Cái này mình giúp rồi nha<3

\(Q=\left(\dfrac{1}{x+1}+\dfrac{3\left(2x+1\right)}{x^3+1}-\dfrac{2}{x^2-x+1}\right):\left(x+2\right)\)\(=\left(\dfrac{x^2-x+1}{x^3+1}+\dfrac{3\left(2x+1\right)}{x^3+1}-\dfrac{2\left(x+1\right)}{x^3+1}\right).\dfrac{1}{x+2}\)\(=\left(\dfrac{x^2-x+1+6x+3-2x-2}{\left(x+1\right)\left(x-x+1\right)}\right)\dfrac{1}{x+2}\)

\(=\left(\dfrac{x^2+3x+2}{\left(x+1\right)\left(x^2-x+1\right)}\right)\dfrac{1}{x+2}\)

\(=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{1}{x+2}=\dfrac{1}{\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}}=\dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)Ta có:

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\) \(\Rightarrow\dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)

Vậy Max Q = \(\dfrac{4}{3}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

\(\frac{13+k}{29+k}=\frac{1}{3}\Rightarrow3.\left(13+k\right)=29+k\Rightarrow39+3k=29+k\)

\(3k-k=29-39\Rightarrow2k=-10\Rightarrow k=-5\)

a, \(A=9x^2-6x+5\)

\(=\left(9x^2-6x+1\right)+4\)

\(=\left(3x-1\right)^2+4\)

ta có:

\(\left(3x-1\right)^2\ge0\forall x\Rightarrow\left(3x-1\right)^2+4\ge4\forall x\)

Vậy Min A = 4

Để A = 4 thì \(3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(b,B=4x^2-5x\)

\(=\left(4x^2-5x+\dfrac{25}{16}\right)-\dfrac{25}{16}\)

\(=\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\)

TA có:

\(\left(2x-\dfrac{5}{4}\right)^2\ge\forall x\Rightarrow\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\ge-\dfrac{25}{16}\forall x\)Vậy Min B = \(-\dfrac{25}{16}\)

Để B = \(-\dfrac{25}{16}\) thì \(2x-\dfrac{5}{4}=0\Rightarrow2x=\dfrac{5}{4}\Rightarrow x=\dfrac{5}{8}\)

\(c,C=3x^2-6x\)

\(=3\left(x^2-2x+1\right)-3\)

\(=3\left(x-1\right)^2-3\)

Ta có:

\(3\left(x-1\right)^2\ge0\forall x\Rightarrow3\left(x-1\right)^2-3\ge-3\)

vậy Min C = -3

Để C = -3 thì x-1=0 => x = 1

\(d,D=5x^2-15x\)

\(=5\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{45}{4}\)

\(=5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\)

Ta có:

\(5\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\ge-\dfrac{45}{4}\)Vậy Min D = \(-\dfrac{45}{4}\)

Để \(D=-\dfrac{45}{4}\) thì \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)

\(e,E=x^2+3x+4\)

\(=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

Vậy Min E = \(\dfrac{7}{4}\) khi \(x+\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)

\(f,F=2x^2-4x+7\)

\(=2\left(x^2-2x+1\right)+5\)

\(=2\left(x-1\right)^2+5\ge5\forall x\)

Vậy Min F = 5 khi x - 1 =0 => x = 1

\(g,2x^2-3x=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{8}\)

\(=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\forall x\)

Vậy Min G = \(\dfrac{-9}{8}\) khi \(x-\dfrac{3}{4}=0\Rightarrow x=\dfrac{3}{4}\)

\(h,H=3x^2-4x=3\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{4}{3}\)

\(=3\left(x-\dfrac{2}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\forall x\)

Vậy Min H = \(-\dfrac{4}{3}\) khi \(x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

Sửa đề:

\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-6^2\)

\(=\left(x^2+5x\right)^2-36\)

Với mọi giá trị của x ta có:

\(\left(x^2+5x\right)^2\ge0\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)

Vậy Min A = -36

Để A = 36 thì \(x^2+5x=0\)

\(\Leftrightarrow x\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(a,x^3-16x=0\)

\(\Leftrightarrow x\left(x^2-16\right)=0\)

\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(b,x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)

\(\Leftrightarrow\left(x-2\right)x\left(x^2+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=0\\x^2+10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\\left[{}\begin{matrix}x^2=10\\x^2=-10\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\\x=\sqrt{10}\\x=-\sqrt{10}\end{matrix}\right.\)\(c,\left(2x-1\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow4x^2-4x+1=x^2+6x+9\)

\(\Leftrightarrow4x^2-4x+1-x^2-6x-9=0\)

\(\Leftrightarrow3x^2-10x-8=0\)

\(\Leftrightarrow3x^2-12x+2x-8=0\)

\(\Leftrightarrow3x\left(x-4\right)+2\left(x-4\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=4\end{matrix}\right.\)

Phần d tương tự

hờ hờ ns với vậy cho nick r h vẫn xin típ ak ==" hazzzz.

Sợ mấy cái bài kiểu này luôn:(

\(x^3+2x^2-4x+\dfrac{8}{3}=0\) (1)

Đặt t = \(x+\dfrac{2}{3}\Rightarrow x=t-\dfrac{2}{3}\)

KHi đó \(\left(1\right)\Leftrightarrow\left(t-\dfrac{2}{3}\right)^3+2\left(t-\dfrac{2}{3}\right)^2-4\left(t-\dfrac{2}{3}\right)+\dfrac{8}{3}=0\)\(t^3-\dfrac{16}{3}t+\dfrac{160}{27}=0\) (2)

Đặt \(y=\dfrac{t}{\dfrac{8}{3}}\Rightarrow t=\dfrac{8}{3}y\)

\(\Rightarrow\left(2\right)\Leftrightarrow4y^3-3y=-\dfrac{5}{32}\)

Đặt \(a=\sqrt[3]{-\dfrac{5}{32}+\sqrt{-\dfrac{5}{32}^2+1}}\) và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) ta được:

\(4\alpha^3-3\alpha=-\dfrac{5}{32}\)

\(\Rightarrow y=\dfrac{1}{2}\left(\sqrt[3]{-\dfrac{5}{32}+\sqrt{-\dfrac{5}{32}^2+1}}+\sqrt[3]{-\dfrac{5}{32}-\sqrt{-\dfrac{5}{32}^2+1}}\right)\)=-0,06155982175

\(\Rightarrow t=\dfrac{8}{3}y=-0,1641595247\)

\(\Rightarrow x=t-\dfrac{2}{3}=-0,8308261913\)