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\(A=\left(x+3y-5\right)^2-6xy+26\)
\(=x^2+9y^2+25+6xy-10x-30y-6xy+26\)
\(=x^2-10x+25+9y^2-30y+25+1\)
\(=\left(x-5\right)^2+\left(3y-5\right)^2+1\)
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 \(\hept{\begin{cases}x=5\\y=\frac{5}{3}\end{cases}}\)
\(A=x^2-3x+5\)
\(=x^2-3x+\frac{9}{4}+\frac{11}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
\(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge\frac{11}{4}\)
Dấu "=" xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
Vậy Min A = \(\frac{11}{4}\Leftrightarrow x=\frac{3}{2}\)
a) \(A=x^2-3x+5\)
\("="\Leftrightarrow x=\frac{11}{4}\Rightarrow x=\frac{3}{2};\frac{11}{4}\)
b) \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\("="\Leftrightarrow x=5\Rightarrow x=0;5\)
c) \(C=4x-x^2+3\)
\("="\Leftrightarrow x=7\Rightarrow x=2;7\)
d) \(D=x^4+x^2+2\)
\("="\Leftrightarrow x=2\Rightarrow x=0;2\)
\(\left(x+7\right)^2+3\left(x+y\right)^2\)
\(=\left(x+7\right)^2+108\)
a, \(A=\left(100+50\right)^2=22500\)
b, \(B=\left(127+73\right)^2=40000\)
c, \(C=-6x+25\)Thay x = 100 ta có :
\(C=-6.100+25=-600+25=-575\)
\(A=100^2+200.50+50^2\)
\(=100^2+2.100.5+50^2\)
\(=\left(100+50\right)^2=150^2\)
\(B=127^2+146.127+73^2\)
\(=127^2+2.73.127+73^2\)
\(=\left(127+73\right)^2=200^2\)
a) \(4x\left(x-5\right)+3y\left(x-5\right)\)
\(=\left(x-5\right)\left(4x+3y\right)\)
b) \(x^2-2x-4y^2-4y\)
\(=\left[x^2-\left(2y\right)^2\right]-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
c) \(x^2+x-y^2+y\)
\(=\left(x^2-y^2\right)+\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+1\right)\)
d) \(3x^2+3y^2-6xy-12\)
\(=3\left(x^2+y^2-2xy-4\right)\)
\(=3\left[\left(x-y\right)^2-2^2\right]\)
\(=3\left(x-y-2\right)\left(x-y+2\right)\)
\(A=x^2+10x-37\)
\(=\left(x+5\right)^2-62\)
Có \(\left(x+5\right)^2\ge0\forall x\in R\)
\(\Rightarrow\left(x+5\right)^2-62\ge-62\forall x\in R\)
Dấu = xảy ra \(\Leftrightarrow x+5=0\Leftrightarrow x=-5\)
Vậy A đạt GTNN là -62 tại x=-5
Trả lời đc câu b chưa bạn
nếu rồi cho mình lời giải nha