a) \(\frac{3}{7}x-1=\frac{1}{7}x\left(3x-7\right)\)

<=> \(3x-7=x\left(3x-7\right)\)

<=> \(\left(3x-7\right)-x\left(3x-7\right)=0\)

<=> \(\left(3x-7\right)\left(1-x\right)=0\)

<=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x=1\end{cases}}\)

Vậy S = { 7/3; 1}

b) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

<=> \(\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

<=> \(\left(3x-1\right)\left(x^2-7x+12\right)=0\)

<=> \(\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)

<=> \(\left(3x-1\right)\left(x\left(x-3\right)-4\left(x-3\right)\right)=0\)

<=> \(\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

<=> x = 1/3 hoặc x = 3 hoặc x = 4.

Vậy S = { 1/3; 3; 4}

a) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)

<=> \(\frac{5x+2\left(3-x\right)}{70}-\frac{5x-4\left(x-1\right)}{24}=\frac{35x+10+9-3x}{60}+\frac{2}{3}\)

<=> \(12\left(5x+6-2x\right)-35\left(5x-4x+4\right)\)

<=> \(14\left(35x+10+9-3x\right)+280.2\) <=> \(12\left(3x+6\right)-35\left(x+4\right)\)

<=> \(14\left(32x+19\right)+560\)

<=> \(36x+72-35x-140=448x+226+560\)

<=> \(-447x=894\)

<=> x = -2

\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)

\(\Leftrightarrow9x-7=7x+5\)

\(\Leftrightarrow9x-7x=5+7\)

\(\Leftrightarrow2x=12\)

\(\Leftrightarrow x=6\)

\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\)

\(\Leftrightarrow x=9\)

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 !

a) Ta có: \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)

\(\Leftrightarrow\frac{63\left(3x-11\right)}{693}-\frac{231x}{693}-\frac{99\left(3x-5\right)}{693}+\frac{77\left(5x-3\right)}{693}=0\)

\(\Leftrightarrow189x-693-231x-297x+495+385x-231=0\)

\(\Leftrightarrow46x-429=0\)

\(\Leftrightarrow46x=429\)

hay \(x=\frac{429}{46}\)

Vậy: \(x=\frac{429}{46}\)

b) Ta có: \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)

\(\Leftrightarrow\frac{9x-0,7}{4}-\frac{5x-1,5}{7}-\frac{7x-1,1}{6}+\frac{5\left(0,4-2x\right)}{5}=0\)

\(\Leftrightarrow105\left(9x-0,7\right)-60\left(5x-1,5\right)-70\left(7x-1,1\right)+420\left(0,4-2x\right)=0\)

\(\Leftrightarrow945x-\frac{147}{2}-300x+90-490x+77+168-840x=0\)

\(\Leftrightarrow-685x+261.5=0\)

\(\Leftrightarrow-685x=-261.5\)

hay \(x=\frac{523}{1370}\)

Vậy: \(x=\frac{523}{1370}\)

c) Ta có: \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)

\(\Leftrightarrow\frac{14\left(5x-3\right)}{84}-\frac{21\left(7x-1\right)}{84}-\frac{24\left(2x-1\right)}{84}+\frac{420}{84}=0\)

\(\Leftrightarrow70x-42-147x+21-48x+24+420=0\)

\(\Leftrightarrow-125x+423=0\)

\(\Leftrightarrow-125x=-423\)

hay \(x=\frac{423}{125}\)

Vậy: \(x=\frac{423}{125}\)

d) Ta có: \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)

\(\Leftrightarrow\frac{435}{30}-\frac{12\left(x+3\right)}{30}-\frac{45x}{30}+\frac{20\left(x-7\right)}{30}=0\)

\(\Leftrightarrow435-12x-36-45x+20x-140=0\)

\(\Leftrightarrow-37x+259=0\)

\(\Leftrightarrow-37x=-259\)

hay \(x=7\)

Vậy: x=7

Trả lời

đưa căn 7x+2 sang vế bên phải rồi mũ 3 lên là đc mầ

hok tốt

ĐKXĐ: x > -²/₇

Đặt \(\hept{\begin{cases}\sqrt[3]{2x-1}=a\\\sqrt{7x+2}=b\ge0\end{cases}}\Rightarrow7a^3-2b^2=14x-7-14x-4=-11\)

Từ đề bài \(\Rightarrow4a-b=1\)

Ta có hệ \(\hept{\begin{cases}7a^3-2b^2=-11\\4a-b=1\end{cases}}\)

   \(\Leftrightarrow\hept{\begin{cases}7a^3-2b^2=-11\\b=4a-1\end{cases}}\)

\(\Rightarrow7a^3-2\left(4a-1\right)^2=-11\)

\(\Leftrightarrow\left(a-1\right)\left(7a^2-25a-9\right)=0\)

Đến đây tìm được a => x

1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)

<=> 21x - 100x + 900 = 80x + 6

<=> -79x - 80x = 6 - 900

<=> -159x = -894

<=> x = 258/53

Vậy S = {258/53}

2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

<=> 12x² + 12x + 3 - 5x² - 10x - 5 = 7x² - 14x - 5

<=> 7x² + 2x - 7x² + 14x = -5 + 2

<=> 16x = 3

<=> x = 3/16

Vậy S  = {3/16}

3) 4(3x - 2) - 3(x - 4) = 7x+  10

<=> 12x - 8 - 3x + 12 = 7x + 10

<=> 9x - 7x = 10 - 4

<=> 2x = 6

<=> x = 3

Vậy S = {3}

4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)

<=> x² + 14x + 40 + 3x² + 6x - 24 = 4x² + 32x - 80

<=> 4x² + 20x - 4x² - 32x = -80 - 16

<=> -12x = -96

<=> x = 8

Vậy S = {8}