1 . Tìm x
2,8 :( 1/5 - 3x ) = \(1\frac{2}{5}\)
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\(2.8:\left(\frac{1}{5}-3x\right)=1\frac{2}{5}\)
\(\frac{14}{5}:\left(\frac{1}{5}-3x\right)=\frac{7}{5}\)
\(\frac{1}{5}-3x=\frac{14}{5}:\frac{7}{5}\)
\(\frac{1}{5}-3x=2\)
\(3x=\frac{1}{5}-2\)
\(3x=-\frac{9}{5}\)
\(x=-\frac{3}{5}\)
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\(2,8:\left(3.x-\frac{5}{8}\right)=2\frac{2}{5}\)
=>\(\frac{14}{5}:\left(3.x-\frac{5}{8}\right)=\frac{12}{5}\)
=>\(\frac{14}{5}:\frac{13}{5}=3.x-\frac{5}{8}\)
=>\(3.x-\frac{5}{8}=\frac{14}{13}\)
=>\(3.x=\frac{14}{13}+\frac{5}{8}\)
=>\(3.x=\frac{177}{104}\)
=>\(x=\frac{59}{104}\)
2,8 :( 3x -5/8)= 2/2/5
2,8:( 3x- 5,8)= 12/5
3x-5/8 = 2,8:12/5
3x -5/8 = 7/6
3x= 7/6 +5/8
3x= 43/ 24
x= 43/24 :3
x= 43/72
nha bn, mknghix vậy
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a) -0,6 . x - 7/3 = 5,4 (sửa dấu phẩy thành dấu nhân)
=> \(-\frac{3}{5}x-\frac{7}{3}=\frac{27}{5}\)
=> \(-\frac{3}{5}x=\frac{27}{5}+\frac{7}{3}=\frac{116}{15}\)
=> \(x=\frac{116}{15}:\left(-\frac{3}{5}\right)=\frac{116}{15}\cdot\left(-\frac{5}{3}\right)=-\frac{116}{9}\)
b) 2,8 : \(\left(\frac{1}{5}-3x\right)=1\frac{2}{5}\)
=> \(\frac{1}{5}-3x=2,8:1\frac{2}{5}\)
=> \(\frac{1}{5}-3x=\frac{14}{5}:\frac{7}{5}=\frac{14}{5}\cdot\frac{5}{7}=2\)
=> 3x = \(\frac{1}{5}-2=-\frac{9}{5}\)
=> \(x=\left(-\frac{9}{5}\right):3=\left(-\frac{9}{5}\right)\cdot\frac{1}{3}=-\frac{3}{5}\)
Nên sửa lại đề câu a đi nhá :))
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\(\frac{1}{5}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)
Đặt \(a=\left|3x-\frac{1}{5}\right|\) ta đc:
\(\frac{1}{5}-\frac{5}{2}a=\frac{2}{3}a-\frac{2}{3}\)
\(\Leftrightarrow-\frac{5}{2}a=\frac{2}{3}a-\frac{13}{15}\)
\(\Leftrightarrow-\frac{5a}{2}=\frac{10a}{15}-\frac{13}{15}\)
\(\Leftrightarrow-\frac{5a}{2}=\frac{10a-13}{15}\Rightarrow-5a\cdot15=2\left(10a-13\right)\)
\(\Leftrightarrow-75a=20a-26\)
\(\Leftrightarrow-95a=-26\Leftrightarrow a=\frac{26}{95}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{26}{95}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{5}=\frac{26}{95}\\3x-\frac{1}{5}=-\frac{26}{95}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x=\frac{9}{19}\\3x=-\frac{7}{95}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{19}\\x=-\frac{7}{285}\end{array}\right.\)
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\(2,8:\left(\frac{1}{5}-3x\right)=1\frac{2}{5}\)
\(\frac{14}{5}:\left(\frac{1}{5}-3x\right)=\frac{7}{5}\)
\(\frac{1}{5}-3x=2\)
\(3x=\frac{-9}{10}\)
\(x=\frac{-3}{10}\)
Vậy \(x=\frac{-3}{10}\)
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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
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a) \(\left|2x\right|-\left|-2,5\right|=\left|-7,5\right|\)
\(\Leftrightarrow\left|2x\right|-2,5=7,5\)
\(\Leftrightarrow\left|2x\right|=10\)
\(\Leftrightarrow\begin{cases}x\ge0\\2x=10\end{cases}\) hoặc \(\begin{cases}x< 0\\2x=-10\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge0\\x=5\left(tm\right)\end{cases}\) hoặc \(\begin{cases}x< 0\\x=-5\left(tm\right)\end{cases}\)
Vậy x={5;-5}
b)\(\left|3x\right|\cdot\left|-3,5\right|=\left|-2,8\right|\)
\(\Leftrightarrow\left|3x\right|\cdot3,5=2,8\)
\(\Leftrightarrow\left|3x\right|=\frac{4}{5}\)
\(\Leftrightarrow\begin{cases}x\ge0\\3x=\frac{4}{5}\end{cases}\) hoặc \(\begin{cases}x< 0\\3x=-\frac{4}{5}\end{cases}\)
\(\Leftrightarrow\begin{cases}x\ge0\\x=\frac{4}{15}\end{cases}\) hoặc \(\begin{cases}x< 0\\x=-\frac{4}{15}\end{cases}\)
Vậy x={4/15;-4/15}
c) \(\left(3x-5\right)\left(\frac{3}{2}x+2\right)\left(0,5x-10\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-5=0\\\frac{3}{2}x+2=0\\0,5x-10=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{3}\\x=-\frac{4}{3}\\x=20\end{array}\right.\)
a)|2x|-|-2,5|=|-7,5|
|2x|-2,5=7,5
|2x|=10
\(\Rightarrow\left[\begin{array}{nghiempt}2x=10\\2x=-10\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=5\\x=-5\end{array}\right.\)
Vậy x=5;-5
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a) \(\frac{3x-5}{x+4}=\frac{5}{2}\)
<=> 2(3x-5) = 5(x+4)
<=> 6x-10 = 5x+20
<=> x = 30
b) \(\frac{3x-1}{2x+1}=\frac{3}{7}\)
<=> 7(3x-1) = 3(2x+1)
<=> 21x-7 = 6x+3
<=>15x = 10
<=> x = \(\frac{2}{3}\)
\(\Rightarrow\) 1/5-3x = 2/8 : 7/5 = 5/28
\(\Rightarrow\) 3x = 1/5 - 5/28 = 3/140
\(\Rightarrow\) x = 3/140 : 3 = 1/140