right)\left(y-z\right)}\)

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

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

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

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

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

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

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

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

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

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

C = (2x-3)²-(x+4)(2x-1) -(x+3)²

(Chuyển đổi các hằng đẳng thức)

    = (4x²-12x+9)-(2x²-x+8x-4)-(x²+6x+9)

    =  4x²-12x+9-2x²+x-8x+4-x²-6x-9

(Ta thu gọn các hạng tử đồng dạng với nhau)

    = x²-25x-14 

 (x - y + z)^2 + (z - y)^2 + (x - y + z)(2y -2z)

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

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

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

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

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

Đặt B = \(bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)

\(=bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2\left(bcyz+acxz+abxy\right)\) (1)

Từ \(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\)

=>\(a^2x^2+b^2y^2+c^2z^2+2\left(bcyz+acxz+abxy\right)=0\)

=>\(a^2x^2+b^2y^2+c^2z^2=-2\left(bcyz+acxz+abxy\right)\) (2)

Thay (2) vào (1) ta được:

\(B=ax^2\left(b+c\right)+by^2\left(a+c\right)+cz^2\left(a+b\right)+a^2x^2+b^2y^2+c^2z^2\)

\(=ax^2\left(a+b+c\right)+by^2\left(a+b+c\right)+cz^2\left(a+b+c\right)\)

\(=\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)\)

Vậy \(A=\frac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)