sao co moi 1 cau vay ban.

\(F=\left|x\right|+\left|x+2\right|=\left|-x\right|+\left|x+2\right|\ge\left|-x+x+2\right|=2\)(Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\))Dấu "=" xảy ra \(\Leftrightarrow-x\left(x+2\right)\ge0\)
\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}-x\ge0\\x+2\ge0\end{cases}}\\\hept{\begin{cases}-x\le0\\x+2\le0\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}x\le0\\x\ge-2\end{cases}\Rightarrow x=0;-1;-2}\\\hept{\begin{cases}x\ge0\\x\le-2\end{cases}\Rightarrow x\in\varnothing}\end{cases}}\)

Vậy x = 0;-1;-2

cái chỗ giải -x(x+2) >=0 bạn tự giải làm 2 trường hợp: (-x>=0 và x+2>=0) hoặc (-x<=0 và x+2<=0)

a) 740:(x+10)=10²-2x13

=>740:(x+10)=100-2x13

=>740:(x+10)=1274

=>x=10,5808477237048(6)
6 là chu kì (có nghĩa là 6soos 6 kéo dài mãi mãi không giới hạn)
 

\(a,x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x-\frac{61}{8}=\frac{5}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)

\(b,x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x+\frac{43}{5}=\frac{37}{4}\)

=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)

\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)

=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)

=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)

d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)

=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)

=> \(\frac{98x}{198}=99\)

=>  98x = 99 . 198

=> 98x = 19602

=> x = 19602 : 98 = 9801/49

a) \(x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{71}{8}\)

b) \(x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x=\frac{37}{4}-\frac{61}{8}\)

=> \(x=\frac{13}{8}\)

c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)

=> \(x-\frac{61}{8}=3.\frac{1}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}\)

=> \(x=\frac{73}{8}\)

d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)

=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)

=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)

=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)

=> \(x.\frac{98}{99}=198\)

=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)

\(a,-12.\left(x-5\right)+7.\left(-x+3\right)=5\)

                  \(-12x+60-7x+21=5\)

                                        \(-19x+81=5\)

                                                   \(-19x=5-81\)

                                                   \(-19x=-76\)

                                                            \(x=4\)

\(b,30.\left(x+2\right)-6.\left(x-5\right)-24.x=100\)

             \(30x+60-6x+30-24x=100\)

                                                  \(0x+90=100\)

                                                             \(0x=100-90\)

                                                             \(0x=10\)

=> ko có giá trị nào thõa mãn x 

cong chua xinh xan oi,cam on nhieu nha

Lời giải:
a. $(x^2-9)(5x+15)=0$

$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$

$\Rightarrow x^2=9=3^2=(-3)^2$

$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$

$\Rightarrow x=-3$
b.

$x^2-8x=0$
$\Rightarrow x(x-8)=0$

$\Rightarrow x=0$ hoặc $x-8=0$

$\Rightarrow x=0$ hoặc $x=8$

c. 

$5+12(x-1)^2=53$

$12(x-1)^2=53-5=48$

$(x-1)^2=48:12=4=2^2=(-2)^2$

$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$

d.

$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$

$\Rightarrow x=11$ hoặc $x=-1$

e.

$(3x-5)^3=64=4^3$

$\Rightarrow 3x-5=4$

$\Rightarrow 3x=9$

$\Rightarrow x=3$

f.

$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$

$2^{4x}(1+8)=144$

$2^{4x}.9=144$

$2^{4x}=144:9=16=2^4$

$\Rightarrow 4x=4\Rightarrow x=1$

ta có: x/1.2+x/2.3+x/3.4+.....+x/9.10=9

x-x/2+x/2-x/3+x/3-......+x/9-x/10=9

x-x/10=9

=>x=10

a)   \(x-\frac{4}{5}=\frac{5}{7}\)

                  \(x=\frac{5}{7}+\frac{4}{5}=\frac{53}{35}\)

b)   \(5x=-\frac{1}{5}\)

        \(x=-\frac{1}{5}:5=-\frac{1}{25}\)

c)   \(\frac{5}{3}-x=7+\frac{4}{5}\)

      \(\frac{5}{3}-x=\frac{39}{5}\)

                  \(x=\frac{5}{3}-\frac{39}{5}=-\frac{92}{15}\)

d)    \(-\frac{5}{11}+2x=\frac{7}{22}\)

                        \(2x=\frac{7}{22}+\frac{5}{11}\)

                        \(2x=\frac{17}{22}\)

                          \(x=\frac{17}{22}:2\)

                           \(x=\frac{17}{44}\)

       \(x=-\frac{1}{5}:5\)

NÈ BẠN!!!

a) \(x-\frac{4}{5}=\frac{5}{7}\)

\(x=\frac{5}{7}+\frac{4}{5}=\frac{25}{35}+\frac{28}{35}=\frac{53}{35}\)

b) \(5x=-\frac{1}{5}+\frac{11}{5}\)

\(5x=2\)

\(x=\frac{2}{5}\)

c)\(\frac{5}{3}-x=7\)

\(x=\frac{5}{3}-7=\frac{5}{3}-\frac{21}{3}=-\frac{16}{3}\)

d) \(-\frac{5}{11}+2x=\frac{7}{22}\)

\(2x=\frac{7}{22}-\frac{-5}{11}=\frac{7}{22}-\frac{-10}{22}=\frac{17}{22}\)

\(x=\frac{17}{22}:2=\frac{17}{22}\cdot\frac{1}{2}=\frac{17}{44}\)

K CHO MÌNH NHA!!!