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bạn vào loigiaihay rồi chọn toán lớp 8 rồi chọn đẳng thức đáng nhớ
dễ mà áp dụng hết hằng đẳng thức nếu bạn thuộc hằng đẳng thức mik chỉ làm mỗi bài 1 ý nha xong dựa vô mà làm
\(1a.\left(2x+3y\right)^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2\)
\(=4y^2+12xy+9y^2\)
\(2a.x^2-6x+9\)
\(=x^2-2.x.3+3^2\)
\(=\left(x-3\right)^2\)
\(P=\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)+2xy}\)
\(=\frac{x^2+5x+y^2+5y+2xy-6}{x^2+6x+y^2+6y+2xy}\)
\(=\frac{\left(x+y\right)^2+5\left(x+y\right)-6}{\left(x+y\right)^2+6\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(x+y+5\right)-6}{\left(x+y\right)\left(x+y+6\right)}\)
\(=\frac{2005\times\left(2005+5\right)-6}{2005\times\left(2005+6\right)}\)
\(=\frac{2005\times2010-6}{2005\times2011}\)
\(=\frac{2004}{2005}\)
\(P=\frac{1}{x^2+y^2+z^2}+\frac{2009}{xy+yz+zx}=\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+zx}+\frac{1}{xy+yz+zx}+\frac{2007}{xy+yz+zx}\)
\(P\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}+\frac{2007}{\frac{1}{3}\left(x+y+z\right)^2}\)
\(P\ge\frac{9}{\left(x+y+z\right)^2}+\frac{6021}{\left(x+y+z\right)^2}=\frac{6030}{\left(x+y+z\right)^2}\ge\frac{6030}{3^2}=670\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Áp dụng BĐT Côsi dưới dạng engel, ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{x+y+z}\)
⇒\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)\ge\left(x+y+z\right).\frac{9}{x+y+z}\) = 9
Dấu "=" xảy ra ⇔ x = y = z
a) Rút gọn :
Ta có : \(A=\frac{y-x}{xy}:\left[\frac{y^2}{\left(x-y\right)^2}-\frac{2x^2y}{\left(x^2-y^2\right)^2}+\frac{x^2}{y^2-x^2}\right]\)
\(=\frac{y-x}{xy}:\left[\frac{y^2\left(x+y\right)^2-2x^2y-x^2\left(x^2-y^2\right)}{\left(x^2-y^2\right)^2}\right]\)
\(=\frac{y-x}{xy}:\left[\frac{y^2\left(x^2+2xy+y^2\right)-2x^2y-x^4+x^2y^2}{\left(x^2-y^2\right)^2}\right]\)
...
TXD : \(\hept{\begin{cases}y\left(x+y\right)\ne0\\\left(x+y\right)x\ne0\\\left(x-y\right)\left(x+y\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne y\\x\ne-y\\xy\ne0\end{cases}}}\)
Câu b :
\(A=\frac{xy-\left(x+y\right)y}{xy\left(x+y\right)}:\frac{y^2+x\left(x-y\right)}{x\left(x^2-y^2\right)}:\frac{x}{y}\)
\(=\frac{x^2-xy+y^2}{xy\left(x+y\right)}.\frac{x\left(x-y\right)\left(x+y\right)}{x^2-xy+y^2}.\frac{y}{x}\)\(=1-\frac{y}{x}\)
Để \(A>1\)mà \(y< 0\)nên \(x\)và \(y\)phải cùng dấu \(\Rightarrow x< 0\)
\(b.=\frac{1\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{1\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{1c-1a+1a-1b+1b-1c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=-\frac{2b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Ta có A = 2018.2020 + 2019.2021
= (2020 - 2).2020 + 2019.(2019 + 2)
= 20202 - 2.2020 + 20192 + 2.2019
= 20202 + 20192 - 2(2020 - 2019) = 20202 + 20192 - 2 = B
=> A = B
b) Ta có B = 964 - 1= (932)2 - 12
= (932 + 1)(932 - 1) = (932 + 1)(916 + 1)(916 - 1) = (932 + 1)(916 + 1)(98 + 1)(98 - 1)
= (932 + 1)(916 + 1)(98 + 1)(94 + 1)(94 - 1)
= (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1)(92 - 1)
(932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1).80
mà A = (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1).10
=> A < B
c) Ta có A = \(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x^2-y^2}{x^2+2xy+y^2}< \frac{x^2-y^2}{x^2+xy+y^2}=B\)
=> A < B
d) \(A=\frac{\left(x+y\right)^3}{x^2-y^2}=\frac{\left(x+y\right)^3}{\left(x+y\right)\left(x-y\right)}=\frac{\left(x+y\right)^2}{x-y}=\frac{x^2+2xy+y^2}{x-y}< \frac{x^2-xy+y^2}{x-y}=B\)
=> A < B