Tìm \(x\), biết :
\((x+\frac{7x}{9})\left(x+\frac{7x}{33}\right)(x+\frac{7x}{33})...\left(x+\frac{7x}{9200}\right)=\frac{186}{25}\)
Giải pt
\(\frac{1}{7x+1}+\frac{1}{\sqrt{\left(7x+11\right)\left(9-7x\right)}}=\frac{7}{24}\left(x\inℝ\right)\)
Đặt \(\hept{\begin{cases}\sqrt{7x+11}=a\\\sqrt{9-7x}=b\end{cases}}\)
\(\Rightarrow a^2-b^2=14x+2\)
\(\Rightarrow\frac{2}{a^2-b^2}+\frac{1}{ab}=\frac{7}{24}\)
\(\Leftrightarrow\left(b+7a\right)\left(7b-a\right)=0\)
Làm nhầm phần phân tích nhân tử giờ làm lại cách khác.
Đặt \(7x+11=a\)
\(\Rightarrow7x=a-11\)
\(\Rightarrow\frac{1}{a-10}+\frac{1}{\sqrt{a\left(20-a\right)}}=\frac{7}{24}\)
\(\Leftrightarrow\frac{1}{\sqrt{a\left(20-a\right)}}=\frac{7}{24}-\frac{1}{a-10}\)
\(\Leftrightarrow\frac{1}{a\left(20-a\right)}=\left(\frac{7}{24}-\frac{1}{a-10}\right)^2\)
\(\Leftrightarrow\left(a-18\right)\left(a-16\right)\left(49a^2-630a+200\right)=0\)
PS: Bài giải trên bỏ đi nha
Tìm x biết :
\(\frac{36}{\left(1-7x\right)}=\frac{\left(1-7x\right)}{49}\)
Giải phương trình:
a) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
b) \(\frac{3}{10}\left(1,2-x\right)-\frac{5+7x}{4}=\frac{1}{20}\left(9x+0,2\right)-\frac{12,5x+4,5}{5}\)
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
1) giải phương trình:
a) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x+5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
b) \(\frac{7x+10}{x+1}\left(x^2-x-2\right)-\frac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
c) \(\frac{2x+5}{x+3}+1=\frac{4}{x^2+2x-3}-\frac{3x-1}{1-x}\)
d) \(\frac{13}{2x^2+x-21}+\frac{1}{2x+7}+\frac{6}{9-x^2}=0\)
e) \(\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
f) \(\frac{1+\frac{x}{x+3}}{1-\frac{x}{x+3}}=3\)
1. \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
2 . \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
3 . \(4\left(3x-2\right)-3\left(x-4\right)=7x+10\)
4. \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
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}
giải phương trình
a)\(\frac{7x+10}{x+1}\left(x^2-x-2\right)=\frac{7x+10}{x+1}\left(2x^2-3x-5\right)\)
b)\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)
c)\(x^2+\frac{1}{x^2}+\frac{9x}{2}-\frac{9}{2x}+7=0\)
Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
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 :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
giải pt: a)\(\left(x^2-3x\right)\left(x^2+7x+10\right)=216\) b)\(\left(2x^2-7x+3\right)\left(2x^2+x-3\right)+9=0\) c)\(\frac{1}{\left(x+29\right)^2}+\frac{1}{\left(x+30\right)^2}=\frac{13}{36}\)
a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
b)\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
c)\(\frac{x +1}{x-2}+\frac{x-1}{x +2}=\frac{2\left(x^{2^{ }}+2\right)}{x^2-4}\)
d)(2x+3)\(\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{1\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
<=> x2+2x-x+2=2
<=> x2+x=2-2
<=> x2+x=0
<=>x(x+1)=0
<=>x=0 hoặc x+1=0
<=>x=0 hoặc x = -1
a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
<=> x-3 =10x-15
<=> x-10x= -15+3
<=> -9x = -12
<=> x = \(\frac{-12}{-9}\)
<=> x = \(\frac{4}{3}\)