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a)
\(B=4x^2+4x+2\)
\(=4x^2+4x+1+1\)
\(=\left(2x+1\right)^2+1\)
Nhận thấy: \(\left(2x+1\right)^2\ge0\)
=> \(\left(2x+1\right)^2+1>0\)
hay B luôn dương
a)
A=\(x^2+5x+7=x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+7=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
C=\(3x^2-6x+5=\left[\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.\sqrt{3}+\left(\sqrt{3}\right)^2\right]-\left(\sqrt{3}\right)^2+5\ge2 \)
b)
C=\(-x^2+4x-5=-\left(x^2-4x+5\right)=-\left(x^2-4x+4+1\right)=-\left[\left(x-2\right)^2+1\right]\)
Ta có :\(\left(x-2\right)^2+1\ge1\Leftrightarrow-\left[\left(x-2\right)^2+1\right]\le\)-1
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
\(•B=4x-9x^2=-\left(9x^2-4x\right)\\ =-\left(9x^2-3x.2.\dfrac{2}{3}+\dfrac{4}{9}\right)+\dfrac{4}{9}\\ =-\left(3x-\dfrac{2}{3}\right)^2+\dfrac{4}{9}\le\dfrac{4}{9}\\dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{2}{9}\\ vậy\: MAX_B=\dfrac{4}{9}\: tại\: x=\dfrac{2}{9}\\ •C=5-2x-4x^2=-\left(4x^2+2x-5\right)\\ =-\left(4x^2+2.2x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{21}{4}\\ =-\left(2x+\dfrac{1}{2}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\\ dấu\: "="\: xảy\: ra\: khi\: x=-\dfrac{1}{4}\\ vậy\: MAX_C=\dfrac{21}{4}\: tại\: x=\dfrac{-1}{4}\\ •D=7+3x-x^2=-\left(x^2-3x-7\right)\\ =-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)+\dfrac{37}{4}\\ =-\left(x-\dfrac{3}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\\ dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{3}{2}\\ vậy\: MAX_D=\dfrac{37}{4}\: tại\: x=\dfrac{3}{2}\)\(•E=1+x-x^2=-\left(x^2-x-1\right)\\ =-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\\ dấu\:"="\:xảy\:ra\:khi\:x=\dfrac{1}{2}\\ vậy\:MAX_E=\dfrac{5}{4}\:tại\:x=\dfrac{1}{2}\\ •F=-5x-6x^2\\ -\dfrac{F}{6}=x^2+\dfrac{5}{6}x=x^2+2.x.\dfrac{5}{12}+\dfrac{25}{144}-\dfrac{25}{144}\\ -\dfrac{F}{6}=\left(x+\dfrac{5}{12}\right)^2-\dfrac{25}{144}\\ F=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{25}{24}\le\dfrac{25}{24}\\ dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{-5}{12}\\ vậy\: MAX_F=\dfrac{25}{24}\: tại\: x=\dfrac{-5}{12}\)
\(B=4x-9x^2=-9\left(x^2-\dfrac{4}{9}x+\dfrac{4}{81}\right)+\dfrac{4}{9}\)
\(=-9\left(x-\dfrac{2}{9}\right)^2+\dfrac{4}{9}\le\dfrac{4}{9}\forall x\)
vậy Max B = \(\dfrac{4}{9}\) khi \(x-\dfrac{2}{9}=0\Rightarrow x=\dfrac{2}{9}\)
\(C=5-2x-4x^2=-4\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)+\dfrac{21}{4}\)\(=-4\left(x+\dfrac{1}{4}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\)
Vậy Max C = \(\dfrac{21}{4}\) khi \(x+\dfrac{1}{4}=0\Rightarrow x=-\dfrac{1}{4}\)
\(D=7+3x-x^2\)
\(=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{37}{4}\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\forall x\)
Vậy Max D = \(\dfrac{37}{4}\) khi \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(E=1+x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{5}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\forall x\)
Vậy Max E = \(\dfrac{5}{4}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
\(F=-5x-6x^2=-6\left(x^2+\dfrac{5}{6}x+\dfrac{25}{144}\right)+\dfrac{25}{24}\)\(=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{25}{24}\le\dfrac{25}{24}\forall x\)
Vậy Max F = \(\dfrac{25}{24}\) khi \(x+\dfrac{5}{12}=0\Leftrightarrow x=-\dfrac{5}{12}\)
a: \(=-\left(x^2+10x-11\right)\)
\(=-\left(x^2+10x+25-36\right)\)
\(=-\left(x+5\right)^2+36< =36\)
Dấu '=' xảy ra khi x=-5
b: \(=-\left(x^2-6x+5\right)\)
\(=-\left(x^2-6x+9-4\right)\)
\(=-\left(x-3\right)^2+4< =4\)
Dấu '=' xảy ra khi x=3
c: \(=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)
Dấu '=' xảy ra khi x=1/2
d: \(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9< =9\)
Dấu '=' xảy ra khi x=-1
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)
Làm nốt
2/
\(T=8x^2-4x+\frac{1}{4x^2}+15\)
\(=\left(4x^2-4x+1\right)+\left(4x^2+\frac{1}{4x^2}-2\right)+16\)
\(=\left(2x-1\right)^2+\left(\frac{4x^2-1}{2x}\right)^2+16\ge16\)
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...