Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài a) bn trên đã dẫn link cho bn r
bài b)
Đặt x-y=a;y-z=b;z-x=c
\(=>a+b+c=x-y+y-z+z-x=0\)
\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=a^3+b^3+c^3\)
Theo câu a)\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) (do a+b+c=0)
\(=>a^3+b^3+c^3=3abc=>\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
a) Ta có :
\(a^3+b^3+c^3-3abc\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b^2\right)-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
P/s tham khảo nha
hok tốt
1) a) \(x^3-2x^2y+xy^2-25x=x\left(x^2-2xy+y^2-25\right)\)
\(=x\left[\left(x-y\right)^2-5^2\right]=x\left(x-y-5\right)\left(x-y+5\right)\)
b)\(x^2-y^2-2x-2y=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left(x-1-y-1\right)\left(x-y+y+1\right)=\left(x-y-2\right)\left(x+1\right)\)
a) Ta có:
x³ + y³ + z³ - 3xyz = (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz).
\(\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
cộng ((x+y)^3 + z^3) vào 1 nhóm, -3xy(x+y)-3xyz vào 1 nhóm dc
\(\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)-3yz\left(x+y+z\right)\)xuất hiện nhân tử chung x+y+z
\(\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2-3xy\right)\)
Kết quả: \(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(=\left(x^3+y^3\right)+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy+yz+zx\right)\)
\(x^3+y^3+z^3-3xyz=\left(x^3+y^3\right)-3xyz+z^3\)
\(=\left(x+y\right)^3-3xy.\left(x+y\right)-3xyz+z^3\)
\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy.\left(x+y\right)+3xyz\right]\)
\(=\left(x+y+z\right).\left(x^2+2xy+y^2-zx-zy+z^2\right)-3xy.\left(x+y+z\right)\)
\(=\left(x+y+z\right).\left(x^2+y^2+z^2-zx-zy+2zy-3xy\right)\)
\(=\left(x+y+z\right).\left(x^2+z^2+y^2-zx-zy-xy\right)\)
Vừa làm xong . Chúc bạn học tốt !
\(=\left(x+y\right)^3+z^z-3x^2y-3xy^2-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
a, x^4 - 5x^2 + 4
= x^4 - 4x^2- x+ 4
= x^2 . (x^2 - 4) - (x^2 - 4)
= (x^2 - 4) . (x^2 - 1)
= (x - 2) . (x + 2) . (x - 1) . (x + 1)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(a,x^3+y^3+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
b) \(f\left(x\right)=12x^3-32x^2+25x-6\)
Thấy \(x=\frac{3}{2}\) là một nghiệm.Vậy đa thức có chứa nhân tử \(\left(x-\frac{3}{2}\right)\)
Ta có: \(f\left(x\right)=\left(x-\frac{3}{2}\right)\left(\frac{12x^3-32x^2+25x-6}{x-\frac{3}{2}}\right)\)
\(=\left(x-\frac{3}{2}\right)\left(12x^2-14x+4\right)\)
\(=\left(x-\frac{3}{2}\right)\left[\left(12x^2-6x\right)-\left(8x-4\right)\right]\)
\(=\left(x-\frac{3}{2}\right)\left(x-\frac{2}{3}\right)\left(12x-6\right)\)
\(=12\left(x-\frac{3}{2}\right)\left(x-\frac{2}{3}\right)\left(x-\frac{1}{2}\right)\)