\(\frac{3x+5}{2}\)+ \(\frac{3x+5}{4}\)+\(\frac{3x+5}{6}\)=\(\frac{3x+5}{8}\)Tìm x
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
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)
\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)
\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)
\(\Rightarrow-16x-272=120x+312\)
\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)
\(a.\frac{4x-7}{12}-x=\frac{3x}{8}\\ b.\frac{5x-8}{3}=\frac{1-3x}{2}\\ c.\left(\frac{x-1}{\frac{2}{5}}-3\right)-\left(\frac{3x-2}{\frac{5}{4}}-2\right)=1\)
ai quen thì kb cả 2 tk nhé
a) \(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Rightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Rightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Rightarrow-56-64x=36x\)
\(\Leftrightarrow100x=-56\Leftrightarrow x=\frac{-14}{25}\)
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!
\(\frac{5x-2}{3}=\frac{5-3x}{2}\)
\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)
\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)
\(< =>10x-4=15-9x\)
\(< =>10x+9x=15+4\)
\(< =>19x=19< =>x=1\)
\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)
\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)
\(< =>30x+9=36+24+32x\)
\(< =>32x-30x=9-36-24\)
\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)
a) \(\frac{7x}{8}-5\left(x-9\right)=\frac{20x+1,5}{6}\)
b) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
c) \(\frac{3\left(x-3\right)}{4}+\frac{4x-10,5}{10}=\frac{3\left(x+1\right)}{5}+6\)
d) \(\frac{2\left(3x+1\right)+1}{4}-5=\frac{2\left(3x-1\right)}{5}-\frac{3x+2}{10}\)
Giải phương tình nha :v
a) \(\frac{7x}{8}-5\left(x-9\right)=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40\left(x-9\right)}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40x-360}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{360-33x}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow2160-198x=160x+12\)
\(\Leftrightarrow358x=2148\)
\(\Leftrightarrow x=6\)
Vậy nghiệm của pt x=6
b) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
\(\Leftrightarrow\frac{10\left(x-1\right)+4}{12}-\frac{21x-3}{12}=\frac{4x+2}{7}-\frac{35}{7}\)
\(\Leftrightarrow\frac{-11x-3}{12}=\frac{4x-33}{7}\)
\(\Leftrightarrow-77x-21=48x-396\)
\(\Leftrightarrow125x=375\)
\(\Leftrightarrow3\)
Vậy nghiệm của pt x=3
c)\(\frac{3\left(x-3\right)}{4}+\frac{4x-10,5}{10}=\frac{3\left(x+1\right)}{5}+6\)
\(\Leftrightarrow\frac{15\left(x-3\right)}{20}+\frac{8x-21}{20}=\frac{3x+3}{5}+\frac{30}{5}\)
\(\Leftrightarrow\frac{23x-66}{20}=\frac{3x+33}{5}\)
\(\Leftrightarrow115x-330=60x+660\)
\(\Leftrightarrow55x=990\)
\(\Leftrightarrow x=18\)
Vậy nghiệm của pt x=18
d) \(\frac{2\left(3x+1\right)+1}{4}-5=\frac{2\left(3x-1\right)}{5}-\frac{3x+2}{10}\)
\(\Leftrightarrow\frac{6x+3}{4}-\frac{20}{4}=\frac{4\left(3x-1\right)}{10}-\frac{3x+2}{10}\)
\(\Leftrightarrow\frac{6x-17}{4}=\frac{9x-6}{10}\)
\(\Leftrightarrow60x-170=36x-24\)
\(\Leftrightarrow24x=146\)
\(\Leftrightarrow x=\frac{73}{12}\)
Vậy nghiệm của pt \(x=\frac{73}{12}\)
Tìm x :
a) \(\frac{3x+2}{2}-\frac{3x+1}{6}=2x+\frac{5}{3}\)
b) \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
c) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
d) \(\left(x+1\right)^2-4\left(x^2-2x+1\right)=0\)
a) Qui đồng rồi khử mẫu ta được:
3(3x+2)-(3x+1)=2x.6+5.2
<=> 9x+6-3x-1 = 12x+10
<=> 9x-3x-12x = 10-6+1
<=> -6x = 5
<=> x = -5/6
Vậy ....
b) ĐKXĐ: \(x\ne\pm2\)
Qui đồng rồi khử mẫu ta được:
(x+1)(x+2)+(x-1)(x-2) = 2(x2+2)
<=> x2+3x+2+x2-3x+2 = 2x2+4
<=> x2+x2-2x2+3x-3x = 4-2-2
<=> 0x = 0
<=> x vô số nghiệm
Vậy x vô số nghiệm với x khác 2 và x khác -2
c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)
Vậy ......
d) (x+1)2-4(x2-2x+1) = 0
<=> x2+2x+1-4x2+8x-4 = 0
<=> -3x2+10x-3 = 0
giải phương trình
mấy chế ai biết giải thì giải dùm mik mấy bài nè vs.
Giải phương trình:
1) \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}=\frac{1}{6}\)
2) \(\frac{1}{x^2-6x+8}+\frac{1}{x^2-10x+24}+\frac{1}{x^2-14x+48}=\frac{1}{9}\)
3) \(\frac{1}{x^2-3x+3}+\frac{2}{x^2-3x+4}=\frac{6}{x^2-3x+5}\)
4) \(\frac{6}{\left(x+1\right)\left(x+2\right)}+\frac{8}{\left(x-1\right)\left(x+4\right)}=1\)
5) \(4\left(x^3+\frac{1}{x^3}\right)=13\left(x+\frac{1}{x}\right)\)
6) \(\frac{4x}{4x^2-8x+7}+\frac{3x}{4x^2-10x+7}=1\)
7) \(\frac{x^2-3x+5}{x^2-4x+5}-\frac{x^2-5x+5}{x^2-6x+5}=\frac{-1}{4}\)
8) \(x\frac{8-x}{x-1}.\left(x-\frac{8-x}{x-1}\right)=15\)
\(\frac{1}{x^2-3x+3}+\frac{2}{x^2-3x+4}=\frac{6}{x^2-3x+5}\)
+ Đặt \(t=x^2-3x+3\) thì pt đã cho trở thành :
\(\frac{1}{t}+\frac{2}{t+1}=\frac{6}{t+2}\)
\(\Leftrightarrow\frac{t+1+2t}{t\left(t+1\right)}=\frac{6}{t+2}\) \(\Leftrightarrow\frac{3t+1}{t^2+t}=\frac{6}{t+2}\)
\(\Leftrightarrow\left(3t+1\right)\left(t+2\right)=6\left(t^2+t\right)\)
\(\Leftrightarrow3t^2+7t+2=6t^2+6t\)
\(\Leftrightarrow3t^2-t-2=0\)
\(\Leftrightarrow3t^2-3t+2t-2=0\)
\(\Leftrightarrow\left(3t+2\right)\left(t-1\right)=0\)
\(\Leftrightarrow t-1=0\) ( do \(3t+2=3x^2-9x+11\)\(=3\left(x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}+\frac{17}{12}\right)=3\left[\left(x-\frac{3}{2}\right)^2+\frac{17}{12}\right]>0\forall x\))
\(\Leftrightarrow x^2-3x+3=1\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{3}{4}=1\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{2}=\frac{1}{2}\\x-\frac{3}{2}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\left(TM\right)\)
Vậy tập nghiệm của pt đã cho là \(S=\left\{1;2\right\}\)
\(\frac{1}{x^2-3x+3}-1+\frac{2}{x^2-3x+4}-1+2-\frac{6}{x^2-3x+5}=0\)
\(\Leftrightarrow\frac{-x^2+3x-2}{x^2-3x+3}+\frac{-x^2+3x-2}{x^2-3x+4}-\frac{2\left(-x^2+3x-2\right)}{x^2-3x+5}=0\)
\(\Leftrightarrow\left(-x^2+3x-2\right)\left(\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+3x-2=0\left(1\right)\\\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}=0\)
Do \(\left\{{}\begin{matrix}\frac{1}{x^2-3x+3}>\frac{1}{x^2-3x+5}\\\frac{1}{x^2-3x+4}>\frac{1}{x^2-3x+5}\end{matrix}\right.\) \(\forall x\Rightarrow\frac{1}{x^2-3x+3}+\frac{1}{x^2-3x+4}-\frac{2}{x^2-3x+5}>0\)
\(\Rightarrow\left(2\right)\) vô nghiệm
Tìm x,y \(\frac{3x-2}{5}=\frac{5y+8}{9}=\frac{3x+5y+6}{7y}\)
ta co 3x-2/5 =5y+8 /9 (1)
ap dung tinh chat day ti so bang nhau ta co
3x-2/5=5y+8/ 9=3x-2+5y+8 /9 =3x+5y+6 /14 =3x+5y+6 /7y
=> 14=7y=>y= 2
thay y=2 vao (1) ta co 3x-2 /5 = 5.2+8 /9= 18/9=2 =>3x-2 /5 =2
=>3x-2 =10 =>3x =12 =>x =4
Giải phương trình:
2x-2=8-3x
x2-3x+1=x+x2
(x2+1)(2x+4)=0
(4x+1)(x2+2)=0
\(\frac{x}{2}=3-\frac{x+4}{3}\)
\(\frac{3-x}{4}=1-\frac{3x-5}{6}\)
\(\frac{2x+5}{9}=2+\frac{x-3}{6}\)
\(\frac{x+5}{3}=1+\frac{x-3}{9}\)
\(\frac{2x-5}{x+5}=3\)
\(\frac{x^2-6}{x}=x+\frac{3}{2}\)
\(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
\(\frac{5}{3x+2}=2x-1\)
\(2x-2=8-3x\)
\(\Leftrightarrow\)\(2x+3x=8+2\)
\(\Leftrightarrow\)\(5x=10\)
\(\Leftrightarrow\)\(x=2\)
Vậy...
\(x^2-3x+1=x+x^2\)
\(\Leftrightarrow\)\(x^2-3x-x-x^2=-1\)
\(\Leftrightarrow\)\(-4x=-1\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy...
mấy cái này bấm máy tính là đc òi. giải mất thời gian lắm :))
\(2x-2=8-3x\)
\(\Leftrightarrow\)\(2x+3x=8+2\)
\(\Leftrightarrow\)\(5x=10\)
\(\Leftrightarrow\)\(x=2\)
Vậy \(x=2\)
\(x^2-3x+1=x+x^2\)
\(\Leftrightarrow\)\(4x-1=\left(x^2+x\right)-\left(x^2+x\right)\)
\(\Leftrightarrow\)\(4x-1=0\)
\(\Leftrightarrow\)\(4x=1\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy \(x=\frac{1}{4}\)
\(\left(x^2+1\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2+1=0\\2x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\2x=-4\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\sqrt{-1}\\x=\frac{-4}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x\in\left\{\varnothing\right\}\\x=-2\end{cases}}}\)
Vậy \(x=-2\)
Chúc bạn học tốt ~