Xe máy thứ nhất 1 giờ đi được 1/4 quảng đường

Xe máy thứ hai 1 giờ đi được 1/3 quảng đường

Sau 1,5 giờ 2 xe đi được:(1/4+1/3)x1,5=7/12x3/2=7/8(quảng đường)

quảng đường AB là:

15x8=120(km)

Xem lại đề đi bạnn

Trả lời đúng giúp mình.

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............................

k tớ trc ik tớ lm cho *hỳ hỳ*

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

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: \(\left|2x+1\right|=\pm22\)

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

     \(2x=21\)

      \(x=\frac{21}{2}\)

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

     \(2x=-23\)

     \(x=\frac{-23}{2}\) 

mk sai đề rồi k cần trl nữa đâu

a/ \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)+\dfrac{4}{5}=x+1\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-x=1+\dfrac{4}{5}\)

\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{9}{5}+\dfrac{5}{2}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{43}{10}\)

\(\Rightarrow x=\dfrac{43}{5}\)

b/ \(\dfrac{1}{6}\left(2x-3\right)=\dfrac{1}{2}\left(-x+\dfrac{1}{4}\right)-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{1}{3}x-\dfrac{1}{2}=-\dfrac{1}{2}x+\dfrac{1}{8}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{1}{3}x+\dfrac{1}{2}x=\dfrac{1}{8}-\dfrac{2}{3}+\dfrac{1}{2}\)

\(\Rightarrow\dfrac{5}{6}x=-\dfrac{1}{24}\Rightarrow x=-\dfrac{1}{20}\)

c/ làm như b

d/ \(\left(x-1\right)^4=\left(x-1\right)^6\)

\(\Rightarrow\left[{}\begin{matrix}x-1=-1\\x-1=0\\x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{0;1;2\right\}\)

Ta có bảng xét dấu:

Với \(x< 2;pt\Leftrightarrow2-x+3-x+4-x=2\)

\(\Leftrightarrow7-3x=0\Leftrightarrow x=\frac{7}{3}\left(l\right)\)

Với \(2\le x< 3;pt\Leftrightarrow x-2+3-x+4-x=2\)

\(\Leftrightarrow5-x=2\Leftrightarrow x=3\left(l\right)\)

Với \(3\le x< 4;pt\Leftrightarrow x-2+x-3+4-x=2\)

\(\Leftrightarrow x-1=2\Leftrightarrow x=3\left(tm\right)\)

Với \(x\ge4;pt\Leftrightarrow x-2+x-3+x-4=2\)

\(\Leftrightarrow3x-11=0\Leftrightarrow x=\frac{11}{3}\left(l\right)\)

Vậy pt có nghiệm duy nhất x = 3.