999 x 222 - 3333 + 21³ = 999 x 222 - 3333 + 9261 = 221778 - 3333 + 9261 = 218445 + 9261 = 209184

999x222-3333+21³=111x(9+2) -(3333-961)

                              =111X11-2372

                              =1221-2372

                              =-1151

Đáp án :

1+1

= 2

1+1

= 2

ti ck mik

Ta có : \(x^2-6x=0\) 

\(\Leftrightarrow x\left(x-6\right)=0\) 

\(\Leftrightarrow x=0\) hoặc \(x-6=0\) 

+, \(x=0\) 

+, \(x-6=0\) 

\(\Leftrightarrow x=6\) 

Vậy \(S=\left\{0;6\right\}\)

cho x^2-6x=0

x^2-6=0

x(x-6)=0

suy ra x=0 hoặc x=6

Ta có : \(\left|x\right|\ge0\) với mọi x

\(\Rightarrow\left|x\right|+2\ge0+2\) với mọi x

\(\Rightarrow\frac{\left|x\right|+2}{3}\ge\frac{2}{3}\) 

Hay \(C\ge\frac{2}{3}\) 

Dấu ''='' xảy ra khi :

 \(\left|x\right|=0\) 

\(\Rightarrow x=0\) 

Vậy GTNN của C là \(\frac{2}{3}\) khi x = 0

Ý tiếp theo làm tương tự

Nhớ t.i.c.k cho mình nha !

b) Ta có: \(|x|+10\ge10\forall x\)

\(\Rightarrow\frac{10}{|x|+10}\le1\forall x\)

\(\Rightarrow\frac{-10}{|x|+10}\ge-1\forall x\)

Dấu"=" xảy ra \(\Leftrightarrow x=0\)

Vậy Min D=-1 \(\Leftrightarrow x=0\)

\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}.\frac{48}{98}\)

\(A=\frac{8}{49}\)

A = \(\frac{1}{3}\).{ \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\)}

A = \(\frac{1}{3}\).{\(\frac{1}{2}-\frac{1}{98}\)}

A = \(\frac{1}{3}.\left\{\frac{49}{98}-\frac{1}{98}\right\}\)

A=\(\frac{1}{3}.\frac{24}{49}\)

A = \(\frac{49}{98}\)

\(1.x^2+x-6>0\)

\(\Leftrightarrow x^2-x+6x-6>0\)

\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(x+6\right)>0\)

TH1:\(\hept{\begin{cases}x-1>0\\x+6>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>1\\x>-6\end{cases}}\Leftrightarrow x>1}\)

TH2:\(\hept{\begin{cases}x-1< 0\\x+6< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 1\\x< -6\end{cases}\Leftrightarrow}x< -6}\)

\(2.x^2+7x+12\le0\)

\(\Leftrightarrow x^2+3x+4x+12\le0\)

\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\le0\)

TH1:\(\hept{\begin{cases}x+3\ge0\\x+4\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-3\\x\le-4\end{cases}\left(l\right)}}\)

TH2:\(\hept{\begin{cases}x+3\le0\\x+4\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le-3\\x\ge-4\end{cases}\Leftrightarrow}-4\le x\le-3\left(n\right)}\)

\(3.\) \(\left(x-2\right)\left(x+6\right)\left(2x+5\right)\le0\)

TH1:\(\hept{\begin{cases}x-2\ge0\\x+6\ge0\\2x+5\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge2\\x\ge-6\\x\le-\frac{5}{2}\end{cases}}}\left(l\right)\)

TH2:(loại)

TH3:\(\hept{\begin{cases}x-2\le0\\x+6\ge0\\2x+5\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le2\\x\ge-6\\x\ge-\frac{5}{2}\end{cases}\Leftrightarrow}-\frac{5}{2}\le x\le2}\)

Và còn nhiều TH khác nữa tự tìm nhé

\(4.\) \(\left(1-x\right)\left(x^2-6\right)>0\)

TH1:\(\hept{\begin{cases}1-x>0\\x^2-6>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 1\\x>\sqrt{6}\end{cases}\left(l\right)}}\)

TH2:\(\hept{\begin{cases}1-x< 0\\x^2-6< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>1\\x< \sqrt{6}\end{cases}\Leftrightarrow}1< x< \sqrt{6}\left(n\right)}\)

\(A=\sqrt{9-4\sqrt{5}}+\frac{1}{\sqrt{5}-2}=\sqrt{\left(\sqrt{5}-2\right)^2}+\frac{1}{\sqrt{5}-2}=\sqrt{5}-2+\frac{1}{\sqrt{5}-2}.\Leftrightarrow\) 

\(A=\frac{\left(\sqrt{5}-2\right)^2+1}{\sqrt{5}-2}=\frac{10-4\sqrt{5}}{\sqrt{5}-2}=\frac{\left(10-4\sqrt{5}\right)\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=10\sqrt{5}+20-20-8\sqrt{5}=\) 

\(=2\sqrt{5}\)