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1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)
⇔ 3(x - 3) = 90 - 5(1 - 2x)
⇔ 3x - 9 = 90 - 5 + 10x
⇔ 3x - 10x = 90 - 5 + 9
⇔ -7x = 94
⇔ x = \(\frac{-94}{7}\)
S = { \(\frac{-94}{7}\) }
b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)
⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)
⇔ 6x - 4 - 60 = 9 - 6x - 42
⇔ 6x + 6x = 9 - 42 + 60 + 4
⇔ 12x = 31
⇔ x = \(\frac{31}{12}\)
S = { \(\frac{31}{12}\) }
c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7
⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210
⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210
⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40
⇔ 13x = 150
⇔ x = \(\frac{150}{13}\)
S = { \(\frac{150}{13}\) }
d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)
⇔ 21x - 120(x - 9) = 4(2x + 1,5)
⇔ 21x - 120x + 1080 = 8x + 6
⇔ 21x - 120x - 8x = 6 - 1080
⇔ -107x = -1074
⇔ x = \(\frac{1074}{107}\)
S = { \(\frac{1074}{107}\) }
e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5
⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840
⇔ 140x -140+56 -294x+42= 96x+48 -840
⇔ 140x -294x -96x = 48 -840 -42 -56+140
⇔ -250x = -750
⇔ x = 3
S = { 3 }
f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)
⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x
⇔ 4x+4+18x+9 = 4x+6x+6+7+12x
⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4
⇔ 0x = 0
S = R
Chúc bạn học tốt !
Bạn ơi giải giúp mình 2 bài này với ạ : https://hoc24.vn/hoi-dap/question/969683.html
Mình cảm ơn trước nhaa
a)
\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)
\(\Leftrightarrow\frac{49-13x}{12}=0\)
\(\Rightarrow49-13x=0\)
\(\Rightarrow x=\frac{-49}{13}\)
b)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)
\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)
\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)
\(\Leftrightarrow\frac{-3x}{4}=0\)
\(\Rightarrow-3x=0\)
\(\Rightarrow x=0\)
a, Ta có : \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}=\frac{x+7}{15}\)
=> \(3\left(2x-1\right)-5\left(x-2\right)=x+7\)
=> \(6x-3-5x+10-x-7=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
b, Ta có : \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
=> \(\frac{3\left(x+3\right)}{6}-\frac{2\left(x-1\right)}{6}=\frac{x+5}{6}+\frac{6}{6}\)
=> \(3\left(x+3\right)-2\left(x-1\right)=x+5+6\)
=> \(3x+9-2x+2-x-5-6=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
=> \(\frac{4\left(x+5\right)}{6}+\frac{3\left(x+12\right)}{6}-\frac{5\left(x-2\right)}{6}=\frac{2x}{6}+\frac{66}{6}\)
=> \(4\left(x+5\right)+3\left(x+12\right)-5\left(x-2\right)=2x+66\)
=> \(4x+20+3x+36-5x+10-2x-66=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
Bài 1:
b) Phương trình đã cho tương đương với phương trình:
\(\frac{8\left(x+22\right)-55\left(7x+149\right)-6\left(x+12\right)}{45}=\frac{9\left(x+35\right)+2\left(x+50\right)}{45}\)
\(\Leftrightarrow44x=-1056\)
\(\Leftrightarrow x=-24\)
Vậy x=-24 là nghiệm của phương trình
c) Phương trình đã cho tương đương với phương trình:
\(\frac{3x+6}{70}-\frac{x+4}{24}=\frac{32x+19}{60}+\frac{2}{3}\)
\(\Leftrightarrow12\left(3x+6\right)-35\left(x+4\right)=14\left(32x+19\right)+560\)
\(\Leftrightarrow-447x=894\)
\(\Leftrightarrow x=-2\)
Vậy x=-2 là nghiệm của phương trình
Bài 2:
a) \(5+\frac{x+4}{5}< x-\frac{x-2}{2}+\frac{x+3}{3}\)
\(\Leftrightarrow\frac{150}{30}+\frac{6\left(x+4\right)}{30}< \frac{30x}{30}-\frac{15\left(x-2\right)}{30}+\frac{10\left(x+3\right)}{30}\)
\(\Leftrightarrow150+6x+24-30x+15x-30-10x-30< 0\)(vì 30>0)
\(\Leftrightarrow-19x+114< 0\)
\(\Leftrightarrow-19x< -114\)
\(\Leftrightarrow x>6\)
vậy \(x\in\left(6;+\infty\right)\)
b) \(x+1-\frac{x-1}{3}< \frac{2x+3}{2}+\frac{x}{3}+5\)
\(\Leftrightarrow\frac{6\left(x+1\right)}{6}-\frac{2\left(x-1\right)}{6}-\frac{3\left(2x+3\right)}{6}-\frac{2x}{6}-\frac{30}{6}< 0\)
\(\Leftrightarrow6x+6-2x+2-6x-9-2x-30< 0\)(vì 6>0)
\(\Leftrightarrow-4x-31< 0\)
\(\Leftrightarrow-4x< 31\)
\(\Leftrightarrow x>\frac{-31}{4}\)
Vậy \(x\in\left(\frac{-31}{4};+\infty\right)\)
c) \(\frac{\left(3x-2\right)^2}{3}-\frac{\left(2x+1\right)^2}{3}\le x\left(x+1\right)\)
\(\Leftrightarrow\frac{\left(3x-2\right)^2}{3}-\frac{\left(2x+1\right)^2}{3}-\frac{3x\left(x+1\right)3}{3}\le0\)
\(\Leftrightarrow9x^2-12x+4-4x^2-4x-1-3x^2-3x\le0\)(vì 3>0)
\(\Leftrightarrow2x^2-19x+3\le0\)
\(\Leftrightarrow\frac{19-\sqrt{337}}{4}\le x\le\frac{19+\sqrt{337}}{4}\)
Vậy \(x\in\left[\frac{19-\sqrt{337}}{4};\frac{19+\sqrt{337}}{4}\right]\)
d) Bạn kiểm tra lại đề bài. Mình nghĩ đề là: \(\frac{2x-3}{4}-\frac{x+1}{3}\ge\frac{1}{2}-\frac{3-x}{5}\)
Bài 1 : Vào TKHĐ quay màn hình đt mà nhìn :))
Bài 2 :
a, \(5+\frac{x+4}{5}< x-\frac{x-2}{2}+\frac{x+3}{3}\)
\(\Leftrightarrow\frac{150}{30}+\frac{6x+24}{30}< \frac{30x}{30}-\frac{15x-30}{30}+\frac{10x+30}{30}\)
\(\Leftrightarrow150+6x+24< 30x-15x+30+10x+30\)
\(\Leftrightarrow174+6x< 25x+60\)
\(\Leftrightarrow6x-25x< 60-174\)
\(\Leftrightarrow-19x< -114\Leftrightarrow x< \frac{114}{19}=6\)
b, \(x+1-\frac{x-1}{3}< \frac{2x+3}{2}+\frac{x}{3}+5\)
\(\Leftrightarrow\frac{30x+30}{30}-\frac{10x-10}{30}< \frac{30x+45}{30}+\frac{10x}{30}+\frac{150}{30}\)
\(\Leftrightarrow30x+30-10x+10< 30x+45+10x+150\)
\(\Leftrightarrow20x+40< 40x+195\)
\(\Leftrightarrow-20x< 155\Leftrightarrow x< -\frac{155}{20}\)
c, \(\frac{\left(3x-2\right)^2}{3}-\frac{\left(2x+1\right)^2}{3}\le x\left(x+1\right)\)
\(\Leftrightarrow\frac{\left(3x-2\right)^2}{3}-\frac{\left(2x+1\right)^2}{3}\le\frac{3x^2+3x}{3}\)
\(\Leftrightarrow9x^2-12x+4-4x^2-4x-1\le3x^2+3x\)
\(\Leftrightarrow5x^2-16x+3-3x^2-3x\le0\)
\(\Leftrightarrow2x^2-19x+3\le0\)
Thẳng tay delta mà giải vì < hoặc ''='' 0 mà :))
d, \(\frac{2x+3}{4}-\frac{x+1}{3}\ge\frac{1}{2}-\frac{3-x}{5}\)
Đề sai rồi, chả thấy có j chung )):