Rút gọn biểu thức
a)\(\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(a+c-b\right)^3-\left(a+b-c\right)^3\)
b)\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Giải CHI TIẾT nha
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Đặt x = a+b , y = b+c , z = c+a
Thì biểu thức trên trở thành \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy-3xyz\)
\(=\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Từ đó thay a,b,c vào rồi rút gọn :)
a,\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a^2b+b^2a+c^2a+ca^2+b^2c+c^2b\right)\)
Tương tự :
\(\left(b+c-a\right)^3=b^3+c^3-a^3+3\left(a^2b-b^2a+ca^2-ac^2+b^2c+c^2b\right)\)
\(\left(b+a-c\right)^3=b^3-c^3+a^3+3\left(a^2b+b^2a-ca^2+ac^2-b^2c+c^2b\right)\)
\(\left(a+c-b\right)^3=c^3+a^3-b^3+3\left(-a^2b+b^2a+ca^2+ac^2+b^2c-c^2b\right)\)
Biểu thức sau khi rút gọn ta được
24abc
b,\(\left(a+b\right)^3=a^3+b^3+3\left(a^2b+b^2a\right)\)
\(\left(c+b\right)^3=c^3+b^3+3\left(c^2b+b^2c\right)\)
\(\left(a+c\right)^3=a^3+c^3+3\left(a^2c+b^2c\right)\)
=>\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3=\)\(2\left(a^2+b^2+c^2\right)+3\left(a^2b+b^2a+c^2a+ca^2+b^2c+c^2b\right)\)
Lại có
\(3\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(3\left(a^2b+b^2a+c^2a+ca^2+b^2c+c^2b+2abc\right)\right)\)
Biểu thức khi đó trở thành
\(2\left(a^2+b^2+c^2\right)-6abc=2\left(a^2+b^2+c^2-3abc\right)\)
Tặng vk iu
Đặt \(a+b=x,b+c=y,c+a=z\) ( đặt cho dễ nhìn ý mà :)) ta có :
\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(=2\left(a+b+c\right)\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-\left(a+b\right)\left(b+c\right)-\left(b+c\right)\left(c+a\right)-\left(c+a\right)\left(a+b\right)\right]\)
\(=2\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Đặt \(\left(a+b;b+c;c+a\right)=\left(x;y;z\right)\)
\(x^3+y^3+z^3-3xyz=x^3+y^3+3xy\left(x+y\right)+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-yz-xz\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(=\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]\)
\(=\frac{1}{2}\left(2a+2b+2c\right)\left[\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\right]\)
\(=\left(a+b+c\right)\left[\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\right]\)
Phân tích mẫu thức thành nhân tử :
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+ac^2-bc^2\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)\)
\(=\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]=\left(b-c\right)\left(a-c\right)\left(a-b\right).\)
Do đó : \(A=\frac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Nhận xét : Nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz.\)
Đặt \(b-c=x,c-a=y,a-b=z\) thì \(x+y+z=0\)
Theo nhận xét trên : \(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3.\)
Tử:
(b - c)3 + (c - a)3 + (a - b)3
= (b - c + c - a + a - b)3 - 3(b - c + c - a)(b - c + a - b)(c - a + a - b)
= 0 - 3(b - a)(a - c)(c - b)
= 3(a - b)(a - c)(c - b)
Mẫu:
a2(b - c) + b2(c - a) + c2(a - b)
= a2(b - c) + b2c - ab2 + ac2 - bc2
= a2(b - c) - a(b2 - c2) + bc(b - c)
= a2(b - c) - a(b - c)(b + c) + bc(b - c)
= (b - c)(a2 - ab - ac + bc)
= (b - c)[a(a - b) - c(a - b)]
= (b - c)(a - b)(a - c)
\(A=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}\)
\(=\frac{3\left(c-b\right)}{b-c}\)
Phân tích mẫu \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-c^2b\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b+c\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)=\left(b-c\right)\left[a\left(a-c\right)-b\left(a-c\right)\right]\)
\(=\left(b-c\right)\left(a-b\right)\left(a-c\right)=-\left(b-c\right)\left(a-b\right)\left(c-a\right)\)
Đặt b - c = x, c - a = y, a - b = z
=> x + y + z = b - c + c - a + a - b = 0
Từ x+y+z=0 => x3+y3+z3=3xyz (tự c/m)
=>\(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3\)
Đặt x = a+b , y = b+c , z = c+a
Khi đó : \(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-yz-xz\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(=\frac{x+y+z}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)
Thay x,y,z bởi a,b,c vào và rút gọn :)
trình bày dài quá ; giờ chỉ cho cách làm thôi nha![hehe hehe](https://hoc24.vn/media/cke24/plugins/smiley/images/hehe.png)
dùng hằng đẳng thức : mũ 3
biền đổi
\(\left(a+b+c\right)^3=\left(a+\left(b+c\right)\right)^3\)
\(\left(b+c-a\right)^3=\left(b+\left(c-a\right)\right)^3\)
\(\left(a+c-b\right)^3=\left(a+\left(c-b\right)\right)^3\)
\(\left(a+b-c\right)^3=\left(a+\left(b-c\right)\right)^3\)
xong áp dụng hằng đẳng thức mũ 3
k có câu b ạ![lolang lolang](https://hoc24.vn/media/cke24/plugins/smiley/images/lolang.png)