\(\left\{{}\begin{matrix}\left|2x-27\right|^{2011}\ge0\\\left(3y+10\right)^{2012}\ge0\end{matrix}\right.\Leftrightarrow\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0\)

Mà \(\left|2x-27\right|^{2017}+\left(3y+10\right)^{2012}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|2x-27\right|^{2011}=0\\\left(3y+10\right)^{2012}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=13,5\\y=\dfrac{-10}{3}\end{matrix}\right.\)

Vậy...

\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\\\left|x+4\right|\ge0\end{matrix}\right.\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\ge0\)

\(\Rightarrow8x\ge0\Rightarrow x\ge0\)

\(\Rightarrow x+1+x+2+x+3+x+4=8x\)

\(\Rightarrow4x+10=8x\)

\(\Rightarrow4x=10\)

\(\Rightarrow x=2,5\)

Vậy x = 2,5

\(A=\dfrac{x^3-7x+6}{x^3+5x^2-2x-24}\)

\(=\dfrac{x^3-2x^2+2x^2-4x-3x+6}{x^3+4x^2+x^2+4x-6x-24}\)

\(=\dfrac{x^2\left(x-2\right)+2x\left(x-2\right)-3\left(x-2\right)}{x^2\left(x+4\right)+x\left(x+4\right)-6\left(x+4\right)}\)

\(=\dfrac{\left(x^2+2x-3\right)\left(x-2\right)}{\left(x^2+x-6\right)\left(x+4\right)}\)

\(=\dfrac{\left(x^2+3x-x-3\right)\left(x-2\right)}{\left(x^2+3x-2x-6\right)\left(x+4\right)}\)

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

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

\(=\dfrac{x-1}{x+4}\)

Bài 2:

Áp dụng bất đẳng thức Cauchy-Shwarz dạng Engel có:

\(\dfrac{4}{x}+\dfrac{1}{4y}=\dfrac{4^2}{4x}+\dfrac{1}{4y}\ge\dfrac{\left(4+1\right)^2}{4\left(x+y\right)}=\dfrac{25}{5}=5\)

Dấu " = " khi x = y = \(\dfrac{5}{8}\)

2 ha

Mình trước !

1,

a, Ta có: \(\left|2,5-x\right|\ge0\Rightarrow A=\left|2,5-x\right|+5,8\ge5,8\)

Dấu " = " khi \(\left|2,5-x\right|=0\Rightarrow x=2,5\)

Vậy \(MIN_A=5,8\) khi x = 2,5

b, Ta có: \(-\left|x+\dfrac{2}{3}\right|\le0\Rightarrow B=2-\left|x+\dfrac{2}{3}\right|\le2\)

Dấu " = " khi \(-\left|x+\dfrac{3}{2}\right|=0\Rightarrow x=\dfrac{-3}{2}\)

Vậy \(MAX_B=2\) khi \(x=\dfrac{-3}{2}\)

2,

Giải:

Oy // AC \(\Rightarrow\widehat{O_1}+\widehat{C_2}=180^o\) ( 2 góc trong cùng phía )

\(\Rightarrow\widehat{C_2}=110^o\left(\widehat{O_1}=70^o\right)\)

AB // Ox \(\Rightarrow\widehat{A}+\widehat{C_2}=180^o\) ( 2 góc trong cùng phía )

\(\Rightarrow\widehat{A}=70^o\left(\widehat{C_2}=110^o\right)\)

Vậy...

a, \(\left(x-3\right)^{10}=\left(x-3\right)^{30}\)

\(\Leftrightarrow\left(x-3\right)^{30}-\left(x-3\right)^{10}=0\)

\(\Leftrightarrow\left(x-3\right)^{10}\left[\left(x-3\right)^{20}-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^{10}=0\\\left(x-3\right)^{20}-1=0\end{matrix}\right.\)

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

+) \(\left(x-3\right)^{20}-1=0\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

Vậy...

c, \(2^{x-1}+5.2^{x-2}=7\)

\(\Leftrightarrow2^{x-2}.2+5.2^{x-2}=7\)

\(\Leftrightarrow2^{x-2}\left(2+5\right)=7\)

\(\Leftrightarrow2^{x-2}=1\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy x = 2

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

\(\Leftrightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)

+) \(\left(x-1\right)^{x+2}=0\Leftrightarrow x=1\)

+) \(\left(x-1\right)^2-1=0\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

Vậy x = 1 hoặc x = 2 hoặc x = 0

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

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

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

\(\Leftrightarrow\dfrac{3x-1}{3x+1}=\dfrac{x+9}{x+3}\)

\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)=\left(x+9\right)\left(3x+1\right)\)

\(\Leftrightarrow3x^2+8x-3=3x^2+28x+9\)

\(\Leftrightarrow-20x=12\)

\(\Leftrightarrow x=\dfrac{-3}{5}\)

Vậy...

\(x^3+9x^2+26x+24\)

\(=x^3+4x^2+5x^2+20x+6x+24\)

\(=x^2\left(x+4\right)+5x\left(x+4\right)+6\left(x+4\right)\)

\(=\left(x^2+5x+6\right)\left(x+4\right)\)