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\(A=x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1>1\)(dương)
\(B=x^2+4x+6=x^2+2.x.2+2^2+2=\left(x+2\right)^2+2>2\)(dương)
\(C=x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>\frac{3}{4}\)(dương)
\(D=x^2+x+1=x^2+2x\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>\frac{3}{4}\)(dương)
\(E=x^2+3x+3=x^2+2.x.\frac{3}{2}+\frac{9}{4}+\frac{3}{4}=\left(x+\frac{3}{4}\right)^2+\frac{3}{4}>\frac{3}{4}\)(dương)
Bạn làm tương tự nhé
\(A=x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\)
Vậy \(A_{min}=1\Leftrightarrow x=-1\)
\(B=x^2+4x=6=x^2+4x+4+2=\left(x+2\right)^2+2\ge2>0\)
Vậy \(B_{min}=2\Leftrightarrow x=-2\)
\(A=3\left(x-\frac{5}{6}\right)^2+\frac{11}{12}\)
\(B=2\left(x-\frac{3}{4}\right)^2+\frac{23}{8}\)
\(C=\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\)
\(D=\left(x-5\right)^2+\left(3y+1\right)^2+4\)
\(E=\left(4x+1\right)^2+\left(y-2\right)^2+1\)
\(M=-\left(x+\frac{7}{2}\right)^2-\frac{11}{4}\)
\(N=-5\left(x-\frac{3}{5}\right)^2-\frac{41}{5}\)
\(C\) đề sai ví dụ \(x=3\Rightarrow C=2>0\)
\(D=-5\left(x-\frac{7}{10}\right)^2-\frac{131}{20}\)
a, \(E=4x^2+6x+5=4\left(x^2+\frac{2.3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+5\)
\(=4\left(x+\frac{3}{4}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)
Vậy ta có đpcm
b, \(F=2x^2-3x+7=2\left(x^2-\frac{2.3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+7\)
\(=2\left(x-\frac{3}{4}\right)^2+\frac{47}{8}\ge\frac{47}{8}>0\forall x\)
Vậy ta có đpcm
c, \(K=5x^2-4x+1=5\left(x^2-\frac{2.2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+1\)
\(=5\left(x-\frac{2}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}>0\forall x\)
Vậy ta có đpcm
d, \(Q=3x^2+2x+5=3\left(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}\right)+5\)
\(=3\left(x+\frac{1}{3}\right)^2+\frac{14}{3}\ge\frac{14}{3}>0\forall x\)
Vậy ta có đpcm
\(G=3x^2-5x+3\)
\(=x^2+x^2+x^2-2x-2x-x+1+1+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x^2-2x+1\right)+\left(x^2-2x+1\right)+\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x-1\right)^2+\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(=2\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có :
\(2\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
=> Biểu thức luôn dương với mọi x
Mk chỉ làm về dạng bình phương cộng( trừ ) một số thôi ,bn lại tự đánh giá nhé !
\(C=x^2-x+1\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(E=x^2+3x+3\)
\(=x^2+3x+\dfrac{9}{4}+\dfrac{3}{4}\)
\(=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{3}{4}\)
\(=\left[x^2+3.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{3}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\)
\(G=3x^2-5x+3\)
\(=x^2+x^2+x^2-2x-2x-x+1+1+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x^2-2x+1\right)+\left(x^2-2x+1\right)+\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x-1\right)^2+\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(K=4x^2+3x+2\)
\(=4x^2+3x+\dfrac{9}{16}+\dfrac{23}{16}\)
\(=\left(4x^2+3x+\dfrac{9}{16}\right)+\dfrac{23}{16}\)
\(=\left(2x+\dfrac{3}{4}\right)^2+\dfrac{23}{16}\)