Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=x2y+xy2+xyz+xyz+y2z+yz2+x2z+xyz+xz2-xyz
A=(x2y+xy2+xyz+y2z)+(yz2+x2z+xyz+xz2)
A=y(x2+xy+xz+yz)+z(yz+x2+xy+xz)
A=(y+z)(x2+xy+xz+yz)
A=(y+z)[x(x+y)+z(x+y)]
A=(y+z)(x+y)(x+z)
Ta có
C = xyz – (xy + yz + zx) + x + y + z – 1
= (xyz – xy) – (yz – y) – (zx – x) + (z – 1)
= xy(z – 1) – y(z – 1) – x(z – 1) + (z – 1)
= (z – 1)(xy – y – x + 1)
= (z – 1).[y(x – 1) – (x – 1)]
= (z – 1)(y – 1)(x – 1)
Với x = 9; y = 10; z = 101 ta có
C = (101 – 1)(10 – 1)(9 – 1) = 100.9.8 = 7200
Đáp án cần chọn là: C
\(=\dfrac{xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)}{xy\left(z+1\right)+y\left(z+1\right)-x\left(z+1\right)-\left(z+1\right)}\\ =\dfrac{\left(z-1\right)\left(xy-y-x+1\right)}{\left(z+1\right)\left(xy+y-x-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)\left(y-1\right)}{\left(z+1\right)\left(x+1\right)\left(y-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)}{\left(z+1\right)\left(x+1\right)}\\ =\dfrac{\left(5003-1\right)\left(5001-1\right)}{\left(5003+1\right)\left(5001+1\right)}=\dfrac{5002\cdot5000}{5004\cdot5002}=\dfrac{5000}{5004}=\dfrac{1250}{1251}\)
a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz
= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]
= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)
= (xy + yz + zx)(x + y + z)
b) Vô câu hỏi tương tự
a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz
= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]
= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)
= (xy + yz + zx)(x + y + z)
b) tương tự
\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)
\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)
xy( x+ y) + yz(y+z) + xz(x+z) + 3xyz
= xy(x+y) + xyz + yz(y+z) + xyz + xz(x+z) + xyz
= zy(x+y+z) + yz(x + y + z) + xz ( x+y+z)
= ( x+ y +z )( xy + yz + zx)