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\(A=5x^2+2y^2+2xy-26x-16y+54\) \(=2\left(y^2+y\left(x-8\right)+\dfrac{\left(x-8\right)^2}{2}\right)-\dfrac{\left(x-8\right)^2}{2}+5x^2-26x+54\)
\(=2\left(y+\dfrac{x-8}{2}\right)^2+\dfrac{9}{2}x^2-18x+22\)
\(=2\left(y+\dfrac{x-8}{2}\right)^2+\dfrac{9}{2}\left(x-2\right)^2+4\ge4\)
Dấu '' = '' xảy ra khi: \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+\dfrac{x-8}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy: Min A = 4 tại \(x=2;y=3.\)
\(A=5x^2+2y^2+2xy-26x-16y+54\)
\(2A=4y^2+10x^2+4xy-52x-32y+108\)
\(2A=4y^2+4xy-32y-52x+10x^2+108\)
\(2A=\left(2y\right)^2+4y\left(x-8\right)+x^2-16x+64+9x^2-36x+44\)
\(2A=\left(2y\right)^2+2.2y\left(x-8\right)+\left(x-8\right)^2+\)\(9\left(x^2-4x+4\right)+8\)
\(2A=\left(2y+x-8\right)^2+9\left(x-2\right)^2+8\ge8\)
\(=>A\ge4\)
Để A nhỏ nhất thì \(x-2=0=>x=2;2y+x-8=0< =>2y-6=0=>y=3\)
Vậy ..................
\(\)
\(A=3x^2-5x+3=3(x^2-\frac{5}{3}x)+3\)
\(=3(x^2-\frac{5}{3}x+\frac{5^2}{6^2})+\frac{11}{12}=3(x-\frac{5}{6})^2+\frac{11}{12}\)
Vì \((x-\frac{5}{6})^2\geq 0, \forall x\Rightarrow A\geq 3.0+\frac{11}{12}=\frac{11}{12}\)
Vậy A(min)$=\frac{11}{12}$ khi $x=\frac{5}{6}$
\(B=2x^2+2x+1=2(x^2+x+\frac{1}{4})+\frac{1}{2}\)
\(=2(x+\frac{1}{2})^2+\frac{1}{2}\geq 2.0+\frac{1}{2}=\frac{1}{2}\)
Vậy \(B_{\min}=\frac{1}{2}\) tại \((x+\frac{1}{2})^2=0\Leftrightarrow x=\frac{-1}{2}\)
C)
\(C=2x^2+y^2+10x-2xy+27\)
\(=(x^2+10x+25)+(x^2+y^2-2xy)+2\)
\(=(x+5)^2+(x-y)^2+2\)
Vì \((x+5)^2\ge 0, (x-y)^2\geq 0\Rightarrow C\geq 0+0+2=2\)
Vậy \(C_{\min}=2\) tại \(\left\{\begin{matrix} (x+5)^2=0\\ (x-y)^2=0\end{matrix}\right.\Leftrightarrow x=y=-5\)
Bài 1 :
\(e,x^2+2xy+y^2-2x-2y+1\)
\(=\left(x+y-1\right)^2\)
Bài 2:
\(b,2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\left(x^2+1>0\right)\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
\(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)
\(=20x^3-10x^2+5x-20x^3+10x^2+4x\)
\(=9x\)
Thay x=15 \(\Rightarrow A=9.15=135\)
\(B=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)
\(=6x^2y^2-6xy^3-8x^3+8x^2y^2+5x^2y^2-5xy^3\)
\(=19x^2y^2-11xy^3-8x^3\)
Thay x=1/2 ; y=2 vào B \(\Rightarrow19.\left(\frac{1}{2}\right)^2.2^2-11\cdot\frac{1}{2}\cdot2^3-8\cdot\left(\frac{1}{2}\right)^3\)
\(=19-44-1\)
\(=-26\)
\(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(\Leftrightarrow\left(x+2\right)3x\)