giải phương trình
a. \(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)
Giải phương trình:
a) \(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)
b) \(\dfrac{x-291}{1700}+\dfrac{x-293}{1698}+\dfrac{x-295}{1696}+\dfrac{x-297}{1694}=4\)
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a,\(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)
=>\(\dfrac{x-3}{113}+1+\dfrac{x-5}{115}+1=\dfrac{x-7}{117}+1+\dfrac{x-9}{119}+1\)
=>\(\dfrac{x+110}{113}+\dfrac{x+110}{115}=\dfrac{x+110}{117}+\dfrac{x+110}{119}\)
=>(x+110)(\(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\))=0
vì 1/113+1/115-1/117-1/119 khác 0
=>x+110=0=>x=-110
\(\dfrac{x-291}{1700}+\dfrac{x-293}{1698}+\dfrac{x-295}{1696}+\dfrac{x-297}{1694}=4\)
=>\(\dfrac{x-291}{1700}-1+\dfrac{x-293}{1698}-1+\dfrac{x-295}{1696}-1+\dfrac{x-297}{1694}-1+4=4\)=>\(\dfrac{x-1991}{1700}+\dfrac{x-1991}{1698}+\dfrac{x-1991}{1696}+\dfrac{x-1991}{1694}=0\)
=>(x-1991)(1/1700+1/1698+1/1696+1/1694)=0
vì 1/1700+1/1698+1/1696+1/1694 khác 0
=>x-1991=0=>x=1991
A= \(\dfrac{1003+1007+\dfrac{2010}{113}+\dfrac{2010}{117}-\dfrac{1003}{119}-\dfrac{1007}{119}}{1003+1008+\dfrac{2011}{113}+\dfrac{2011}{117}-\dfrac{1003}{119}-\dfrac{1007}{119}}\)
Giải nhanh nha, mình sắp đi học rồi. Hứa tick hết cho :)
Ta có:
\(A=\dfrac{1003+1007+\dfrac{2010}{113}+\dfrac{2010}{117}-\dfrac{1003}{119}-\dfrac{1007}{119}}{1003+1008+\dfrac{2011}{113}+\dfrac{2011}{117}-\dfrac{1003}{119}-\dfrac{1007}{119}}\)
\(A=\dfrac{2010+\dfrac{2010}{113}+\dfrac{2010}{117}-\dfrac{2010}{119}}{2011+\dfrac{2011}{113}+\dfrac{2011}{117}-\dfrac{2011}{119}}\)
\(A=\dfrac{2010.\left(\dfrac{1}{1}+\dfrac{1}{113}+\dfrac{1}{117}-\dfrac{1}{119}\right)}{2013.\left(\dfrac{1}{1}+\dfrac{1}{113}+\dfrac{1}{117}+\dfrac{1}{119}\right)}\)
\(\Rightarrow A=\dfrac{2010}{2013}\)
= \(\dfrac{2028,076341}{2029,093738}\) hoặc 0,9994985954.
Giải phương trình sau :
a,\(\dfrac{7-3x}{12}+\dfrac{5x+2}{7}=x+13\)
b,\(\dfrac{3\left(x+3\right)}{4}-\dfrac{1}{2}=\dfrac{5x+9}{7}-\dfrac{7x-9}{4}\)
c,\(\dfrac{2x+1}{3}-\dfrac{5x+2}{7}=x+3\)
d,\(\dfrac{2x-3}{3}-\dfrac{2x+3}{7}=\dfrac{4x+3}{5}-17\)
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
a, msc 12.7=84
Chuyển vế về =0 rồi làm
b,msc 28
c,làm tương tự
a, \(\Rightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow-45x=1019\Leftrightarrow x=-\dfrac{1019}{45}\)
b, \(\Rightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
\(\Leftrightarrow21x+63-14=20x+36-49x+63\)
\(\Leftrightarrow50x=50\Leftrightarrow x=1\)
c, \(\Rightarrow14x+7-15x-6=21x+63\Leftrightarrow-22x=62\Leftrightarrow x=-\dfrac{31}{11}\)
d, \(\Rightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-105.17\)
\(\Leftrightarrow70x-105-30x-45=84x+63-1785\)
\(\Leftrightarrow-44x=-1572\Leftrightarrow x=\dfrac{393}{11}\)
Giải các phương trình sau:
a) 2,3 - 2(0,7 + 2) = 3,6 - 1,7x
b) \(\dfrac{5x+7}{4}-\dfrac{3x+5}{8}=\dfrac{4x+9}{5}-\dfrac{x-9}{3}\)
c) \(\dfrac{2x-1}{4}+\dfrac{x-3}{3}=\dfrac{4x-2}{3}-\dfrac{6x+7}{12}\)
d) (x - 1)(x + 2) - x(x + 3) = 8
a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
Tính giá trị biểu thức sau :
a) A = \(3\dfrac{1}{117}.4\dfrac{1}{119}-1\dfrac{116}{117}.5\dfrac{118}{119}-\dfrac{5}{119}\)
b) B = \(4\dfrac{1}{115}.3\dfrac{1}{225}-5\dfrac{114}{115}.1\dfrac{224}{225}-\dfrac{10}{225}\)
Giải các phương trình sau:
\(a.\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(b.\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(c.2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)\)
\(d.\dfrac{7}{8}x-5\left(x-9\right)=\dfrac{20x+1,5}{6}\)
\(e.\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)
\(f.\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
Tính
\(A=\dfrac{1003+1007+\dfrac{2010}{113}+\dfrac{2010}{117}-\dfrac{1003}{119}-\dfrac{1007}{119}}{1003+1008+\dfrac{2011}{113}+\dfrac{2011}{117}-\dfrac{1003}{119}-\dfrac{1008}{119}}\)
\(tuA=1003+1007+\dfrac{2010}{113}+\dfrac{2010}{117}-\dfrac{2010}{119}=2010\left(1+\dfrac{1}{113}+\dfrac{1}{117}-\dfrac{1}{119}\right)\)\(mauA=1003+1008+\dfrac{2011}{113}+\dfrac{2011}{117}-\dfrac{2011}{119}=2011\left(1+\dfrac{1}{113}+\dfrac{1}{117}-\dfrac{1}{119}\right)\)có \(\left(1+\dfrac{1}{113}+\dfrac{1}{117}-\dfrac{1}{119}\right)\ne0=>A=\dfrac{2010}{2011}\)
giải các phương trình sau
a)\(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
b)\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{x^2-9}\)
\(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
\(\Leftrightarrow\dfrac{20\left(2x-1\right)}{60}+\dfrac{15\left(3x-2\right)}{60}=\dfrac{12\left(4x-3\right)}{60}\)
`<=> 20(2x-1) +15(3x-2) =12(4x-3)`
`<=> 40x - 20 + 45x - 30 = 48x - 36`
`<=> 85x -50 = 48x - 36`
`<=> 85x-48x = -36+50`
`<=> 37x =14`
`<=> x= 14/37`
Vậy phương trình có nghiệm `x=14/37`
__
\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{x^2-9}\)
\(\Leftrightarrow\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
Ta có : \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
`=> 5x + 15 + 4x -12=x-6`
`<=> 9x + 3=x-6`
`<=> 9x-x=-6-3`
`<=> 8x = -9`
`<=>x=-9/8(tm)`
Vậy phương trình có nghiệm `x=-9/8`
` @ yngoc`
Giải các phương trình:
\(\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}=4\)
\(\frac{x-3}{2011}+\frac{x-5}{2009}+\frac{x-7}{2007}+\frac{x-9}{2005}=4\)
\(\Leftrightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)+\left(\frac{x-9}{2005}-1\right)=0\)
\(\Leftrightarrow\frac{x-2014}{2011}+\frac{x-2014}{2009}+\frac{x-2014}{2007}+\frac{x-2014}{2005}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2009}+\frac{1}{2007}+\frac{1}{2005}\right)=0\)
|________________A________________|
Do A > 0
nên x - 2014 = 0
<=> x = 2014