\(\left|2-x\right|+\left|x+1\right|=5\)

TH1 : \(\left|2-x\right|=\pm5\)

+ ) \(2-x=5\)

            \(x=2-5\)

           \(x=-3\)

+ ) \(2-x=\left(-5\right)\)

           \(x=2-\left(-5\right)\)

          \(x=7\)

TH2 : \(\left|x+1\right|=\pm5\)

+ ) \(x+1=5\)

             \(x=5-1\)

            \(x=4\)

+ ) \(x+1=\left(-5\right)\)

       \(x=\left(-5\right)-1\)

      \(x=-6\)

2 ) \(\left|x+1\right|+\left|2x+1\right|=22\)

TH1 : \(\left|x+1\right|=\pm22\)

+ ) \(x+1=22\)

      \(x=22-1\)

      \(x=21\)

+ ) \(x+1=-22\)

      \(x=-22-1\)

     \(x=-23\)

 TH2: \(\left|2x+1\right|=\pm22\)

+ ) \(2x+1=22\)

     \(2x=21\)

      \(x=\frac{21}{2}\)

+ ) \(2x+1=-22\)

     \(2x=-23\)

     \(x=\frac{-23}{2}\) 

+ Với x < -5 thì |x + 5| = -(x + 5) = -x - 5

=> -x - 5 = 4x + 1

=> -x - 4x = 1 + 5

=> -5x = 6

=> \(x=-\frac{6}{5}\), không thỏa mãn x < -5

+ Với \(x\ge-5\) thì |x + 5| = x + 5

=> x + 5 = 4x + 1

=> 4x - x = 5 - 1

=> 3x = 4

=> \(x=\frac{4}{3}\), thỏa mãn \(x\ge-5\)

Vậy \(x=\frac{4}{3}\)

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

\(=>\left[\begin{array}{nghiempt}x+5=4x+1\\x+5=-\left(4x+1\right)=-4x-1\end{array}\right.\)

\(=>\left[\begin{array}{nghiempt}3x=4\\5x=-6\end{array}\right.\)

\(=>\left[\begin{array}{nghiempt}x=\frac{4}{3}\\x=-\frac{6}{5}\end{array}\right.\)

với x>hoặc= -5

thì giả trị tuyệt đối của x+5=x+5

suy ra x+5=4x+1

4x-x=5-1

3x=4

suy rax=4/3 thỏ mãn x> hoặc bằng 5

ta rút gọn và nhân thì ta được (4\2-7\12-9\15):(4\3-1\2-5\3)
                                            =(120\60-35\60-36\60):(8\6-3\6-10\6)
                                            =49\60:-5\6
                                            = 49\-300
chúc bạn học tốt ! 

a) Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0\forall x\in Q\)

\(\left|y+\dfrac{2017}{2018}\right|\ge0\forall y\in Q\)

\(\left|z-2019\right|\ge0\forall x\in Q\)

\(\Rightarrow\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|\ge0\forall x,y,z\in Q\)

Dấu \("="\) xảy ra khi \(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\).

b) Lại có:

\(\left|x-\dfrac{9}{5}\right|\ge0\forall x\in Q\)

\(\left|y+\dfrac{3}{4}\right|\ge0\forall y\in Q\)

\(\left|z+\dfrac{7}{2}\right|\ge0\forall z\in Q\)

\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,zQ\)

Mà theo đề bài:

\(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\forall\)

\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{9}{5}\right|=0\\\left|y+\dfrac{3}{4}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)

Vậy .....

a) \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\)

Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0;\left|y+\dfrac{2017}{2018}\right|\ge0;\left|z-2019\right|\ge0\)

Để \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\) thì:

\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\)

Vậy............................

b) Ta có: \(\left|x-\dfrac{9}{5}\right|\ge0;\left|y+\dfrac{3}{4}\right|\ge0;\left|z+\dfrac{7}{2}\right|\ge0\)

Mà \(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\) thì:

\(\left|x-\dfrac{9}{5}\right|=\left|y+\dfrac{3}{4}\right|=\left|z+\dfrac{7}{2}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)

Vậy............................

help

:")))

a) \(x>3\Leftrightarrow\left|x-3\right|=x-3\)

\(x< 3\Leftrightarrow\left|x-3\right|=3-x\)

b) \(x>0\Leftrightarrow\left|-2x\right|=2x\)

\(x< 0\Leftrightarrow\left|-2x\right|=-2x\)

dễ mà

=\(\left(\dfrac{3}{2}-\dfrac{7}{12}-\dfrac{3}{5}\right):\left(\dfrac{4}{3}-\dfrac{1}{2}-\dfrac{5}{3}\right)\)

= \(\left(\dfrac{90}{60}-\dfrac{35}{60}-\dfrac{36}{60}\right):\left(\dfrac{80}{60}-\dfrac{30}{60}-\dfrac{100}{60}\right)\)

= \(\dfrac{19}{60}:\dfrac{-50}{60}\)

= \(\dfrac{19.60}{60.\left(-50\right)}\)

\(\dfrac{-19}{50}\)

TH1: x<-1

Pt sẽ là \(2-x-2\left(-x-1\right)=x\)

\(\Leftrightarrow2-x+2x+2=x\)

=>x+4=x(loại)

TH2: -1<=x<2

Pt sẽ là \(2-x-2\left(x+1\right)=x\)

\(\Leftrightarrow2-x-2x-2=x\)

=>x=0(nhận)

TH3: x>=2

Pt sẽ là \(x-2-2\left(x+1\right)=x\)

=>x-2-2x-2=x

=>-x-4=x

=>-2x=4

hay x=-2(loại)

a: \(3^{x+1}\cdot3=9^4\)

\(\Leftrightarrow3^{x+2}=3^8\)

=>x+2=8

hay x=6

c: \(\left|x+\dfrac{1}{2}\right|-\dfrac{5}{3}=1\)

=>|x+1/2|=8/3

=>x+1/2=8/3 hoặc x+1/2=-8/3

=>x=13/6 hoặc x=-19/6