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\(A=-x^2+2x+3=-\left(x^2-2x-3\right)\)
\(=-\left(x^2-2x+1-4\right)\)
\(=-\left[\left(x-1\right)^2-4\right]=-\left(x-1\right)^2+4\le4\)
Vậy \(A_{max}=4\Leftrightarrow x-1=0\Leftrightarrow x=1\)
\(B=-2x^2-4x=-2\left(x^2+2x\right)\)
\(=-2\left(x^2+2x+1-1\right)\)
\(=-2\left[\left(x+1\right)^2-1\right]=-\left(x+1\right)^2+2\le2\)
Vậy \(B_{max}=2\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
\(C=-x^2-6x+12=-\left(x^2+6x-12\right)\)
\(=-\left(x^2+6x+9-21\right)\)
\(=-\left[\left(x+3\right)^2-21\right]=-\left(x+3\right)^2+21\le21\)
Vậy \(C_{max}=21\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
\(D=-x^2+3x-1==-\left(x^2-3x+1\right)\)
\(=-\left(x^2-3x+\frac{9}{4}-\frac{5}{4}\right)\)
\(=-\left[\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\right]=-\left(x-\frac{3}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\)
Vậy \(D_{max}=\frac{5}{4}\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)
Bạn tự xét dấu "=" nhé, mình chỉ hướng dẫn cách tách thôi
a) \(A=5x^2-4x+1\)
\(A=5\left(x^2-\frac{4}{5}x+\frac{1}{5}\right)\)
\(A=5\left[x^2-2\cdot x\cdot\frac{2}{5}+\left(\frac{2}{5}\right)^2+\frac{1}{25}\right]\)
\(A=5\left[\left(x-\frac{2}{5}\right)^2+\frac{1}{25}\right]\)
\(A=5\left(x-\frac{2}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}\forall x\)
b) Tương tự đặt -9 ra ngoài rồi khai triển như câu a)
c) \(F=-2x^2-y^2+2xy+4x-40\)
\(F=-x^2-x^2-y^2+2xy+4x-40\)
\(F=-\left(x^2-2xy+y^2\right)-\left(x^2-4x+4\right)-36\)
\(F=-36-\left(x-y\right)^2-\left(x-2\right)^2\)
\(F=-36-\left[\left(x-y\right)^2+\left(x-2\right)^2\right]\le-36\forall x;y\)
2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
lê thị hương giang e ko nghĩ câu F đề sai đâu ạ! Chị check giúp xem em có tính sai hay ko nha!
2/ Ta có:
\(F=\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4+x^2+8x+16-20\)
\(=\left(x-y+2\right)^2+\left(x+4\right)^2-20\ge-20\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=-4\\y=-2\end{matrix}\right.\)
\(f\left(x\right)=\left(2x-3\right)^2-\left(3x+1\right)\left(3x-1\right)+5x+3\)
\(=4x^2-12x+9-9x^2+1+5x+3=-5x^2-7x+12\)
\(=-5\left(x^2+\frac{7}{5}x+\frac{49}{100}\right)+\frac{289}{20}\)
\(=-5\left(x+\frac{7}{10}\right)^2+\frac{289}{20}>0\)
=> Đa thức f(x) ko có nghiệm
Bài 2:
\(a,A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\)
Dấu = xảy ra \(\Leftrightarrow x=3\)
\(b,B=9x^2-6x-5=\left(9x^2-6x+1\right)-6\)
\(=\left(3x-1\right)^2-6\ge-6\)
Dấu = xảy ra \(\Leftrightarrow x=\frac{1}{3}\)
\(c,C=2x^2-5x+5\)
\(=2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{15}{8}\)
\(=2\left(x-\frac{5}{4}\right)^2+\frac{15}{8}\ge\frac{15}{8}\)
Dấu = xảy ra \(\Leftrightarrow x=\frac{5}{4}\)
\(d,D=x^2-4x^2+2023=\left(x^4-4x^2+4\right)+2019=\left(x^2-2\right)^2+2019\ge2019\)
Dấu = xảy ra \(\Leftrightarrow x^2=2\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
MinD = 2019 <=> \(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
\(e,E=5x^2-4xy+y^2+8x+1=\left(x^2+8x+16\right)\left(4x^2-4xy+y^2\right)-15\)
\(=\left(x+4\right)^2+\left(2x-y\right)^2-15\ge-15\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2x=2.\left(-4\right)=-8\end{matrix}\right.\)
Vậy \(Min_E=-15\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-8\end{matrix}\right.\)
\(f,F=2x^2-2xy+y^2+12x-4y\)
Bn kiểm tra lại đề
Cảm ơn các bạn nhé