a. Nếu \(x\ge1\)thì: \(\hept{\begin{cases}x+3>0\\x-1\ge0\end{cases}}\)\(\Rightarrow x+3+x-1< 6\Leftrightarrow2x< 4\Leftrightarrow x< 2\)(Loại)

  nếu \(x\le-3\)thì \(\hept{\begin{cases}x+3\le0\\x-1< 0\end{cases}}\)\(\Rightarrow-x-3+1-x< 6\Leftrightarrow-2x< 8\Leftrightarrow x>-4\)\(\Rightarrow-4< x\le-3\)

Nếu \(-3< x< 1\)thì: \(\hept{\begin{cases}x+3>0\\x-1< 0\end{cases}}\)\(\Rightarrow x+3+1-x< 6\Leftrightarrow4< 6\)(luôn đúng)

5+6=6+5

hổng có ai nhỉ ?

buồn......

:(

2x(x - 1) - (1 - x)

= 2x² - 2x - 1 + x

= 2x² - x - 1

Hok tốt~

\(2x\left(x-1\right)-\left(1-x\right)\)

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

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

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

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\(=\frac{x^4-x^2-3x^2+3}{x^4-x^2+7x^2-7}=\frac{x^2\left(x^2-1\right)-3\left(x^2-1\right)}{x^2\left(x^2-1\right)+7\left(x^2-1\right)}=\frac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\frac{x^2-3}{x^2+7}\)

HELP ME

a.\(\Leftrightarrow\left(x-1\right)^3+8-x^3+3x\left(x+2\right)=17\)

    \(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)

   \(\Leftrightarrow9x+7=17\)

   \(\Leftrightarrow9x=10\Leftrightarrow x=\frac{10}{9}\)

Lấy - 2(1) + 5(2) ta được

x + y = (√m) - 2 < - 1

<=> √m < 1

<=> m < 1

Còn m >= 0 thì là ĐKXĐ rồi nên tự làm nhá

a, Đặt \(A=2+x-x^2=-\left(x^2-x-2\right)=-\left(x^2-x+\frac{1}{4}-\frac{9}{4}\right)=-\left(x-\frac{1}{2}\right)^2+\frac{9}{4}\)

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow-\left(x-\frac{1}{2}\right)^2\le0\Rightarrow A=-\left(x-\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)

Dấu "=" xảy ra khi x = 1/2

Vậy Amax=9/4 khi x=1/2

b, Đặt \(B=4x^2-20x+26=\left(2x\right)^2-2.2x.5+25+1=\left(2x-5\right)^2+1\)

Vì \(\left(2x-5\right)^2\ge0\Rightarrow B=\left(2x-5\right)^2+1\ge1\)

Dấu "=" xảy ra khi x = 5/2

Vậy Bmin=1 khi x=5/2