Câu 3: Giải các phương trình sau bằng cách đưa về dạng ax+b=0
1. a, \(\frac{5x-2}{3}=\frac{5-3x}{2}\); b, \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
c, \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\); d, \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
e, \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\); f, 4 (0,5-1,5x)=\(\frac{5x-6}{3}\)
g, \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\); h, \(\frac{x+4}{5}.x+4=\frac{x}{3}-\frac{x-2}{2}\)
i, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\); k, \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)
m, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\); n, \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right).\frac{1}{3}\left(x+2\right)\)
p, \(\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x\); q, \(\frac{2+x}{5}-0,5x=\frac{1-2x}{4}+0,25\)
r, \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\); s, \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{6}\)
t, \(\frac{2x-8}{6}.\frac{3x+1}{4}=\frac{9x-2}{8}+\frac{3x-1}{12}\); u, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{3}+\frac{2x-1}{12}\)
v, \(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\); w, \(\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x\frac{x-3}{2}}{5}-x+1\)
Đây là những bài cơ bản mà bạn!
\(\frac{5x-2}{3}=\frac{5-3x}{2}\)
\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)
\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)
\(< =>10x-4=15-9x\)
\(< =>10x+9x=15+4\)
\(< =>19x=19< =>x=1\)
\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)
\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)
\(< =>30x+9=36+24+32x\)
\(< =>32x-30x=9-36-24\)
\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)
c) \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
d) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
e) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
ĐK: ...
c) \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=\frac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)
\(\Leftrightarrow5x+25=0\)
\(\Leftrightarrow x=-5\)( ko t/m )
d) tương tự, ngại tính lắm
e) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1}{x^3-1}-\frac{3x^2}{x^3-1}=\frac{2x\left(x-1\right)}{x^3-1}\)
\(\Leftrightarrow4x^2-3x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=\frac{-1}{4}\left(c\right)\end{matrix}\right.\)
Ai giúp vs !!!
\(a.\frac{3x-7}{5}=\frac{2x-1}{3}\\ b.\frac{4x-7}{12}-x=\frac{3x}{8}\\ c.\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\\ d.\frac{5x-8}{3}=\frac{1-3x}{2}\\ e.\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\\ f.\frac{x-1}{\frac{2}{5}}-3-\frac{3x-2}{\frac{5}{4}}-2=1\)
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)
\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)
\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)
\(\Rightarrow-16x-272=120x+312\)
\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)
Bài 2. Giải các phương trình sau
a, \(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
b, \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
c, \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
d,\(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
e, \(\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)\)
f, \(\frac{x+5}{2}+\frac{3-2x}{4}=x-\frac{7+x}{6}\)
g, \(\frac{x-3}{11}+\frac{x+1}{3}=\frac{x+7}{9}-1\)
h, \(\frac{3x-0,4}{2}+\frac{1,5-2x}{3}=\frac{x+0,5}{5}\)
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)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)
\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)
\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)
\(\Leftrightarrow\frac{-3x}{4}=0\)
\(\Rightarrow-3x=0\)
\(\Rightarrow x=0\)
e)
\(\frac{3\cdot\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\cdot\left(x-7\right)\)
\(\Leftrightarrow\frac{45x-18-24-28x+60x-420}{12}=0\)
\(\Leftrightarrow\frac{77x-462}{12}=0\)
\(\Rightarrow77x-462=0\)
\(\Rightarrow x=\frac{462}{77}=6\)
Bài 1:Giải Phương trình
d) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)
e)\(\frac{6}{x^{2^{ }}-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)
g) \(\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\frac{1}{2}\)
h) \(\frac{x+4}{x^2-3x+2}-\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
Bài 1:
d)ĐKXĐ: \(x\ne8\)
Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)
\(\Leftrightarrow\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3x-24}=0\)
\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)
MTC=24(x-8)
\(\Leftrightarrow\frac{36}{24\left(x-8\right)}+\frac{72x-480}{24\left(x-8\right)}+\frac{3x-24}{24\left(x-8\right)}-\frac{104x-816}{24\left(x-8\right)}=0\)
\(\Leftrightarrow36+72x-480+3x-24-104x+816=0\)
\(\Leftrightarrow348-29x=0\)
\(\Leftrightarrow-29x+348=0\)
\(\Leftrightarrow x=\frac{-348}{-29}=12\)
Vậy: x=12
e) ĐKXĐ: \(x\ne\pm1\)
Ta có: \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)
\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4x+4}+\frac{12x-1}{4-4x}=0\)
\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}+\frac{12x-1}{4\left(1-x\right)}=0\)
MTC=4(x+1)(x-1)
\(\Leftrightarrow\frac{24}{4\left(x-1\right)\left(x+1\right)}+\frac{20x^2-20}{4\left(x-1\right)\left(x+1\right)}-\frac{8x^2-9x+1}{4\left(x-1\right)\left(x+1\right)}-\frac{12x^2-11x-1}{4\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2+11x+1=0\)
\(\Leftrightarrow20x+4=0\)
\(\Leftrightarrow20x=-4\)
\(\Leftrightarrow x=-\frac{4}{20}=-0,2\)(loại)
Vậy: x không có giá trị
g) Ta có: \(\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\frac{1}{2}\)
\(\Leftrightarrow\frac{\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}}{\frac{x-1}{x-1}+\frac{x+1}{x-1}}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}}{\frac{2x}{x-1}}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-1}{2x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{4x\cdot\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\cdot2x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{2}=0\)
MTC=2(x+1)
\(\Leftrightarrow\frac{2}{2\left(x+1\right)}-\frac{x+1}{2\left(x+1\right)}=0\)
\(\Leftrightarrow2-x+1=0\)
\(\Leftrightarrow1-x=0\)
\(\Leftrightarrow x=1\)(loại vì không thỏa mãn ĐKXĐ)
Vậy: x không có giá trị
giải các hệ BPT sau:
a) \(\left\{{}\begin{matrix}5x-2>4x+5\\5x-4< x+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x+1>3x+4\\5x+3\ge8x-9\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\frac{5x+2}{3}\ge4-x\\\frac{6-5x}{13}< 3x+1\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}6x+\frac{5}{7}< 4x+7\\\frac{8x+3}{2}< 2x+5\end{matrix}\right.\)
f) \(\left\{{}\begin{matrix}15x-2>2x+\frac{1}{3}\\2\left(x-4\right)< \frac{3x-14}{2}\end{matrix}\right.\)
g) \(\left\{{}\begin{matrix}x-1\le2x-3\\3x< x+5\\5-3x\le2x-6\end{matrix}\right.\)
h) \(\left\{{}\begin{matrix}2x+\frac{3}{5}>\frac{3\left(2x-7\right)}{3}\\x-\frac{1}{2}< \frac{5\left(3x-1\right)}{2}\end{matrix}\right.\)
j) \(\left\{{}\begin{matrix}\frac{3x+1}{2}-\frac{3-x}{3}\le\frac{x+1}{4}-\frac{2x-1}{3}\\3-\frac{2x+1}{5}>x+\frac{4}{3}\end{matrix}\right.\)
1) Giải các phương trình.
a) x(x-1)=x(x+2)
b) -3x+2=2x+8
c) \(\frac{3x-3}{4}=2-\frac{x-2}{8}\)
d) \(\left(x-\frac{1}{2}\right)\left(x+\frac{4}{3}\right)=0\)
e) \(\frac{2}{5}x+x-\frac{1}{3}=-\frac{4}{7}x+\frac{5}{6}\)
f) \(\frac{3x-5}{9}=1+\frac{2x+4}{6}\)
g) \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
h) \(\frac{2x+1}{x-2}=3\)
k) \(\frac{2}{x-3}+\frac{x-5}{x-1}=1\)
a, x( x - 1) = x ( x + 2)
<=> x2 - x = x2 + 2x
<=> x2 - x - x2 - 2x = 0
<=> -3x = 0
<=> x = 0
b, tương tự câu a
c,\(\Leftrightarrow\frac{3x-3}{4}=2-\frac{x-2}{8}\)
\(\Leftrightarrow\frac{\left(3x-3\right)2}{8}=\frac{16}{8}-\frac{x-2}{8}\)
\(\Leftrightarrow\frac{6x-6}{8}=\frac{16}{8}-\frac{x-2}{8}\)
=> 6x - 6 = 16 - x + 2
<=> 6x + x = 16 + 2 + 6
<=> 7x = 24
<=> x=\(\frac{24}{7}\)
Các câu còn lại làm tương tự
a,\(\frac{3}{x}+\frac{1}{x+3}+\frac{3}{x+6}+\frac{1}{x+7}=\frac{1}{1-x}\)
b, \(\frac{1}{x-5}+\frac{1}{x-2}+\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+3}=\frac{3x-3}{4}\)
c,\(\frac{1}{x-3}+\frac{1}{3x+1}+\frac{10x-13}{4x-6}=\frac{1}{x+1}+\frac{1}{2x-1}+\frac{1}{3x+7}\)
d,\(\frac{x^2+x+1}{2x-1}\left(\frac{3x^2-x+5}{4x-2}-3\right)=8\)
e,\(\frac{2x^2-3}{3x-1}\left(2x-\frac{7+4x}{3x-1}\right)=2\)
f,\(\frac{x\left(3x-1\right)\left(3x^2+1\right)\left(6x^2-3x-1\right)}{\left(x+1\right)^3}=\frac{1}{2}\)
g, \(x\left(x^2+2\right)\left(x^2+2x+8+\frac{12}{x-2}\right)=3\left(x-2\right)\)
bài 1 giải phương trình
a) (2x+3)\(^2\)-3(x-4)(x+4)=\(\left(x-2\right)^2\)+1
b)(3x-2) (9x\(^2\)+6x+4)-(3x-1) (9x\(^2\)+3x+1)=x-4
c)x (x-1) -(x-3) (x+4)=5x
d) (2x+1)(2x-1)=4x(x-7)-3x
bài 2 giải phương trình
a)\(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
b)\(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}+\frac{12-x}{9}\)
c)\(\frac{10x+3}{8}=\frac{7-8x}{12}\)
d)\(\frac{x+4}{5}-x-5=\frac{x+3}{3}-\frac{x-2}{2}\)
