Tìm x :
(x+5)-(x-9)=x+2
\(\frac{x-1}{3}+\frac{3x-5}{2}+\frac{2x}{9}+\frac{-5x+3}{9}=\frac{210}{420}\)
Tìm x biết: \(\frac{x-1}{3}+\frac{3x-5}{2}+\frac{2x}{9}+\frac{-5x+3}{9}=\frac{210}{420}\)
=>\(\frac{6\left(x-1\right)}{18}+\frac{9\left(3x-5\right)}{18}+\frac{2\left(2x\right)}{18}+\frac{2\left(-5x+3\right)}{18}=\frac{1}{2}\)
=>\(\frac{6x-6}{18}+\frac{27x-45}{18}+\frac{4x}{18}+\frac{-10x+6}{18}=\frac{1}{2}\)
=>\(\frac{\left(6x-6\right)+\left(27x-45\right)+4x+\left(-10x+6\right)}{18}=\frac{1}{2}\)
=>\(\frac{27x-45}{18}=\frac{1}{2}\)
=>54x-90=18
54x =108
=>x=2
\(\frac{X-1}{3}\) +\(\frac{3X-5}{2}\) +\(\frac{2.X}{9}\) +\(\frac{-5X+3}{9}\) =\(\frac{210}{420}\)
MSC:1260
(x - 1).420\1260 + (3x - 5).630\1260 + 2.x.140/1260 + (-5x + 3).140\1260 =630\1260
=>(x - 1).420 + (3x - 5).630 + 2.x.140 + (-5x + 3).140 = 630
x.420 - 420 + 1890.x - 3150 + x.280 + (-700).x +420 = 630
x.( 420 + 1890 + 280 + -700) - (420 + 3150 -420) = 630
x.1890 - 3150 = 630
x.1890 = 630 + 3150
x.1890 =3780
x =3780 : 1890
x =2
Tíck cho mình nhé và chúc bạn học giỏi !
Sai chỗ nào thì bạn sửa cho mình nhé!
1. /x-5/=\(2^4\)-6
2.\(\frac{x-1}{3}+\frac{3x-5}{2}+\frac{2x}{3}+\frac{-5x+3}{9}=\frac{210}{420}\)
3.\(\left(4x-3\right)^4=\left(4^x-3\right)^2\)
4.\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2017}\left(x\in N^{ }SAO\right)\)
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
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)
\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)
\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)
\(\Rightarrow-16x-272=120x+312\)
\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)
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!
\(\frac{5x-2}{3}=\frac{5-3x}{2}\)
\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)
\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)
\(< =>10x-4=15-9x\)
\(< =>10x+9x=15+4\)
\(< =>19x=19< =>x=1\)
\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)
\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)
\(< =>30x+9=36+24+32x\)
\(< =>32x-30x=9-36-24\)
\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)
Thực hiện phép tính
a) \(\frac{\text{x + 9}}{x^2 - 9}-\frac{\text{3}}{\text{x^2 + 3x}}\)
b) \(\frac{\text{3x + 5 }}{\text{x^2 - 5x }}+\frac{\text{ 25 - x }}{\text{25 - 5x }}\)
c) \(\frac{\text{3 }}{\text{2x }}+\frac{\text{3x - 3 }}{\text{2x - 1 }}+\frac{ 2x^2 + 1 }{\text{4x^2 - 2x }}\)
d) \(\frac{\text{1}}{\text{3x - 2 }}-\frac{1}{\text{3x + 2 }}- \frac{\text{3x - 6}}{\text{4 - 9x^2}}\)
e) \(\frac{\text{18 }}{\text{(x - 3)(x^2 - 9) }}-\frac{\text{3 }}{\text{x^2 - 6x + 9 }}-\frac{\text{x}}{\text{x^2 - 9}}\)
g) \(\frac{\text{x + 2 }}{\text{x + 3 }}-\frac{\text{5 }}{\text{x^2 + x - 6 }}+\frac{\text{1}}{\text{2 - x}}\)
h) \(\frac{\text{4x }}{\text{x + 2 }}-\frac{\text{3x }}{\text{x - 2 }}+\frac{\text{12x}}{\text{x^2 - 4}}\)
i) \(\frac{\text{ x + 1 }}{\text{ x - 1 }}-\frac{\text{ x - 1 }}{\text{ x + 1 }}-\frac{\text{4}}{\text{1 - x^2}}\)
k) \(\frac{\text{
3x + 21
}}{\text{
x^2 - 9
}}+\frac{\text{2 }}{\text{x + 3 }}-\frac{\text{3}}{\text{x - 3}}\)
1,Giải Pt
a,\(\frac{3x-7}{2}+\frac{x+1}{3}=-16\)
b,\(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
c,\(\frac{7-3x}{12}+\frac{3}{4}=2\left(x-2\right)+\frac{5\left(5-2x\right)}{6}\)
e,\(\frac{3\left(x+3\right)}{4}+\frac{1}{2}=\frac{5x+9}{3}-\frac{7x-9}{4}\)
$\frac{4x+3}{5}$ -$\frac{6x-2}{7}$ =$\frac{5x+4}{3}$ +3
b.
$\frac{x+4}{5}$ -x+4=$\frac{x}{3}$ -$\frac{x-2}{2}$
c.$\frac{5x+2}{6}$ -$\frac{8x-1}{3}$ =$\frac{4x+2}{5}$ -5
d.$\frac{2x+3}{3}$ =$\frac{5-4}{2}$
e. $\frac{5x+3}{12}$ =$\frac{1+2x}{9}$
f.$\frac{7x-1}{6}$ =$\frac{16-x}{5}$
g. $\frac{x-3}{5}$ =6-$\frac{1-2x}{3}$
h. $\frac{3x-2}{6}$ -5=$\frac{3-2(x+7)}{4}$
giúp vs ạ, cần gấp
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm