A+B+C= x²yz+xy²z+xyz² =xyz (x+y+z)=xyz.1=xyz

b) x²+4x+4+1=x²+2x+2x+2+1=x(x+2)+(x+1)+1=(x+1)(x+1)+1=(x+1)²+1

cho (x+1)² +1=0

-> (x+1)²=-1 (vô lý )

da thuc k co nghiem

c) f(x)=g(x)

-3x²+2x+1=-3x²-2+x

-3x²+3x²+2x-x=-1

x=-1

Đợi tí nhé, đừng off, mk giải ra ròi, mình sẽ chép lên cho bạn

theo bài ra ta có 
n = 8a +7=31b +28 
=> (n-7)/8 = a 
b= (n-28)/31 
a - 4b = (-n +679)/248 = (-n +183)/248 + 2 
vì a ,4b nguyên nên a-4b nguyên => (-n +183)/248 nguyên 
=> -n + 183 = 248d => n = 183 - 248d (vì n >0 => d<=0 và d nguyên ) 
=> n = 183 - 248d (với d là số nguyên <=0) 
vì n có 3 chữ số lớn nhất => n<=999 => d>= -3 => d = -3 
=> n = 927

\(A+B+C=x^2yz+xy^2z+xyz^2\) 

                         \(=xxyz+xyyz+xyzz\)

                         \(=x\left(xyz+yyz+yzz\right)\)

                          \(=x\left[y\left(xz+yz+zz\right)\right]\)

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

                          \(=x\left[y\left(z\cdot1\right)\right]\)

                          \(=x\cdot y\cdot z\)

\(A+B+C=x^2yz+xy^2z+xyz^2=xyz\left(x+y+z\right)=xyz\)

\(A=x^2yz\) \(B=xy^2z\) \(C=xyz^2\)

\(A+B+C=x^2yz+xy^2z+xyz^2\)

                    \(=xyz\left(x+y+z\right)=xyz.1=xyz\)

help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

hộ caiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

A=x^2yz
B=xy^2z
C=xyz^2
=>A+B+C=x^2yz+xy^2z+xyz^2=xyz(x+y+z)=xyz

\(A+B+C=xyz\)

\(VT=A+B+C\)

\(\Leftrightarrow VT=x^2yz+xy^2z+xyz^2\)

\(\Leftrightarrow VT=xyz\left(x+y+z\right)\)

\(\Leftrightarrow VT=xyz\)

\(\Rightarrow VT=VP\)

\(\Rightarrow A+B+C=xyz\left(dpcm\right)\)

A + B + C = x².y.z + x.y².z + x.y.z² = x.y.z.(x + y + z) = x.y.z .1 = xyz (Vì x+ y + z = 1)

ta có A+B+C=x²yz+xy²z+xyz²

^{=x(xyz)+y(xyz)+z(xyz)}

^=x.1+y.1+z.1

^=x+y+z(dpcm)

\(A=x^2yz=x.\left(xyz\right)=x.1=x\)

\(B=xy^2z=y.\left(xyz\right)=y.1=y\)

\(C=xyz^2=z.\left(xyz\right)=z.1=z\)

\(\Rightarrow A+B+C=x+y+z\)

Ta có \(A+B+C=x^2yz+xy^2z+xyz^2\)

\(A+B+C=\left(xyz\right)x+\left(xyz\right)y+\left(xyz\right)z\)

\(A+B+C=\left(x+y+z\right)xyz\)

Mà xyz=1 thay vào A+B+C ta có 

\(A+B+C=x+y+z\) (đpcm)

\(Ta\) \(có:\)

\(A+B+C=x^2yz+xy^2z+xyz^2=xyz\left(x+y+z\right)=xyz.1=xyz\)