Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu a : \(\left(2a-3b\right)^2-\left(2a+3b\right)^2\)
\(=\left(2a-3b+2a+3b\right)\left(2a-3b-2a-3b\right)\)
\(=4a.-6b=-24ab\)
Câu b : \(\left(a-2b-3c\right)^2-\left(a-2b+3c\right)^2\)
\(=\left(a-2b-3c+a-2b+3c\right)\left(a-2b-3c-a+2b-3c\right)\)
\(=\left(2a-4b\right).\left(-6c\right)\)
\(=2\left(a-2b-3c\right)\)
Lời giải:
a)
\((2a-5b)^2+(2a+5b)^2\)
\(=4a^2-2.2a.5b+25b^2+4a^2+2.2a.5b+25b^2\)
\(=8a^2+50b^2=2(4a^2+25b^2)\)
b)
\((a-2b-3c)^2-(a-2b+3c)^2\)
\(=[(a-2b-3c)-(a-2b+3c)][(a-2b-3c)+(a-2b+3c)]\)
\(=-6c(2a-4b)=12c(2b-a)\)
\(=\left(a-2b-3c-a+2b-3c\right)\left(a-2b-3c+a-2b+3c\right)\)
\(=-6c\cdot\left(2a-4b\right)\)
\(=-12c\left(a-2b\right)\)
\(8VT=4\left(a^2b+b^2c+c^2a+abc\right)\left(2ab^2+2bc^2+2ca^2+2abc\right)\le\left(a^2b+b^2c+c^2a+2ab^2+2bc^2+2ca^2+3abc\right)^2\)
\(\Rightarrow VT\le\frac{1}{32}\left(2a^2b+2b^2c+2c^2a+4ca^2+4ab^2+4bc^2+6abc\right)^2\)
\(\Rightarrow VT\le\frac{1}{32}\left(2a^2b+2b^2c+2c^2a+4ca^2+4ab^2+4bc^2+9abc\right)^2\)
\(\Rightarrow VT\le\frac{1}{32}\left[\left(a+2b\right)\left(b+2c\right)\left(c+2a\right)\right]^2\)
\(\Rightarrow VT\le\frac{1}{512}\left[\left(a+2b\right)\left(4b+8c\right)\left(c+2a\right)\right]^2\)
\(\Rightarrow VT\le\frac{1}{512}\left(\frac{a+2b+4b+8c+c+2a}{3}\right)^6=\frac{1}{512}\left(a+2b+3c\right)^6=\frac{4^6}{512}=8\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;1;0\right)\)
Do -2\(\le a\le5\)
=>(a+2)(a-5)\(\le0\)
=> a2-3a-10\(\le0\)
=>\(a^2\le3a+10\)
Tuong tu =>\(2b^2\le6b+20\)
\(3c^2\le9c+30\)
=>A\(\le3\left(a+2b+c\right)+60\le3.2+60=66\)
Dau ''='' xảy ra khi a=-2,b=5,c=-2 hoặc a=2,b=-5,c=2