Giải pt
\(\dfrac{5}{3x+2}=2x-1\)
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giải pt:
\(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
\(\Leftrightarrow\dfrac{15\left(2x-1\right)-2\left(3x+1\right)+20}{20}=\dfrac{8\left(3x+2\right)}{20}\)
\(\Rightarrow30x-15-6x-2+20=24x+16\)
\(\Leftrightarrow0x=13\) (Vô lí)
Vậy phương trình vô nghiệm
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GIẢI PT SAU:
\(\dfrac{2x+1}{3x+2}=5\)
\(\Rightarrow2x+1=5\left(3x+2\right)\)
=> 2x + 1 = 15x + 10
=> -13x - 9 = 0
\(\Rightarrow x=-\dfrac{9}{13}\)
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\(a.x+\dfrac{2x+\dfrac{x-1}{5}}{3}\)
\(b.\dfrac{3x-1-\dfrac{x-1}{2}}{3}-\dfrac{2x+\dfrac{1-2x}{3}}{2}=\dfrac{\dfrac{3x-1}{2}-6}{5}\) giải pt
GIẢI CÁC PT SAU:
\(\dfrac{2x+1}{3x+2}=5\)
\(\dfrac{2x^2-5x+2}{x-1}=\dfrac{2x^2+x+15}{x-3}\)
\(\dfrac{2x+3}{x-3}-\dfrac{4}{x+3}=\dfrac{24}{x^2-9}+2\)
Giải các pt sau:1)dfrac{2x+1}{x^2-4}+dfrac{2}{x+1}dfrac{3}{2-x}2)dfrac{3x+1}{1-3x}+dfrac{3+x}{3-x}23)dfrac{8x-2}{3}1+dfrac{5-2x}{4}4)dfrac{x}{x+1}-dfrac{2x+3}{x}dfrac{-3}{x+1}-dfrac{3}{x}5)dfrac{x+1}{x-1}-dfrac{x-1}{x+1}dfrac{4}{x^2-1}6)dfrac{2x+5}{2x}-dfrac{x}{x+5}0giúp mình với cám ơn
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Giải các pt sau:
1)\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+1}=\dfrac{3}{2-x}\)
2)\(\dfrac{3x+1}{1-3x}+\dfrac{3+x}{3-x}=2\)
3)\(\dfrac{8x-2}{3}=1+\dfrac{5-2x}{4}\)
4)
\(\dfrac{x}{x+1}-\dfrac{2x+3}{x}=\dfrac{-3}{x+1}-\dfrac{3}{x}\)
5)\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
6)\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
giúp mình với cám ơn
1: Sửa đề: 2/x+2
\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)
=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=>4x-3=-3x-6
=>7x=-3
=>x=-3/7(nhận)
2: \(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3-x\right)+\left(3+x\right)\left(1-3x\right)}{\left(1-3x\right)\left(3-x\right)}=2\)
=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)
=>-6x^2+6=2(3x^2-10x+3)
=>-6x^2+6=6x^2-20x+6
=>-12x^2+20x=0
=>-4x(3x-5)=0
=>x=5/3(nhận) hoặc x=0(nhận)
3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)
=>x*19/6=35/12
=>x=35/38
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Giải PT sau:a, 3x - 7 0b, 8 - 5x 0c, 3x - 2 5x + 8d, dfrac{3x-2}{3} dfrac{1-x}{2}e, ( 5x + 1)(x - 3) 0f, (x + 1)(2x - 3) 0g, 4x(x + 3) - 5(x + 3) 0h, 8(x - 6) - 2x(6 - x) 0i, dfrac{2}{x-1} + dfrac{1}{x} dfrac{2x+5}{x^2-x}k, dfrac{3}{x+2} - dfrac{2}{x-2} dfrac{2-x}{x^2-4}m, dfrac{3}{x} - dfrac{2}{x-3} dfrac{4-x}{x^2-3}n,dfrac{3}{2x+10}+ dfrac{2x}{x^2-25} dfrac{3}{x-5}u, dfrac{2}{x+3} - dfrac{3}{x-2} dfrac{x+4}{left(x+3right)left(x-2right)}
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Giải PT sau:
a, 3x - 7 = 0
b, 8 - 5x = 0
c, 3x - 2 = 5x + 8
d, \(\dfrac{3x-2}{3}\) = \(\dfrac{1-x}{2}\)
e, ( 5x + 1)(x - 3) = 0
f, (x + 1)(2x - 3) = 0
g, 4x(x + 3) - 5(x + 3) = 0
h, 8(x - 6) - 2x(6 - x) = 0
i, \(\dfrac{2}{x-1}\) + \(\dfrac{1}{x}\) = \(\dfrac{2x+5}{x^2-x}\)
k, \(\dfrac{3}{x+2}\) - \(\dfrac{2}{x-2}\) = \(\dfrac{2-x}{x^2-4}\)
m, \(\dfrac{3}{x}\) - \(\dfrac{2}{x-3}\) = \(\dfrac{4-x}{x^2-3}\)
n,\(\dfrac{3}{2x+10}\)+ \(\dfrac{2x}{x^2-25}\) = \(\dfrac{3}{x-5}\)
u, \(\dfrac{2}{x+3}\) - \(\dfrac{3}{x-2}\) = \(\dfrac{x+4}{\left(x+3\right)\left(x-2\right)}\)
a, 3x - 7 = 0
<=> 3x = 7
<=> x = 7/3
b, 8 - 5x = 0
<=> -5x = -8
<=> x = 8/5
c, 3x - 2 = 5x + 8
<=> -2x = 10
<=> x = -5
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e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)
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`a ) 3x - 7 = 0`
`\(\Leftrightarrow \) 3x = 7`
`\(\Leftrightarrow \) x = 7/3`
Vậy `S = {-7/3}`
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giải pt :
a, (x+5)(2-x)=3\(\sqrt{x^2+3x}\)
b, \(\sqrt[3]{\dfrac{2x}{x+1}}+\sqrt[3]{\dfrac{1}{2}+\dfrac{1}{2x}}=2\)
c,\(\sqrt[5]{\dfrac{16x}{x-1}}+\sqrt[5]{\dfrac{x-1}{16x}}=\dfrac{5}{2}\)
d, \(\sqrt{5x^2+10x+1}=7-2x-x^2\)
e, \(\sqrt{2x^2+4x+1}=1-2x-x^2\)
Giải bất pt:
\(\dfrac{2x}{5}+\dfrac{3-2x}{3}\ge\dfrac{3x+2}{2}\)
\(\dfrac{2x}{5}+\dfrac{3-2x}{3}\ge\dfrac{3x+2}{2}\)
\(\Leftrightarrow12x+10\left(3-2x\right)\ge15\left(3x+2\right)\)
\(\Leftrightarrow12x+30-20x-45x-30\ge0\)
\(\Leftrightarrow-53x\ge0\Leftrightarrow x\le0\)
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Giải pt: \(\dfrac{3}{5x-1}+\dfrac{2}{3-5x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\dfrac{5+96}{x^2-16}=\dfrac{2x—1}{x+4}-\dfrac{3x-1}{4-x}\)
a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)
Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)
Suy ra: \(9-3x+10x-2=4\)
\(\Leftrightarrow7x+7=4\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)
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