kb lqmb vs mk ko mk là P.A.D8a1

Làm hết thì đã lên " THÁNH "

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) \(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)

\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)

\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)

\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)

\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)

\(7-4x-3x^2=25x-25\)

\(7-4x-3x^2-25x+25=0\)

\(32-29x-3x^2=0\)

\(3x^2+29x-30=0\)

\(3x^2+32x-3x-32=0\)

\(x\left(3x+32\right)-\left(3x+32\right)=0\)

\(\left(3x+32\right)\left(x-1\right)=0\)

\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)

\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)

lên mạng chứ 0 tự làm ak

\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)

=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)

=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)

=> \(-\frac{3}{4}+\left(-2x\right)=-2\)

=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)

=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)

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 \(x\in\left\{\frac{1}{4}\right\}\)

Đây là những bài cơ bản mà bạn!

\(\frac{5x-2}{3}=\frac{5-3x}{2}\)

\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)

\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)

\(< =>10x-4=15-9x\)

\(< =>10x+9x=15+4\)

\(< =>19x=19< =>x=1\)

\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)

\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)

\(< =>30x+9=36+24+32x\)

\(< =>32x-30x=9-36-24\)

\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)

1. \(\frac{3x-7}{5}=\frac{2x-1}{3}\)

<=> 3(3x-7)=5(2x-1)

<=> 9x-21=10x-5

<=> -21+5=10x-9x

<=> x=-16

2. \(\frac{3x-7}{2}+\frac{2x-1}{3}=-16\)

<=> \(\frac{3\left(3x-7\right)}{6}+\frac{2\left(2x-1\right)}{6}=\frac{-96}{6}\)

=> 9x-21+4x-2=-96

<=> 13x-23=-96

<=> 13x=-73

<=> x=\(\frac{-73}{13}\)

3. \(x-\frac{x+1}{3}=\frac{2x+1}{5}\)

<=> \(\frac{15x}{15}-\frac{5\left(x+1\right)}{15}=\frac{3\left(2x+1\right)}{15}\)

=> 15x-5x-5=6x+3

<=> 15x-5x-6x=3+5

<=> 4x=8

<=> x=2

4. \(\frac{7-3x}{12}+\frac{3}{4}=2\left(x-2\right)+\frac{5-\left(5-2x\right)}{6}\)

<=>\(\frac{7-3x}{12}+\frac{9}{12}=\frac{24\left(x-2\right)}{12}+\frac{2\left[5-\left(5-2x\right)\right]}{12}\)

=> 7-3x+9=24x-48+4x

<=> -3x-24x-4x=-48-7

<=> -31x=-55

<=> x= \(\frac{55}{31}\)

5. \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)

<=> \(\frac{7\left(2x-1\right)}{21}-\frac{3\left(5x+2\right)}{21}=\frac{21\left(x+13\right)}{21}\)

=> 14x-7-15x-6=21x+273

<=> 14x-15x-21x=273+7+6

<=> -22x=286

<=> x= -13

a/\(\Leftrightarrow3\left(3x-7\right)=5\left(2x-1\right)\Leftrightarrow9x-21=10x-5\Leftrightarrow x=-16\)

b/\(\Leftrightarrow\frac{9x-21+4x-2}{6}=-16\)\(\Leftrightarrow13x-23=-96\Leftrightarrow x=x=-\frac{73}{13}\)

c/\(\Leftrightarrow\frac{3x-x+1}{3}-\frac{2x+1}{5}=0\Leftrightarrow\left(2x+1\right)\left(\frac{1}{3}-\frac{1}{5}\right)=0\Leftrightarrow x=-\frac{1}{2}\)

GDCD mà lại là Toán

Sai môn rồi bạn ơi!Vu Thanhh Dat

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