Cho tam giác ABC. M là trung điểm AB. Trên cạnh AC lấy N sao cho AN = 4NC. MN cắt BC tại P. Tính tỉ số PC/PB
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a3 + b3=(a+b)(a2-ab+b2)
(a + b)3 =a3+b3+3ab(a+b)
a2 + b2=a2+2ab+b2
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}-\frac{1}{\left(a-b\right)\left(a^2+ab-c^2-ac\right)}\)
\(=\frac{1}{\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]}+\frac{1}{\left(c-a\right)\left[\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\right]}-\frac{1}{\left(a-b\right)\left[\left(a-c\right)\left(a+c\right)-b\left(a-c\right)\right]}\)
\(=\frac{c-a}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{a-b}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}-\frac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(a+b+c\right)}\)
\(=\frac{c-a}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{a-b}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{b-c}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)\(=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)
\(=\frac{0}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)
\(=0\)
\(2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)
\(=\left[\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]-4y^2\)
\(=\left(x+y-x+y\right)^2-\left(2y\right)^2\)
\(=\left(2y\right)^2-\left(2y\right)^2=0\)
Sửa:
\(2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)
\(=2\left(x^2-y^2\right)-\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)-4y^2\)
\(=2x^2-2y^2-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)
\(=2x^2-6y^2+4xy\)
\(=2\left(x^2-3y^2+2xy\right)\)
\(=2\left(x^2-3y^2+3xy-xy\right)\)
\(=2\left[x\left(x-y\right)+3y\left(x-y\right)\right]\)
\(=2\left(x-y\right)\left(3y+x\right)\)