a) \(3\left(2x-1\right)-2\left(1-x\right)=x+9\)

\(\Leftrightarrow6x-3-2+2x=x+9\)

\(\Leftrightarrow6x+2x-x=9+2+3\)

\(\Leftrightarrow7x=14\)

\(\Leftrightarrow x=2\)

b) \(-3\left(2x-1\right)-2\left(1-x\right)=x+9\left(1-x\right)\)

\(\Leftrightarrow-6x+3-2+2x=x+9-9x\)

\(\Leftrightarrow-6x+2x+9x-x=9+2-3\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\)

c) \(\left(1-x\right)\left(2x-1\right)-2\left(2-x\right)\left(2+x\right)=x+9\)

\(\Leftrightarrow2x-1-2x^2+x-8+2x^2=x+9\)

\(\Leftrightarrow-2x^2+2x^2+2x+x-x=9+8+1\)

\(\Leftrightarrow2x=18\)

\(\Leftrightarrow x=9\)

\(5x+3\frac{1}{5}-2=4x-21\frac{1}{2}\)

\(\Leftrightarrow5x+\frac{16}{5}-2=4x-\frac{43}{2}\)

\(\Leftrightarrow5x-4x=-\frac{43}{2}+2-\frac{16}{5}\)

\(\Leftrightarrow x=-\frac{227}{10}\)

\(\frac{x^2-x-6}{x-3}=\frac{x^2-3x+2x-6}{x-3}=\frac{x\left(x-3\right)+2\left(x-3\right)}{\left(x-3\right)}=x+2=0\Leftrightarrow x=-2\)

\(\frac{x^2+2x-\left(3x+6\right)}{x+2}=\frac{x\left(x+2\right)-3\left(x+2\right)}{x+2}=x-3=0\Leftrightarrow x=3\)

\(\frac{4}{x-2}-\left(x-2\right)=0\Leftrightarrow\frac{4}{a}-a=0\left(a=x-2\right)\Leftrightarrow\frac{4}{a}=a\Leftrightarrow a^2=4\Leftrightarrow a=\pm2\Leftrightarrow x=4\text{ hoặc 0}\)

a) ĐKXĐ: x \(\ne\)3

Ta có: \(\frac{x^2-x-6}{x-3}=0\)

<=> x² - x - 6 = 0

<=> x² - 3x + 2x - 6 = 0

<=> (x + 2)(x - 3) = 0

<=> \(\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-2\\x=3\left(vn\right)\end{cases}}\)

Vậy S = {-2}

b) ĐKXĐ: x \(\ne\)-2

Ta có: \(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x+2}=0\)

<=> \(x\left(x+2\right)-3\left(x+2\right)=0\)

<=> \(\left(x-3\right)\left(x+2\right)=0\)

<=> \(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\x=-2\left(vn\right)\end{cases}}\)

Vậy S = {3}

c) ĐKXĐ: x \(\ne\)2

Ta có: \(\frac{4}{x-2}-x+2=0\)

<=> \(\frac{4-\left(x-2\right)^2}{x-2}=0\)

<=> \(\left(2-x+2\right)\left(2+x-2\right)=0\)

<=> \(x\left(4-x\right)=0\)

<=> \(\orbr{\begin{cases}x=0\\4-x=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Vậy S = {0; 4}

1,2−(x−1,4)=−6(x+0,9)

<=> 1,2 - x + 1,4 = -6x -6.0,9

<=> 2,6 - x = -6x - 5,4

<=> 6x - x  + 2,6 + 5,4 =0 

<=> 5x + 8 = 0 

a) Với a = 5 thì b = 8

b) Nghiệm của phương trình là -8/5

\(\left(x^2-1+x\right)\left(x2-1+3x\right)-x\)

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

\(\Leftrightarrow5x^3-x^2-5x+1+5x^2-x-x\)

\(\Leftrightarrow5x^3+4x^2-7x+1\)

Mình đã rút gọn ngắn nhất có thể rồi đấy!

a)A=\(x^2-6x+2=x^2-2.3x+9-7\)\(=\left(x-3\right)^2-7\ge-7\)với mọi x\(\inℝ\)

Dấu bằng xảy ra\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy minA = - 7 tại x = 3

b)\(B=4x^2-x+2=4x^2-2.2x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+2\)

  \(=\left(2x-\frac{1}{4}\right)^2+\frac{31}{16}\ge\frac{31}{16}\)với mọi x\(\inℝ\)

Dấu bằng xảy ra \(\Leftrightarrow2x-\frac{1}{4}=0\Leftrightarrow x=\frac{1}{8}\)

Vậy minB = \(\frac{31}{16}\)tại \(x=\frac{1}{8}\)

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