`(x+1)/(x^{3}+2x^{2}-4x-5)`

`=(x+1)/(x^{3}+x^{2}+x^{2}+x-5x-5)`

`=(x+1)/(x^{2}(x+1)+x(x+1)-5(x+1))`

`=(x+1)/((x+1)(x^{2}+x-5)`

`=1/(x^{2}+x-5)`

Chỗ này sao tìm đc Amax?

Đề thiếu rồi

\(A=4\left(x-1\right)+\dfrac{25}{x-1}+4\ge2\sqrt{\dfrac{100\left(x-1\right)}{x-1}}+4=24\)

\(A_{min}=24\) khi \(x=\dfrac{7}{2}\)

Hoặc là:

\(A=\dfrac{4x^2-4x+25}{x-1}=\dfrac{4x^2-28x+49+24\left(x-1\right)}{x-1}=\dfrac{\left(2x-7\right)^2}{x-1}+24\ge24\)

Biểu thức này chỉ có GTNN, không có GTLN

\(A=4\left(x-1\right)+\dfrac{25}{x-1}+4\ge2\sqrt{\dfrac{100\left(x-1\right)}{x-1}}+4=24\)

Dấu "=" xảy ra khi \(x=\dfrac{7}{2}\)

a) \(\text{5x(x-2)+(2-x)=0}\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\text{x(2x-5)-10x+25=0}\)

\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)

\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)

\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)

a) \(5x\left(x-2\right)+\left(2-x\right)=0\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(x\left(2x-5\right)-10x+25=0\)

\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(2x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{2}\end{matrix}\right.\)

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(x-\dfrac{9}{8}\right)\left(x+\dfrac{1}{8}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{9}{8}=0\\x+\dfrac{1}{8}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=-\dfrac{1}{8}\end{matrix}\right.\)

d) \(x^4+2x^2-8=0\)

\(\Rightarrow\left(x^4+2x^2+1\right)-9=0\)

\(\Rightarrow\left(x^2+1\right)^2-3^2=0\)

\(\Rightarrow\left(x^2+1-3\right)\left(x^2+1+3\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\end{matrix}\right.\) \(\Rightarrow x^2=2\) \(\Rightarrow x=\pm\sqrt{2}\)

vì \(\left(4x^2-4x+1\right)^{2022}\ge0\left(\forall x\right)\),\(\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}\ge0\left(\forall y\right)\),\(\left|x+y+z\right|\ge0\)

mà \(\left(4x^2-4x+1\right)^{2022}+\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)

=>\(\left\{{}\begin{matrix}4x^2-4x+1=0\\y^2+\dfrac{4}{5}y+\dfrac{4}{25}=0\\x+y-z=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-1=0\\y+\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\\dfrac{1}{2}-\dfrac{2}{5}-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

KL: vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

a  x = \(\dfrac{-1}{12}\)

b  x = \(\dfrac{-4}{3}\)

c  x = \(\dfrac{-1}{6}\)

d  x = \(\dfrac{-1}{4}\)

\(\left(4x+1\right)^2=\dfrac{4}{9}\)

\(\left(4x+1\right)=\perp\left(\dfrac{2}{3}\right)^2\)

\(\text{Vậy }4x+1=\dfrac{2}{3}\)

       \(4x\)        \(=\dfrac{2}{3}+\left(-1\right)=\dfrac{-1}{3}\)

        \(x\)         \(=\left(\dfrac{-1}{3}\right).\dfrac{1}{4}=\dfrac{-1}{12}\)

\(\text{hoặc }4x+1=\dfrac{-2}{3}\)

        \(4x\)        \(=\left(\dfrac{-2}{3}\right)+\left(-1\right)=\dfrac{-5}{3}\)

         \(x\)         \(=\left(\dfrac{-5}{3}\right).\dfrac{1}{4}=\dfrac{-5}{12}\)

\(\Rightarrow x\in\left\{\dfrac{-1}{12};\dfrac{-5}{12}\right\}\)

\(\left(3x-1\right)^2=25\)

\(\left(3x-1\right)^2=\perp\left(5\right)^2\)

\(\text{Vậy }3x-1=5\)

       \(3x\)        \(=5+1=6\)

        \(x\)         \(=6:3=2\)

\(\text{hoặc }3x-1=-5\)

        \(3x\)       \(=\left(-5\right)+1=-4\)

         \(x\)        \(=\left(-4\right):3=\dfrac{-4}{3}\)

\(\Rightarrow x\in\left\{2;\dfrac{-4}{3}\right\}\)

\(\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)

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

\(\text{Vậy }x-\dfrac{1}{3}=\dfrac{1}{2}\)

       \(x\)         \(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

\(\text{hoặc }x-\dfrac{1}{3}=\dfrac{-1}{2}\)

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

\(\Rightarrow x\in\left\{\dfrac{5}{6};\dfrac{-1}{6}\right\}\)

\(\left(4x-3\right)^2=16\)

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

\(\text{Vậy }4x-3=4\) 

        \(4x\)       \(=4+3=7\)

          \(x\)       \(=7:4=\dfrac{7}{4}\)

\(\text{hoặc }4x-3=-4\)

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

          \(x\)        \(=\left(-1\right):4=\dfrac{-1}{4}\)

\(\Rightarrow x\in\left\{\dfrac{7}{4};\dfrac{-1}{4}\right\}\)

a.

\(A=x^2+\dfrac{2021}{x}=x^2+\dfrac{2021}{2x}+\dfrac{2021}{2x}\ge3\sqrt[3]{\dfrac{2021^2}{4x^2}}=3\sqrt[3]{\dfrac{2021^2}{4}}\)

Dấu "=" xảy ra khi \(x=\sqrt[3]{\dfrac{2021}{3}}\)

b.

\(B=4\left(x-1\right)+\dfrac{25}{x-1}+4\ge2\sqrt{\dfrac{100\left(x-1\right)}{x-1}}+4=24\)

Dấu "=" xảy ra khi \(x=\dfrac{7}{2}\)

c.

\(C=3x+\dfrac{16}{x^3}=x+x+x+\dfrac{16}{x^3}\ge4\sqrt[4]{\dfrac{16x^3}{x^3}}=8\)

\(A_{min}=8\) khi \(x=2\)

d.

\(D=x+\dfrac{1}{x}=\left(\dfrac{x}{4}+\dfrac{1}{x}\right)+\dfrac{3}{4}.x\ge2\sqrt{\dfrac{x}{4x}}+\dfrac{3}{4}.2=\dfrac{5}{2}\)

Dấu "=" xảy ra khi \(x=2\)

e.

\(E=\dfrac{9\left(x-2\right)+18}{2-x}+\dfrac{2}{x}=2\left(\dfrac{1}{x}+\dfrac{9}{2-x}\right)-9\ge\dfrac{2.\left(1+3\right)^2}{x+2-x}-9=7\)

\(E_{min}=7\) khi \(x=\dfrac{1}{5}\)

f.

\(F=\dfrac{3}{1-x}+\dfrac{4}{x}\ge\dfrac{\left(\sqrt{3}+2\right)^2}{1-x+x}=7+4\sqrt{3}\)

Dấu "=" xảy ra khi \(x=4-2\sqrt{3}\)

1: Khi x=2 thì \(A=\dfrac{4\cdot2+1}{2-1}=9\)

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