\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\)
K
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NN
7 tháng 4 2020
a) \(\frac{5x-2}{3}=\frac{5-3x}{2}\)\(\Leftrightarrow2\left(5x-2\right)=3\left(5-3x\right)\)\(\Leftrightarrow10x-4=15-9x\)
\(\Leftrightarrow10x+9x=15+4\)\(\Leftrightarrow19x=19\)\(\Rightarrow x=1\)
Vậy tập nghiệm của phương trình là: \(S=\left\{1\right\}\)
b) \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)\(\Leftrightarrow\frac{3\left(10x+3\right)}{36}=\frac{36}{36}+\frac{4\left(6+8x\right)}{36}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)\(\Leftrightarrow30x+9=36+24+32x\)\(\Leftrightarrow32x-30x=9-36-24\)\(\Leftrightarrow2x=-51\)\(\Leftrightarrow x=\frac{-51}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{-51}{2}\right\}\)
c) \(2\left(x+\frac{3}{5}\right)=5\left(\frac{13}{5}+x\right)\)\(\Leftrightarrow2\left(\frac{5x}{5}+\frac{3}{5}\right)=5\left(\frac{13}{5}+\frac{5x}{5}\right)\)\(\Leftrightarrow\frac{2\left(5x+3\right)}{5}=\frac{5\left(13+5x\right)}{5}\)
\(\Leftrightarrow2\left(5x+3\right)=5\left(13+5x\right)\)\(\Leftrightarrow10x+6=65+25x\)\(\Leftrightarrow25x-10x=6-65\)\(\Leftrightarrow15x=-59\)\(\Leftrightarrow x=\frac{-59}{15}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{-59}{15}\right\}\)
PN
24 tháng 5 2020
\(a,\frac{5x-2}{3}=\frac{5-3x}{2}\)
\(< =>\frac{\left(5x-2\right)2}{3.2}=\frac{\left(5-3x\right)3}{2.3}\)
\(< =>\frac{10x-4}{6}=\frac{15-9x}{6}\)
\(< =>10x-4=15-9x\)
\(< =>10x+9x=15+4=19\)
\(< =>19x=19< =>x=1\)
KN
18 tháng 8 2020
1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow95x-5=96-6x\)
\(\Leftrightarrow95x+6x=96+5\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9=32x+60\)
\(\Leftrightarrow30x-32x=60-9\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\frac{51}{2}\)
3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
19 tháng 8 2020
4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)
=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)
=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)
=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)
=> 27 - 9x + 80 - 16x = 12 - 12x - 48
=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0
=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0
=> 143 - 13x = 0
=> 13x = 143
=> x = 11
5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)
=> 6x - 18 + 7x - 35 - 13x - 4 = 0
=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0
=> -57 = 0(vô nghiệm)
6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)
=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)
=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)
=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)
=> \(12x+10-10x-3=12x+2\)
=> \(2x+10-3=12x+2\)
=> 2x + 10 - 3 - 12x - 2 = 0
=> (2x - 12x) + (10 - 3 - 2) = 0
=> -10x + 5 = 0
=> -10x = -5
=> x = 1/2
7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)
=> 6x - 3 - 5x + 10 - x - 7 = 0
=> (6x - 5x - x) + (-3 + 10 - 7) = 0
=> 0x + 0 = 0
=> 0x = 0
=> x tùy ý
Bài 8 tự làm nhé
28 tháng 3 2020
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
\( 2\left( {x + \dfrac{3}{5}} \right) = 5 - \left( {\dfrac{{13}}{5} + x} \right)\\ \Leftrightarrow 2x + \dfrac{6}{5} = \dfrac{{12}}{5} - x\\ \Leftrightarrow 3x = \dfrac{6}{5}\\ \Leftrightarrow x = \dfrac{2}{5} \)
\( \dfrac{{10x + 3}}{{12}} = 1 + \dfrac{{6 + 8x}}{9}\\ \Leftrightarrow 3\left( {10x + 3} \right) = 36 + 4\left( {6 + 8x} \right)\\ \Leftrightarrow 30x + 9 = 36 + 24 + 32x\\ \Leftrightarrow - 2x = 51\\ \Leftrightarrow x = - \dfrac{{51}}{2} \)