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làm tiếp cái trước(ấn nhầm)
\(x=\frac{5}{42}-\frac{15}{28}\)
\(x=\frac{5.4}{6.4.7}-\frac{15.6}{4.7.6}\)
\(x=\frac{20}{168}-\frac{90}{168}\)
\(x=\frac{-70}{168}\)
\(x=\frac{-5}{12}\)
2.
1.
\(\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{15}{28}-\frac{11}{13}\right)\)
\(\frac{11}{13}-\frac{5}{42}+x=-\frac{15}{28}+\frac{11}{13}\)
\(\frac{11}{13}-\frac{11}{13}-\frac{5}{42}+\frac{15}{28}=-x\)
\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
\(\left|x+\frac{4}{15}\right|=1,6\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{4}{15}=1,6\\x+\frac{4}{15}=-1,6\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{-28}{15}\end{cases}}}\)
│x+14/15│-3,75=-2,15
│x+14/15│=-2,15+3,75=1,6
Suy ra :
TH1 :x+14/15=1,6
x=2/3
TH2 :x+14/15=-1,6
x=-38/15
\(\left|x+\dfrac{14}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\left|x+\dfrac{14}{15}\right|-3,75=-2,15\)
\(\Leftrightarrow\left|x+\dfrac{14}{15}\right|=-2,15+3,75\)
\(\Leftrightarrow\left|x+\dfrac{14}{15}\right|=1,6\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{14}{15}=-1,6\\x+\dfrac{14}{15}=1,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-38}{15}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy....
Chúc các bạn học tốt !!!
\(B=\left|x+3\right|+\left|2-x\right|\ge\left|x+3+2-x\right|=\left|5\right|=5\)
Dấu "=" xảy ra khi \(x=0\)
Vậy \(B_{min}=5\Leftrightarrow x=0\)
a) \(\left|2,5-x\right|-1,3=0\)
th1: \(2,5-x\ge0\Leftrightarrow x\le2,5\)
\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow2,5-x-1,3=0\Leftrightarrow x=1,2\left(tmđk\right)\)
th2: \(2,5-x< 0\Leftrightarrow x>2,5\)
\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow x-2,5-1,3=0\Leftrightarrow x=3,8\left(tmđk\right)\)
vậy \(x=1,2;x=3,8\)
b) \(1,6.\left|x-0,2\right|=0\Leftrightarrow\left|x-0,2\right|=0\Leftrightarrow x-0,2=0\Leftrightarrow x=0,2\) vậy \(x=0,2\)
c) \(\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\)
th1: \(\dfrac{1}{3}-x\ge0\Leftrightarrow x\le\dfrac{1}{3}\)
\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow\dfrac{1}{3}-x-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{-2}{21}\left(tmđk\right)\)
th2: \(\dfrac{1}{3}-x< 0\Leftrightarrow x>\dfrac{1}{3}\)
\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow x-\dfrac{1}{3}-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{16}{21}\left(tmđk\right)\)
vậy \(x=\dfrac{-2}{21};x=\dfrac{16}{21}\)
d) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
th1: \(x+\dfrac{4}{15}\ge0\Leftrightarrow x\ge\dfrac{-4}{15}\)
\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow x+\dfrac{4}{15}-3,75=-2,15\)
\(\Leftrightarrow x=\dfrac{4}{3}\left(tmđk\right)\)
th2: \(x+\dfrac{4}{15}< 0\Leftrightarrow x< \dfrac{-4}{15}\)
\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow-x-\dfrac{4}{15}-3,75=-2,15\)
\(\Leftrightarrow x=\dfrac{-28}{15}\left(tmđk\right)\)
vậy \(x=\dfrac{4}{3};x=\dfrac{-28}{15}\)
e) ta có : \(\left|x-1,5\right|\ge0\forall x\) và \(\left|2,5-x\right|\ge0\forall x\)
\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|=0\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\) 2 giá trị này khác nhau \(\Rightarrow\) phương trình vô nghiệm
\(70:\frac{4x+720}{x}=\frac{1}{2}\)
\(\frac{4x+720}{x}=70:\frac{1}{2}\)
\(\frac{4x+720}{x}=140\)
\(4x+720=140x\)
\(4x-140x=-720\)
\(-136x=-720\)
\(x=\frac{-720}{-136}\)
\(x=\frac{90}{17}\)
\(\frac{70x}{4x+720}=\frac{1}{2}\)
\(140x=4x+720\)
\(136x=720\)
\(x=720:136\)
\(x=\frac{90}{17}\)
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=1.6=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
c: =>x-y=0 và y+9/25=0
=>x=y=-9/25
d: =>-1/3<x-3/5<1/3
=>4/15<x<14/15
e: =>|x+5,5|>5,5
=>x+5,5>5,5 hoặc x+5,5<-5,5
=>x>0 hoặc x<-11
\(\left|x+\frac{9}{2}\right|\ge0\)
\(\left|y+\frac{3}{4}\right|\ge0\)
\(\left|z+\frac{7}{2}\right|\ge0\)
Mà \(\left|x+\frac{9}{2}\right|+\left|y+\frac{3}{4}\right|+\left|z+\frac{7}{2}\right|\le0\)
\(\Rightarrow\hept{\begin{cases}x+\frac{9}{2}=0\\y+\frac{3}{4}=0\\z+\frac{7}{2}=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{9}{2}\\y=-\frac{3}{4}\\z=-\frac{7}{2}\end{cases}}}\)
\(\left|x+\frac{4}{15}\right|-\left|-3.75\right|=-\left|-2,15\right|\)
\(\Rightarrow\left|x+\frac{4}{15}\right|-\frac{15}{4}=-\frac{43}{20}\)
\(\Rightarrow\left|x+\frac{4}{15}\right|=-\frac{43}{20}+\frac{15}{4}\)
\(\Rightarrow\left|x+\frac{4}{15}\right|=\frac{8}{5}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=-\frac{8}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{28}{15}\end{cases}}\)
\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
\(\left|x+\frac{4}{15}\right|=1,6\)
=> \(x+\frac{4}{15}=1,6\) hoặc \(x+\frac{4}{15}=-1,6\)
=> \(x=\frac{4}{3}\) hoặc \(x=\frac{-28}{15}\)
Vậy..