(x+2)² +x(x-1)<2x²+1
x²+4x+4+x²-x<2x²+1
3x+4<1
x< -1

a: =>a/b+2/25=1

=>a/b=23/24

b: =>a/b-5/6=1

=>a/b=11/6

a, \(\dfrac{a}{b}+\dfrac{2}{25}=1\Leftrightarrow\dfrac{a}{b}=1-\dfrac{2}{25}=\dfrac{23}{25}\)

b, \(\dfrac{a}{b}-\dfrac{5}{6}=1\Leftrightarrow\dfrac{a}{b}=1+\dfrac{5}{6}=\dfrac{11}{6}\)

a. 9x² - 6x - 3 = 0

<=> 3(3x² - 2x - 1) = 0

<=> 3(3x² - 3x + x - 1) = 0

<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)

<=> 3(3x + 1)(x - 1) = 0

<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)

b. (2x + 1)² - 4(x + 2)² = 9

<=> (2x + 1)² - \(\left[2\left(x+2\right)\right]^2=9\)

<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9

<=> -3(4x + 5) = 9

<=> 4x + 5 = -3

<=> 5 + 3 = -4x

<=> -4x = 8

<=> -x = 2

<=> x = -2

a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)

\(\Leftrightarrow\left(3x-1\right)^2-4=0\)

\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)

c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)

b) Thay x=-1 vào biểu thức \(B=\dfrac{2x^2+5x+4}{x^2-4x+3}\), ta được:

\(B=\dfrac{2\cdot\left(-1\right)^2+5\cdot\left(-1\right)+4}{\left(-1\right)^2-4\cdot\left(-1\right)+3}=\dfrac{2\cdot1-5+4}{1+4+3}=\dfrac{1}{8}\)

Vậy: Khi x=-1 thì \(B=\dfrac{1}{8}\)

Ta có:

|x| = \(\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{1}{3};x=-\dfrac{1}{3}\)

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

a.x^3-1^3

b.x^3-5^3

c)(2x)^3+3^3

d)x^3+1/2^3

câu a chưa đủ đề em hấy

c, \(x\)(\(x\) - 2022) + 4.(2022 - \(x\)) = 0

       (\(x\) - 2022).(\(x\) - 4) = 0

         \(\left[{}\begin{matrix}x-2022=0\\x+4=0\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=2022\\x=4\end{matrix}\right.\)

b, (\(x\) - 1)(\(x^2\) + \(x\) + 1) - \(x\)(\(x\) - 2)(\(x\) + 2)  = 7

     \(x^3\) - 1 - \(x\).(\(x^2\) - 4) = 7

      \(x^3\) - 1 - \(x^3\) + 4\(x\)  = 7

        (\(x^3\) - \(x^3\)) - 1 + 4\(x\) = 7

                       - 1 + 4\(x\) = 7

                               4\(x\) = 7 + 1

                               4\(x\) = 8

                                  \(x\) = 8:4

                                   \(x\) = 2

Câu 2:

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

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

a) Tính A(x) là sao em?

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

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

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

\(=3x^2+x-2\)

Câu 1:

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

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

\(=x+2\)

Bậc của M(x) là 1

Câu 3

a) \(3x\left(x^2-x+1\right)\)

\(=3x.x^2-3x.x+3x.1\)

\(=3x^3-3x^2+3x\)

b) \(\left(2x^2+x-1\right):\left(x+1\right)\)

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

\(=\left[\left(2x^2+2x\right)-\left(x+1\right)\right]:\left(x+1\right)\)

\(=\left[2x\left(x+1\right)-\left(x+1\right)\right]:\left(x+1\right)\)

\(=\left(x+1\right)\left(2x-1\right):\left(x+1\right)\)

\(=2x-1\)