Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(5x\left(x-4y\right)-4y\)
Thay vào ta được:
\(5\left(\frac{-1}{5}\right)[\left(\frac{-1}{5}\right)-4\left(\frac{-1}{2}\right)]-4\left(\frac{-1}{2}\right)\)
\(=-[\left(\frac{-1}{5}\right)-2]-2\)
\(=\left(\frac{1}{5}-2\right)-2\)
\(=\frac{11}{5}-2\)
\(=\frac{1}{5}\)
\(5x\left(x-4y\right)-4y=5x^2-20xy-4y\)
thay x= -1/5; y= -1/2 vào ta có:
\(5\left(-\frac{1}{5}\right)^2-20\left(-\frac{1}{5}\right)\left(-\frac{1}{2}\right)-4\left(-\frac{1}{2}\right)^2=\frac{5}{25}-\frac{20}{10}-\frac{4}{4}=\frac{1}{5}-2-1=\frac{1}{5}-\frac{15}{5}=-\frac{14}{5}\)
Ta có:(x-2y).(x2+2xy+4y2)-(x+y).(x2-xy-y2)
=x3-2x2y+2x2y+4xy2-8y3-x3-x2y+x2y+xy2+xy2
=6xy2-7y3.
1) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)\)
\(=x^3-16x-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^3-16x-x^4+1\)
b) \(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
\(=28xy-7x^2+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)
\(=-7x^2+7x\)
c) \(\left(3x-1\right)\left(2x-5\right)-4\left(2x^2-5x+2\right)\)
\(=6x^2-17x+5-4\left(2x^2-5x+2\right)\)
\(=6x^2-17x+5-8x^2+20x-8\)
\(=-2x^2+3x-3\)
a) x(x+4)(x-4)-(x2+1)(x2-1)
=>x(x2-42)-(x4-12)
=>x3-16x-x4+1
=>-x4-x3-15x
b) 7x(4y-x)+4y(y-7x)-2(2y2-3.5x)
=>28xy-7x2+4y2-28xy-4y2+30x
=>-7x2+30x
c) (3x+1)(2x-5)-4(2x2-5x+2)
=>6x2-15x+2x-5-8x2+20x-8
=>-2x2+7x-13
\(A=4x^2-y^2-2y-1\)
\(=\left(2x\right)^2-\left(y+1\right)^2\)
\(=\left(2x+y+1\right)\left(2x-y-1\right)\)
\(=-197\)
Vậy....
\(\left(X^2+2x+1\right)+\left(4y^2+\frac{4.1y}{4}+\frac{1}{16}\right)+2-\frac{1}{16}.\)
\(\left(x+1\right)^2+\left(2y+\frac{1}{4}\right)^2+\frac{15}{16}\ge\frac{15}{16}\)
\(x^2+4y^2+2x-y+2\)
\(=\left(x^2+2x+1\right)+\left[\left(2y\right)^2-2.2y.\frac{1}{4}+\left(\frac{1}{4}\right)^2\right]+\frac{15}{16}\)
\(=\left(x+1\right)^2+\left(2y-\frac{1}{4}\right)+\frac{15}{16}\)
Ta có: \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(2y-\frac{1}{4}\right)\ge0\forall y\end{cases}\Rightarrow\left(x+1\right)^2+\left(2y-\frac{1}{4}\right)+\frac{15}{16}\ge\frac{15}{16}}\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(2y-\frac{1}{4}\right)=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1=0\\2y-\frac{1}{4}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=\frac{1}{8}\end{cases}}}\)
Vậy GTNN của \(x^2+4y^2+2x-y+2=\frac{15}{16}\Leftrightarrow\hept{\begin{cases}x=-1\\y=\frac{1}{8}\end{cases}}\)
Tham khảo nhé~
x^2 + 4y^2 - 2x + 10+ 4xy - 4y
= (x^2 + 4xy + 4y^2) - (2x + 4y) + 10
= (x + 2y)^2 - 2 (x + 2y) + 10
Thay x + 2y = 5 vào biểu thức trên ,ta được :
5^2 - 2 . 5 + 10
= 25 - 10 + 10
= 25