Vì \(\left|x+1\right|\ge0\); \(\left|x+2\right|\ge0\); \(\left|x+3\right|\ge0\); \(\left|x+4\right|\ge0\)\(\forall x\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\ge0\)\(\forall x\)

mà \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=5x\)

\(\Rightarrow\)Phương trình có nghiệm \(\Leftrightarrow5x\ge0\)\(\Leftrightarrow x\ge0\)

Vì \(x\ge0\)\(\Rightarrow x+1>0\)\(\Rightarrow\left|x+1\right|=x+1\)

                          \(x+2>0\)\(\Rightarrow\left|x+2\right|=x+2\)

                         \(x+3>0\)\(\Rightarrow\left|x+3\right|=x+3\)

                         \(x+4>0\)\(\Rightarrow\left|x+4\right|=x+4\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=5x\)

\(\Leftrightarrow x+1+x+2+x+3+x+4=5x\)

\(\Leftrightarrow4x+10=5x\)

\(\Leftrightarrow x=10\)( thỏa mãn \(x\ge0\))

Vậy \(x=10\)

Làm ơn ghi đề giùm ạ , mình không nhìn được

\(\frac{5x-2}{-3}=\frac{2x+1}{2}\Leftrightarrow10x-4=-6x-3\)

\(\Leftrightarrow16x-1=0\Leftrightarrow x=\frac{1}{16}\)

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

\(\Leftrightarrow\frac{2\left(5x-2\right)}{-6}=\frac{-3\left(2x+1\right)}{-6}\)

\(\Leftrightarrow10x-4=-6x-3\)

\(\Leftrightarrow10x+6x=4-3\)

\(\Leftrightarrow16x=1\Leftrightarrow x=\frac{1}{16}\)

Vậy x = 1/16

\(0,\left(45\right)=0,45454545454545...=\frac{5}{11}\)    

\(0,4\left(54\right)=0,454545454545...=\frac{5}{11}\)   

Khi viết ra liên tục thì cả hai bên đều giống nhau 

còn nếu muốn dùng máy tính thì bấm alpha nút căn nha 

Vậy 0,(45) = 0,4(54) 

\(0,\left(45\right)=0.\left(01\right).45=\frac{1}{99}.45=\frac{45}{99}\)(1)

\(0,4\left(54\right)=\frac{1}{10}.4,\left(54\right)=\frac{1}{10}.\left(4+0,\left(54\right)\right)=\frac{1}{10}.4+\frac{1}{10}.0,\left(54\right)=\frac{2}{5}+\frac{1}{10}.0,\left(01\right).54\)\(=\frac{2}{5}+\frac{1}{10}.\frac{1}{99}.54=\frac{2}{5}+\frac{54}{990}=\frac{396}{990}+\frac{54}{990}=\frac{450}{990}=\frac{45}{99}\)(2)

+)Từ (1) và (2)

\(\Rightarrow0,\left(45\right)=0,4\left(54\right)\)

Chúc bạn học tốt

( x - 1 )² + 2x - 10 = 0

<=> x² - 2x + 1 + 2x - 10 = 0

<=> x² - 9 = 0

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

<=> x - 3 = 0 hoặc x + 3 = 0

<=> x = ±3

\(\left(x-1\right)^2+2x-10=0\)

\(x^2-2x+1+2x-10=0\)

\(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)

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

Ta có: \(\left|x-3\right|+\left|x-1\right|=\left|x-3\right|+\left|1-x\right|\ge\left|x-3+1-x\right|=\left|-2\right|=2\)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-3\right)\left(1-x\right)\ge0\)

TH1: \(\hept{\begin{cases}x-3\ge0\\1-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge3\\1\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le1\end{cases}}\)( vô lý )

TH2: \(\hept{\begin{cases}x-3\le0\\1-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le3\\1\le x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge1\end{cases}}\Leftrightarrow1\le x\le3\)

Vậy \(\left|x-3\right|+\left|x-1\right|=2\)\(\Leftrightarrow1\le x\le3\)