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\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
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 \(x\in\left\{\frac{1}{4}\right\}\)
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)
\(=\frac{1}{x+3}-\frac{1}{x+34}\)
\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)
\(\Rightarrow x=31\)
Vậy, x = 31
Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với \(x,k\inℝ;x\ne0;x\ne-k\)
Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
1.a) \(\left(31\frac{6}{13}+5\frac{9}{41}\right)-36\frac{6}{13}=\left(31+\frac{6}{13}+5+\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)\)
\(=\left(36+\frac{6}{13}-\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)=\left(36+\frac{6}{13}\right)-\left(36+\frac{6}{13}\right)-\frac{9}{41}=-\frac{9}{41}\)
b) \(\frac{5}{3}+\left(-\frac{2}{7}\right)-\left(-1,2\right)-\left|1.4-0,2\right|\)
\(=\frac{5}{3}-\frac{2}{7}+1,2-1,2=\frac{29}{21}\)
c) \(0,25+\frac{3}{5}-\left(\frac{1}{8}-\frac{2}{5}+1\frac{1}{4}\right)+\left|\frac{3}{5}\right|\)
\(=\frac{1}{4}+\frac{3}{5}-\frac{1}{8}+\frac{2}{5}-1-\frac{1}{4}+\frac{3}{5}\)
\(=\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{3}{5}+\frac{2}{5}-1\right)+\frac{3}{5}-\frac{1}{8}=\frac{19}{40}\)
2) \(-\frac{3}{5}-x=0,75\)
=> \(-\frac{3}{5}-x=\frac{3}{4}\)
=> \(x=-\frac{3}{5}-\frac{3}{4}=\frac{-27}{20}\)
b) \(x+\frac{1}{3}=\frac{2}{5}-\left(-\frac{1}{3}\right)\)
=> \(x+\frac{1}{3}=\frac{2}{5}+\frac{1}{3}\)
=> \(x=\frac{2}{5}\)
c) |2x - 4| + 1 = 5
=> |2x - 4| = 4
<=> \(\orbr{\begin{cases}2x-4=4\\2x-4=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)
Giúp mình với nha cả nhả :<
Cả nhà làm vài ý thui cx được ạ :<
Vế trái: 4/(x+2).(x+6)+7/(x+6).(x+13)
<=>1/x+2 -1/x+6 +1/x+6 -1/x+13
<=>1/x+2-1/x+13
=> 1/x+2-1/x+13=2x+1/(x+2).(x+16) -3/(x+13).(x+16)
<=>1/x+2 - 1/x+13 + 1/x+13 - 1/x+16=2x+1/(x+2).(x+16)
<=>1/x+2 - 1/x+16=2x+1/(x+2).(x+16)
<=> 14/(x+2).(x+16)= 2x+1/(x+2).(x+16)
<=> 2x+1=14
<=> 2x=14-1
<=> 2x=13
<=> x=13:2
<=> x=13/2
Vậy x=13/2
Chúc bạn học tốt
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) \(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)
\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)
\(7-4x-3x^2=25x-25\)
\(7-4x-3x^2-25x+25=0\)
\(32-29x-3x^2=0\)
\(3x^2+29x-30=0\)
\(3x^2+32x-3x-32=0\)
\(x\left(3x+32\right)-\left(3x+32\right)=0\)
\(\left(3x+32\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)
b)\(\frac{x-1}{x+5}=\frac{6}{7}\)
\(\Leftrightarrow7x-7=6x+30\)
\(\Leftrightarrow7x-6x=30+7\)
\(\Leftrightarrow x=37\)