a, \(\frac{3x+2}{7}\)= \(\frac{4x-5}{6}\)
b,\(\frac{x+1}{x+3}\)=\(\frac{x+2}{x+4}\)
K
Khách
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KN
5 tháng 3 2020
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
KN
5 tháng 3 2020
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
HH
3 tháng 7 2020
a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)
\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)
Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\)
Ta được :
\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)
Từ đó ta lắp vào ( 1 ) tính được x
16 tháng 12 2022
a: \(\Leftrightarrow4\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3x^2\)
\(\Leftrightarrow4\cdot\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]=3x^2\)
=>4(x^2+60)^2+132x(x^2+60)+1085x^2=0
=>4(x^2+60)^2+62x(x^2+60)+70x(x^2+60)+1085x^2=0
=>2(x^2+60)(2x^2+120+31x)+35x(2x^2+120+31x)=0
=>(2x^2+120+35x)(2x^2+31x+120)=0
=>\(x\in\left\{\dfrac{-35\pm\sqrt{265}}{4};-\dfrac{15}{2};-8\right\}\)
b: Đặt x^2-3x=a
Phương trình sẽ là \(\dfrac{1}{a+3}+\dfrac{2}{a+4}=\dfrac{6}{a+5}\)
\(\Leftrightarrow\dfrac{a+4+2a+6}{\left(a+3\right)\left(a+4\right)}=\dfrac{6}{a+5}\)
=>(3a+10)(a+5)=6(a^2+7a+12)
=>6a^2+42a+72=3a^2+15a+10a+50
=>3a^2+17a+22=0
=>x=-2 hoặc x=-11/3
4 tháng 5 2019
b, \(\frac{1}{x-1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\left(ĐKXĐ:x\ne\pm1;x\ne2\right)\)
\(\Leftrightarrow\)\(\frac{1}{x-1}+\frac{5}{2-x}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow\)\(\frac{\left(x+1\right)\left(2-x\right)+5\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(2-x\right)\left(x-1\right)}=\frac{15\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(2-x\right)}\)
Suy ra:
\(\Leftrightarrow\)(x+1)(2-x)+5(x-1)(x+1) = 15(x-1)
\(\Leftrightarrow\)2x-x2-x+2+5x2-5 = 15x-15
\(\Leftrightarrow\)2x-x2-x+5x2-15x = -15+5-2
\(\Leftrightarrow\)4x2-14x = -12
\(\Leftrightarrow4x^2-14x+12=0\)
\(\Leftrightarrow4x^2-8x-6x+12=0\)
\(\Leftrightarrow\)4x(x-2) - 6(x-2) = 0
\(\Leftrightarrow\left(x-2\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(kotm\right)\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất x = \(\frac{3}{2}\)
a)\(\frac{3x+2}{7}=\frac{4x-5}{6}\)
\(\Leftrightarrow\frac{6\left(3x+2\right)}{42}=\frac{7\left(4x-5\right)}{42}\)
\(\Leftrightarrow6\left(3x+2\right)=7\left(4x-5\right)\)
\(\Leftrightarrow18x+12=28x-35\)
\(\Leftrightarrow18x-28x=-12-35\)
\(\Leftrightarrow-10x=-47\Leftrightarrow x=\frac{47}{10}\)
b) \(\frac{x+1}{x+3}=\frac{x+2}{x+4}\left(đkxđ:x\ne-3,-4\right)\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=\left(x+2\right)\left(x+3\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)-\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow x^2+4x+x+4-x^2-3x+2x+6=0\)
\(\Leftrightarrow\left(x^2-x^2\right)+\left(4x+4\right)+\left(4+6\right)+\left(x-3x\right)=0\)
\(\Leftrightarrow4x+4+10-2x=0\)
\(\Leftrightarrow4x+14-2x=0\)
\(\Leftrightarrow2x=-14\Leftrightarrow x=-7\)
a) \(\frac{3x+2}{7}=\frac{4x-5}{6}\)
=> \(6\left(3x+2\right)=7\left(4x-5\right)\)
=> \(18x+12=28x-35\)
=> \(18x+12-28x+35=0\)
=> \(\left(18x-28x\right)+\left(12+35\right)=0\)
=> \(-10x+47=0\)
=> \(-10x=-47\Rightarrow x=\frac{47}{10}\)
b) \(\frac{x+1}{x+3}=\frac{x+2}{x+4}\)
=> (x + 1)(x + 4) = (x + 2)(x + 3)
=> x(x + 4) + 1(x + 4) = x(x + 3) + 2(x + 3)
=> x2 + 4x + x + 4 = x2 + 3x + 2x + 6
=> x2 + 5x + 4 = x2 + 5x +6
=> x2 + 5x + 4 - x2 - 5x - 6 = 0
=> (x2 - x2) + (5x - 5x) + (4 - 6) = 0
=> -2 \(\ne\)0
=> không tìm được x thỏa mãn
hay cách khác : \(\frac{x+1}{x+3}=\frac{x+2}{x+4}\)
=> \(\frac{x+3-2}{x+3}=\frac{x+4-2}{x+4}\)
=> \(1-\frac{2}{x+3}=1-\frac{2}{x+4}\)
=> \(1-\frac{2}{x+3}-\left(1-\frac{2}{x+4}\right)=0\)
=> \(1-\frac{2}{x+3}-1+\frac{2}{x+4}=0\)
=> \(\left(1-1\right)+\left(-\frac{2}{x+3}+\frac{2}{x+4}\right)=0\)
=> \(-\frac{2}{x+3}+\frac{2}{x+4}=0\)
=> \(\frac{-2\left(x+4\right)+2\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}=0\)
=> \(-2x-8+2x+6=0\)
=> \(\left(-2x+2x\right)+\left(-8+6\right)=0\)
=> \(-2\ne0\)
=> không tìm được x thỏa mãn