CMR:
a) \(\left(a+b\right)^2=a^2+b^2+2ab\)
b)\(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\)
c) \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
a)\(\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3\)
b)\(\left(x+y\right)^5-x^5-y^5\)
c)\(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)
d)\(3abc+a^2\left(a-b-c\right)+b^2\left(b-a-c\right)+c^2\left(c-a-b\right)-c\left(b-c\right)\left(a-c\right)\)
e) 2bc(b+2c)+2ac(c-2a)-2ab(a+2b)-7abc
f)3bc(3b-c)-3ac(3c-a)-3ab(3a+b)+28abc
1. Cho a,b,c ≠0 thỏa mãn: (a+b+c)2=a2+b2+c2
Rút gọn:
\(M=\dfrac{a^2}{a^2+2bc}+\dfrac{b^2}{b^2+2ca}+\dfrac{c^2}{c^2+2ab}\)
2. Cho a+b+c=0
Rút gọn:
\(A=\dfrac{a^3+b^3+c^3-3abc}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}\)
Bài 1:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)
\(\dfrac{a^2}{a^2+2bc}=\dfrac{a^2}{a^2+bc-ab-ac}=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}\)
CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\end{matrix}\right.\)
\(M=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
Bài 2:
\(a^3+b^3+c^3-3abc=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3abc-3a^2b-3ab^2\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)(do \(a+b+c=0\))
\(\Rightarrow A=\dfrac{0}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}=0\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc\right)\)
\(=3\left[\left(a^2b+ab^2\right)+\left(a^2c+abc\right)+\left(ac^2+bc^2\right)+\left(b^2c+abc\right)\right]\)
\(=3\left[ab\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)+bc\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+b\right)\)
Châu ơi!đăng làm j z
PTĐTTNT:\(3abc+a^2\left(a-b-c\right)+b^2\left(b-a-c\right)+c^2\left(c-b-a\right)-c\left(b-c\right)\left(a-c\right)\)
\(=3abc+a^3-a^2b-a^2c+b^3-b^2a-b^2c+c^3-c^2b-c^2a-\left(abc-bc^2-c^2a+c^3\right)\)
\(=2abc+a^3-a^2b-a^2c+b^3-b^2c-b^2a\)
\(=\left(a^3+a^2b-a^2c\right)-\left(2a^2b+2ab^2-2abc\right)+\left(ab^2+b^3-b^2c\right)\)
\(=a^2\left(a+b-c\right)-2ab\left(a+b-c\right)+b^2\left(a+b-c\right)\)
\(=\left(a+b-c\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a+b-c\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a+b-c\right)\left(a-b\right)^2\) nha !
P/S:Ko có mục đích xấu,đăng lên cho bạn thôi.
Trả lời
Ở phần kết quả bạn vẫn chưa thu gọn hết đâu nha
\(=\left(a+b+c\right).\left(a-b\right)^2\)
Mk góp ý thôi mong mọi người đừng có đáp gạch đáp đá nha
Study well
\(P=\frac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}+\frac{b}{\sqrt{\left(c+1\right)\left(c^2-c+1\right)}}+\frac{c}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\)
\(\ge\frac{2a}{b^2+2}+\frac{2b}{c^2+2}+\frac{2c}{a^2+2}=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+2}+\frac{bc^2}{c^2+2}+\frac{ca^2}{a^2+2}\right)\)
\(=6-\left(\frac{2ab^2}{b^2+4+b^2}+\frac{2bc^2}{c^2+4+c^2}+\frac{2ca^2}{a^2+4+a^2}\right)\ge6-\left(\frac{2ab}{b+4}+\frac{2bc}{c+4}+\frac{2ca}{a+4}\right)\)
\(=6-\left(2a+2b+2c-\frac{8a}{b+4}-\frac{8b}{c+4}-\frac{8c}{a+4}\right)\)
\(=\frac{8a}{b+4}+\frac{8b}{c+4}+\frac{8c}{a+4}-6=\frac{8a^2}{ab+4a}+\frac{8b^2}{bc+4b}+\frac{8c^2}{ca+4c}-6\)
\(\ge\frac{8\left(a+b+c\right)^2}{\left(ab+bc+ca\right)+4\left(a+b+c\right)}-6\ge\frac{288}{\frac{\left(a+b+c\right)^2}{3}+24}-6=2\)
Cho a+b+c =1/2 Tính \(\dfrac{2ab+c}{\left(a+b\right)^2}.\dfrac{2bc+a}{\left(b+c\right)^2}.\dfrac{2ca+b}{\left(c+a\right)^2}\)
1/Cho các số thực dương chứng minh:\(\frac{3\left(a^4+b^4+c^4\right)}{\left(a^2+b^2+c^2\right)^2}+\frac{ab+bc+ca}{a^2+b^2+c^2}\ge2\)
2/Cho a,b dương.Chứng minh:\(\left(\frac{a}{b}+\frac{b}{a}\right)+4\sqrt{2}\frac{a+b}{\sqrt{a^2+b^2}}\ge10\)
3/ Cho các số thực dương. Chứng minh: \(\left(a^2+2bc\right)\left(b^2+2ca\right)\left(c^2+2ab\right)\ge abc\left(a+2b\right)\left(b+2c\right)\left(c+2a\right)\)
Phân tích đa thức thành nhân tử:
a) \(2bc\left(b+2c\right)+2ac\left(c-2a\right)-2ab\left(a+2b\right)-7abc\)
b) \(3bc\left(3b-c\right)-3ac\left(3c-a\right)-3ab\left(3a+b\right)+28abc\)
c) \(a\left(b^2+c^2\right)+b\left(a^2+c^2\right)+c\left(a^2+b^2\right)+2abc\)
Cô quản lí làm nhanh giúp e với ạ!!! BẠn e nhờ e gửi cho cô!!!
Cho a+b+c=0 và a,b,c khác 0.Rút gọn biểu thức
M=\(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b
ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
M=\(\frac{2ab}{a^2-a\left(b-c\right)}+\frac{2bc}{b^2-b\left(c-a\right)}+\frac{2ca}{c^2-c\left(a-b\right)}\)
M=\(\frac{2ab}{a\left(a-b+c\right)}+\frac{2bc}{b\left(b-c+a\right)}+\frac{2ca}{c\left(c-a+b\right)}\)
M=\(\frac{2ab}{-ab+\left(a+c\right)}+\frac{2bc}{-bc+\left(a+b\right)}+\frac{2ac}{-ac+\left(b+c\right)}\)
M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)
M=-1-1-1=-3
Vậy với a+b+c=0 thì M=-3