Tìm MIN
\(A=\left(x+3y-5\right)^2-6xy+26\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NT
0
26 tháng 8 2017
Ta có:\(A=x^2-4x\)
\(A=x^2-4x+4-4\)
\(A=\left(x-2\right)^2-4\le-4\)
Dấu = xảy ra khi x - 2 = 0 ; x = 2
Vậy Min A = - 4 khi x = 2
Ta có:\(B=x^2+x+1\)
\(B=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)
\(B=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu = xảy ra khi \(x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)
Vậy MIn B = 3/4 khi x=-1/2
26 tháng 8 2017
Ta có:\(C=\left(x+3y-5\right)^2-6xy+26\)
\(C=x^2+9y^2+25+6xy-10x-30y-6xy+26\)
\(C=x^2+9y^2-10x-30y+51\)
\(C=x^2-10x+25+9y^2-30y+25+1\)
\(C=\left(x-5\right)^2+\left(3y-5\right)^2+1\ge1\)
Dấu = xảy ra khi \(x-5=0;3y-5=0\Rightarrow x=5;y=\frac{5}{3}\)
Vậy Min C = 1 khi x=5;y=5/3
11 tháng 7 2017
a. \(2x\left(x-5\right)-x\left(2x+3\right)=26\Rightarrow2x^2-10x-2x^2-3x=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b. \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\frac{5}{2}\)
\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3=\frac{5}{2}\)\(\Rightarrow2y=\frac{7}{2}\Rightarrow y=\frac{7}{4}\)
c. \(2x^2+3\left(x+1\right)\left(x-1\right)=5x^2+5x\Rightarrow5x^2-3=5x^2+5x\)
\(\Rightarrow x=-\frac{3}{5}\)
PA
6 tháng 11 2016
\(A=\left(x+3y-5\right)^2-6xy+27\)
\(=x^2+9y^2+25+6xy-30y-10x-6xy+27\)
\(=x^2-10x+25+9y^2-30y+25+2\)
\(=\left(x-5\right)^2+\left(3y-5\right)^2+2\)
\(\left(x-5\right)^2\ge0\)
\(\left(3y-5\right)^2\ge0\)
\(\left(x-5\right)^2+\left(3y-5\right)^2+2\ge2\)
\(MinA=2\Leftrightarrow x=5;y=\frac{5}{3}\)
6 tháng 11 2016
\(A=\left(x+3y-5\right)^2-6xy+27\)
\(=x^2+9y^2+25+6xy-10x-30y-6xy+27\)
\(=\left(x^2-10x+25\right)+\left(9y^2-30y+25\right)+2\)
\(=\left(x-5\right)^2+\left(3y-5\right)^2+2\ge2\)
Dấu = khi \(\begin{cases}\left(x-5\right)^2=0\\\left(3y-5\right)^2=0\end{cases}\)\(\Leftrightarrow\)\(\begin{cases}x=5\\y=\frac{5}{3}\end{cases}\)
Vậy MinA=2 khi \(\begin{cases}x=5\\y=\frac{5}{3}\end{cases}\)
22 tháng 12 2023
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
NN
0
\(A=\left(x+3y-5\right)^2-6xy+26\)
\(=x^2+9y^2+25+6xy-10x-30y-6xy+26\)
\(=x^2-10x+25+9y^2-30y+25+1\)
\(=\left(x-5\right)^2+\left(3y-5\right)^2+1\)
Vì :
\(\left(x-5\right)^2\ge0\forall x\)
\(\left(3y-5\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-5\right)^2+\left(3y-5\right)^2+1\ge1\)
Dấu bằng xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-5\right)^2=0\\\left(3y-5\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{5}{3}\end{cases}}\)
Vậy \(A_{min}=1\) tại \(\hept{\begin{cases}x=5\\y=\frac{5}{3}\end{cases}}\)