Tìm x biết : \(\frac{2x-1}{-3}=\frac{3x+2}{5}\)
Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
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
tìm x biết: a)\(\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}\)
b)
\(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
c)\(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{2}{9-6x}-\frac{3}{2}\)
a) Đặt \(x-1=a\)
\(pt\Leftrightarrow\frac{13}{a}+\frac{5}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2}=2\)(vô lí)
Vậy pt vô nghiệm
a) \(\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{6}{3\left(x-1\right)}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2}=2\)
=> không có x thỏa mãn đề bài.
b) \(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)
\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)
\(7-4x-3x^2=25x-25\)
\(7-4x-3x^2-25x+25=0\)
\(32-29x-3x^2=0\)
\(3x^2+29x-30=0\)
\(3x^2+32x-3x-32=0\)
\(x\left(3x+32\right)-\left(3x+32\right)=0\)
\(\left(3x+32\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)
Tìm x,y biết:
\(\frac{x+1}{4}=\frac{2x-3}{5}=\frac{3x-2}{9y}\)
Ta có:\(\frac{x+1}{4}=\frac{2x-3}{5}=\frac{3x-2}{9y}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x+1}{4}=\frac{2x-3}{5}=\frac{3x-2}{9y}=\frac{x+1+2x-3}{4+5}=\frac{3x-2}{9}\)
Vì \(\frac{3x-2}{9y}=\frac{3x-2}{9}\Rightarrow9y=9\Rightarrow y=1\)
\(\Rightarrow\frac{x+1}{4}=\frac{3x-2}{9}\)
\(\Rightarrow9x+9=12x-8\)
\(9x-12x=-8-9\)
\(-3x=-17\)
\(x=\frac{17}{3}\)
Tìm x biết \(\frac{3}{1-2x}=\frac{-5}{3x-2}\)
\(\Rightarrow\)3 (3x-2) = -5 (1-2x)
\(\Rightarrow\)9x-6 = -5+10x
\(\Rightarrow\)-6+5 = 10x-9x
\(\Rightarrow\)-1 = x
Vậy x = -1
Tìm x, biết: \(7-\frac{1}{2}\left(2x-5\right)-\frac{2}{3}\left(3x+1\right)=5\)
=> 7-x+5/2-2x-2/3 = 5
=> 53/6 - 3x = 5
=> 3x = 53/6 - 5 = 23/6
=> x = 23/6 : 3 = 23/18
Vậy x = 23/18
Tk mk nha
Tìm x biết:
\(\frac{-1}{2}\times\left(3x-1\right)+\frac{3}{4}\left(3-2x\right)=-3\times\left(\frac{x}{2}-1\right)-\left(\frac{4}{5}\right)^{-1}\)
tìm x\(\in\) Q,biết:
a,\(\frac{x}{2}\left(\frac{3x}{5}\right)-\frac{13}{5}=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
b,\(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
c,\(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
d,\(\frac{13}{x-1}+\frac{5}{2x+2}-\frac{6}{3x-3}\)
e,\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-2\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
Ai bt câu nào làm câu đấy nhé ! Làm hết cũng đc nhé!
Tìm x,y,z,biết:
1)2x=3y=4z và 2x-5z=-6
2)\(\frac{x+3}{5}=\frac{y-2}{3}=\frac{z-1}{7}\) và 3x+5y-7z=32
TÌM X BIẾT \(\frac{X-1}{X^2-9X+20}+\frac{2X-2}{X^2-6X+8}+\frac{3X-3}{X^2-X-2}+\frac{4X-4}{X^2+6X+5}=0\)
\(\frac{x-1}{x^2-9x+20}+\frac{2x-2}{x^2-6x+8}+\frac{3x-3}{x^2-x-2}+\frac{4x-4}{x^2+6x+5}=0\)
\(\Leftrightarrow\frac{x-1}{\left(x-5\right)\left(x-4\right)}+\frac{2\left(x-1\right)}{\left(x-4\right)\left(x-2\right)}+\frac{3\left(x-1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{4\left(x-1\right)}{\left(x+1\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{10}{x^2-25}\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
PS: Điều kiện xác đinh bạn tự làm nhé