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Ix-1,7I = 2,3
TH1: x - 1,7 = 2,3
=> x = 2,3 + 1,7
=> x = 4
TH2 : x - 1,7 = -2,3
=> x = -2,3 + 1,7
=> x = -0,6
b) Ix + 3/4I - 1/3 = 0
=> Ix + 3/4I = 0 + 1/3
=> x + 3/4 = 1/3
=> x = 1/3 - 3/4
=> x = -5/12
a.
\(\left|x-1,7\right|=2,3\)
\(x-1,7=\pm2,3\)
TH1:
\(x-1,7=2,3\)
\(x=2,3+1,7\)
\(x=4\)
TH2:
\(x-1,7=-2,3\)
\(x=-2,3+1,7\)
\(x=-0,6\)
Vậy x = 4 hoặc x = -0,6
b.
\(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)
\(\left|x+\frac{3}{4}\right|=\frac{1}{3}\)
\(x+\frac{3}{4}=\pm\frac{1}{3}\)
TH1:
\(x+\frac{3}{4}=\frac{1}{3}\)
\(x=\frac{1}{3}-\frac{3}{4}\)
\(x=\frac{4-9}{12}\)
\(x=-\frac{5}{12}\)
TH2:
\(x+\frac{3}{4}=-\frac{1}{3}\)
\(x=-\frac{1}{3}-\frac{3}{4}\)
\(x=\frac{-4-9}{12}\)
\(x=-\frac{13}{12}\)
Vậy x = -5/12 hoặc x = -13/12.
a, => |5/3.x| = 1/6
=> 5/3.x = -1/6 hoặc 5/3.x = 1/6
=> x = -1/10 hoặc x = 1/10
Tk mk nha
a) |x -1,7| = 2,3 => x - 1,7 = 2,3 hoặc x - 1,7 = -2,3
Với x - 1,7 = 2,3 => x = 4
Với x - 1,7 = -2,3 => x= -0,6
Vậy x = 4 hoặc x = -0,6
b) 
=> 
Suy ra: 
Với 
Với 
1. a) Ta có: M = |x + 15/19| \(\ge\)0 \(\forall\)x
Dấu "=" xảy ra <=> x + 15/19 = 0 <=> x = -15/19
Vậy MinM = 0 <=> x = -15/19
b) Ta có: N = |x - 4/7| - 1/2 \(\ge\)-1/2 \(\forall\)x
Dấu "=" xảy ra <=> x - 4/7 = 0 <=> x = 4/7
Vậy MinN = -1/2 <=> x = 4/7
2a) Ta có: P = -|5/3 - x| \(\le\)0 \(\forall\)x
Dấu "=" xảy ra <=> 5/3 - x = 0 <=> x = 5/3
Vậy MaxP = 0 <=> x = 5/3
b) Ta có: Q = 9 - |x - 1/10| \(\le\)9 \(\forall\)x
Dấu "=" xảy ra <=> x - 1/10 = 0 <=> x = 1/10
Vậy MaxQ = 9 <=> x = 1/10
a) |x -1,7| = 2,3 => x - 1,7 = 2,3 hoặc x - 1,7 = -2,3
Với x - 1,7 = 2,3 => x = 4
Với x - 1,7 = -2,3 => x= -0,6
Vậy x = 4 hoặc x = -0,6
b) =>
Suy ra:
Với
Với
=>x-1,7=2,3 hoặc x-1,7=-2,3
=>x=2,3+1,7 hoặc x=-2,3+1,7
=>x=4 hoặc x= -0,6
vậy x=4 hoặc x=-0,6
b,\(|x+\dfrac{3}{4}|-\dfrac{1}{3}=0\)
\(|x+\dfrac{3}{4}|=\dfrac{1}{3}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-\dfrac{3}{4}\\x=-\dfrac{1}{3}-\dfrac{3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
vậy x=-5/12 hoặc x= -13/12
\(x+\left(\frac{1}{2}\right)^3=\frac{1}{4}\)
\(x+\frac{1}{8}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{8}\)
\(x=\frac{4}{16}-\frac{2}{16}\)
\(x=\frac{1}{8}\)
Vậy \(x=\frac{1}{8}\)
b) \(\left(\frac{2}{3}\right)^3-x=\frac{1}{3}\)
\(\frac{8}{27}-x=\frac{1}{3}\)
\(x=\frac{8}{27}-\frac{1}{3}\)
\(x=\frac{8}{27}-\frac{9}{27}\)
\(x=-\frac{1}{27}\)
Vậy \(x=-\frac{1}{27}\)
c) \(x.\left(-\frac{1}{2}\right)^4=\frac{3}{8}\)
\(x.\frac{1}{16}=\frac{3}{8}\)
\(x=\frac{3}{8}:\frac{1}{16}\)
\(x=\frac{3}{8}.16\)
\(x=6\)
c) \(\left(\frac{1}{2}\right)^3.x=\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^5:\left(\frac{1}{2}\right)^3\)
\(x=\left(\frac{1}{2}\right)^2\)
\(x=\frac{1}{4}\)
Vậy \(x=\frac{1}{4}\)
Chúc bạn học tốt !!!
a) \(x+\left(\frac{1}{2}\right)^3=\frac{1}{4}\Leftrightarrow x+\frac{1}{8}=\frac{1}{4}\Leftrightarrow x=\frac{1}{4}-\frac{1}{8}\Leftrightarrow x=\frac{1}{8}\)
b) \(\left(\frac{2}{3}\right)^3-x=\frac{1}{3}\Leftrightarrow\frac{8}{27}-x=\frac{1}{3}\Leftrightarrow-x=\frac{1}{3}-\frac{8}{27}\Leftrightarrow-x=\frac{1}{27}\Leftrightarrow x=-\frac{1}{27}\)
c) \(x.\left(\frac{-1}{2}\right)^4=\frac{3}{8}\Leftrightarrow x.\frac{1}{16}=\frac{3}{8}\Leftrightarrow x=\frac{3}{8}:\frac{1}{16}\Leftrightarrow x=6\)
d) \(\left(\frac{1}{2}\right)^2.x=\left(\frac{1}{2}\right)^5\Leftrightarrow\frac{1}{8}.x=\frac{1}{32}\Leftrightarrow x=\frac{1}{32}:\frac{1}{8}\Leftrightarrow x=\frac{1}{4}\)
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2\)
\(x=\frac{9}{49}\)
Vậy...
b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)
\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)
\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)
\(x=-\frac{1}{3}\)
Vậy...
c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
=>\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
Vậy...
d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)
=>\(x+\frac{1}{4}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
Vậy...
Phù, mãi mới xong, tk cho mk nha bn
a) \(\left|x-1,7\right|=2,3\)
\(\Rightarrow\orbr{\begin{cases}x-1,7=2,3\\x-1,7=-2,3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x=-0,6\end{cases}}\)
b) \(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{3}\\x+\frac{3}{4}=-\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{5}{12}\\x=-\frac{13}{12}\end{cases}}\)