a) (\(3\frac{1}{2}\)+2x)\(2\frac{2}{3}\)=\(5\frac{1}{5}\)
x=???
b) \(\frac{5}{13}\)+2x=\(\frac{3}{13}\)
x=???
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C
24 tháng 9 2019
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
24 tháng 9 2019
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) \(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)
\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)
\(7-4x-3x^2=25x-25\)
\(7-4x-3x^2-25x+25=0\)
\(32-29x-3x^2=0\)
\(3x^2+29x-30=0\)
\(3x^2+32x-3x-32=0\)
\(x\left(3x+32\right)-\left(3x+32\right)=0\)
\(\left(3x+32\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)
6 tháng 7 2019
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)
L
1
19 tháng 3 2018
\(a,x-\frac{x+1}{3}=\frac{2x+1}{5}\)
\(\Leftrightarrow\frac{15x}{15}-\frac{5\left(x+1\right)}{15}=\frac{3\left(2x+1\right)}{15}\)
\(\Leftrightarrow15x-5x-5=6x+3\)
\(\Leftrightarrow15x-5x-6x=3+5\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
Vậy pt có No là x = 2
b,\(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
\(\Leftrightarrow\frac{7\left(2x-1\right)}{21}-\frac{3\left(5x+2\right)}{21}=\frac{21\left(x+3\right)}{21}\)
\(\Leftrightarrow14x-7-15x-6=21x+273\)
\(\Leftrightarrow14x-5x-21x=273+7+6\)
\(\Leftrightarrow-22x=286\)
\(\Leftrightarrow x=-13\)
Vậy pt có No là x= -13
PT
1
8 tháng 4 2020
Bài làm
Ta có: x2 - 2x - 3 khác 0
<=> x2 + x - 3x - 3 khác 0
<=> x( x + 1 ) - 3( x + 1 ) khác 0
<=> ( x - 3 )( x + 1 ) khác 0
ĐKXĐ: \(x\ne3;x\ne-\frac{1}{2};x\ne-1\)
\(\frac{5}{x-3}-\frac{2x-1}{2x+1}-1=\frac{13}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{5\left(2x+1\right)\left(x+1\right)}{\left(x-3\right)\left(2x+1\right)\left(x+1\right)}-\frac{\left(2x-1\right)\left(x-3\right)\left(x+1\right)}{\left(x-3\right)\left(2x+1\right)\left(x+1\right)}-\frac{\left(x-3\right)\left(2x+1\right)\left(x+1\right)}{\left(x-3\right)\left(2x+1\right)\left(x+1\right)}=\frac{13\left(2x+1\right)}{\left(x-3\right)\left(2x+1\right)\left(x+1\right)}\)
\(\Rightarrow5\left(2x^2+2x+x+1\right)-\left(2x-1\right)\left(x^2+x-3x-3\right)-\left(x-3\right)\left(2x^2+2x+x+1\right)=26x+13\)
\(\Leftrightarrow10x^2+10x+5x+5-\left(2x^3+2x^2-6x^2-6-x^2-x+3x+3\right)-\left(2x^3+2x^2+x^2+x-6x^2-6x-3x-3\right)=26x+13\)
\(\Leftrightarrow10x^2+10x+5x+5-2x^3+2x^2-6x^2-6-x^2-x+3x+3-2x^3-2x^2-x^2-x+6x^2+6x+3x+3-26x-13=0\)
\(\Leftrightarrow\left(-2x^3-2x^3\right)+\left(10x^2-6x^2-x^2-2x^2-x^2+6x^2\right)+\left(10x+5x-x+3x-x+6x+3x-26x\right)+\left(5-6+3+3-13\right)=0\)
\(\Leftrightarrow-4x^3+8x^2+x-8=0\)
giải nốt đi ....
a)x=-31/40
b)x=-1/13