|x-1|=2x-5

x= 4 nha bạn 

chúc bạn học tốt 

\(\left|x-1\right|=2x-5\)ĐK : x >= 5/2 

TH1 : \(x-1=2x-5\Leftrightarrow x=4\)(tm)

TH2 : \(x-1=5-2x\Leftrightarrow3x=6\Leftrightarrow x=2\)(ktm)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{a}{-5}=\frac{b}{-7}=\frac{c}{2}=\frac{a-b+c}{-5+7+2}=-\frac{28}{4}=-7\)

\(\Rightarrow a=35;b=49;c=-14\)

Trả lời:

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{a}{-5}=\frac{b}{-7}=\frac{c}{2}\)\(=\frac{a-b+c}{-5-\left(-7\right)+2}=\frac{-28}{4}=-7\)

\(\Rightarrow\hept{\begin{cases}a=-7.\left(-5\right)=35\\b=-7.\left(-7\right)=49\\c=-7.2=-14\end{cases}}\)

Vậy a = 35; b = 49; c = - 14

Ta có: 

x^4+2x^3+2x^2+1

=x^2(x^2+2x+2)+1

Ta thấy x^2(x^2+2x+2)> hoặc =0 nên 

x^2(x^2+2x+2)+1>0 nên ko có nghiệm

Chúc học tốt

= 11/16 . (107/13 - 23/13 + 7)

= 11/16 . 175/13 = 1925/208

\(8x-\left|4x+1\right|=x+2\Leftrightarrow\left|4x+1\right|=7x-2\)

ĐK : x >= 2/7 

TH1 : \(4x+1=7x-2\Leftrightarrow3x=3\Leftrightarrow x=1\)(tm) 

TH2 : \(4x+1=2-6x\Leftrightarrow10x=1\Leftrightarrow x=\frac{1}{10}\)(ktm)

\(8x-\left|4x+1\right|=x+2\)

\(\Leftrightarrow\left|4x+1\right|=7x+2\left(ĐK:x\ge-\frac{2}{7}\right)\)

\(\Leftrightarrow\orbr{\begin{cases}4x+1=7x+2\\4x+1=-7x-2\end{cases}\Leftrightarrow\orbr{\begin{cases}-3x=1\\11x=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\left(L\right)\\x=-\frac{3}{11}\left(TM\right)\end{cases}}}\)

Vậy:....

#H

a. ta có :\(\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{9}{9}=1\Rightarrow x^2=25\)

\(\orbr{\begin{cases}x=5\Rightarrow y=4\\x=-5\Rightarrow y=-4\end{cases}}\)

2.\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x^3}{27}=\frac{y^3}{64}=\frac{z^3}{125}=\frac{x^3+y^3-z^3}{27+64-125}=\frac{26}{17}\)

Vậy \(x=3\sqrt[3]{\frac{26}{17}},y=4\sqrt[3]{\frac{26}{17}},z=5\sqrt[3]{\frac{26}{17}}\)

3.\(\frac{x}{\frac{1}{8}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{2}}=\frac{x+y-z}{\frac{1}{8}+\frac{1}{3}-\frac{1}{2}}=-\frac{9}{-\frac{1}{24}}=216\) vậy \(\hept{\begin{cases}x=\frac{216}{8}=27\\y=\frac{216}{3}=72\\z=\frac{216}{2}=108\end{cases}}\)

4.\(\frac{x}{3}=\frac{1-y}{4}=\frac{z}{2}=\frac{3x+1-y-z}{3\times3+4-2}=\frac{11}{11}=1\)

Vậy \(x=3,y=-3,z=2\)

\(1.\)  \(P=15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)

       \(=\left(15\frac{1}{4}-25\frac{1}{4}\right)\cdot\left(-\frac{7}{5}\right)\)

       \(=\left(-10\right)\cdot\left(-\frac{7}{5}\right)\)

       \(=14\)

vậy P=14

\(2.\)   \(\left(\frac{21}{10}-|x+2|\right):\left(\frac{19}{10}-\frac{7}{5}\right)+\frac{4}{5}=1\)

           \(\Rightarrow\left(\frac{21}{10}-|x+2|\right):\frac{1}{2}+\frac{4}{5}=1\)

           \(\Rightarrow\left(\frac{21}{10}-|x+2|\right)\cdot2+\frac{4}{5}=1\)

          \(\Rightarrow\left(\frac{21}{5}-|x+2|\right)+\frac{4}{5}=1\)

         \(\Rightarrow\frac{21}{5}-|x+2|=\frac{1}{5}\)

         \(\Rightarrow|x+2|=4\)

         \(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}}\)

          \(\Rightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)

vậy  \(x\in\left\{2;-6\right\}\)

bài 1

ta có \(P=\left(15\frac{1}{4}-25\frac{1}{4}\right):\left(-\frac{5}{7}\right)=-10:\left(-\frac{5}{7}\right)=-10\times-\frac{7}{5}=14\)

2.\(\left(\frac{21}{10}-\left|x+2\right|\right):\left(\frac{19}{10}-\frac{14}{10}\right)+\frac{4}{5}=1\)

\(\Leftrightarrow\left(\frac{21}{10}-\left|x+2\right|\right):\frac{5}{10}=\frac{1}{5}\Leftrightarrow\frac{21}{10}-\left|x+2\right|=\frac{2}{5}\)

\(\Leftrightarrow\left|x+2\right|=\frac{21}{10}-\frac{2}{5}=\frac{17}{10}\Leftrightarrow\orbr{\begin{cases}x+2=\frac{17}{10}\\x+2=-\frac{17}{10}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{10}\\x=-\frac{37}{10}\end{cases}}}\)

A = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2017.2019

A = 1/2 (1 - 1/3 + 1/3 - 1/5 + 1/5 - ... - 1/2019)

A = 1/2 (1 - 1/2019)

A = 1/2 . 2018/2019

A = 1009/2019

@Cỏ

\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2017\cdot2019}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)=\frac{1}{2}\cdot\frac{2018}{2019}\)

\(=\frac{1009}{2019}\)