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\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
a) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{1}{3x}+\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4}{12x}+\dfrac{6}{12x}=\dfrac{3x}{12x}\)
Suy ra: \(3x=10\)
\(\Leftrightarrow x=\dfrac{10}{3}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{10}{3}\right\}\)
b) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}\)
\(\Leftrightarrow\dfrac{3x}{8x^2}-\dfrac{4x}{8x^2}=\dfrac{8}{8x^2}\)
Suy ra: \(3x-4x=8\)
\(\Leftrightarrow-x=8\)
hay x=-8(thỏa ĐK)
Vậy: S={-8}
c)ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{1}{2x}+\dfrac{3}{4x}=\dfrac{5}{2x^2}\)
\(\Leftrightarrow\dfrac{2x}{4x^2}+\dfrac{3x}{4x^2}=\dfrac{10}{4x^2}\)
Suy ra: 2x+3x=10
\(\Leftrightarrow5x=10\)
hay x=2(thỏa ĐK)
Vậy: S={2}
d, \(\dfrac{2a}{x+a}=1\) (x \(\ne\) -a)
\(\Leftrightarrow\) \(\dfrac{2a}{x+a}-\dfrac{x+a}{x+a}=0\)
\(\Leftrightarrow\) \(\dfrac{a-x}{x+a}=0\)
\(\Leftrightarrow\) a - x = 0 (x + a \(\ne\) 0)
\(\Leftrightarrow\) x = a (TM)
Vậy S = {a}
Chúc bn học tốt!
d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)
\(\Leftrightarrow x^2-3x-2x^2-6x+3x=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow-x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)
a: \(\Leftrightarrow4\left(2x+1\right)-3\left(6x-1\right)=2x+1\)
=>8x+4-18x+3=2x+1
=>-10x+7=2x+1
=>-12x=-6
hay x=1/2
b: \(\Leftrightarrow4x^2-12x+7x-21-x^2=3x^2+6x\)
=>5x-21=6x
=>-x=21
hay x=-21
1: Ta có: \(\dfrac{3}{x+2}-\dfrac{x-1}{x^2-4}=\dfrac{2}{x-2}\)
Suy ra: \(3x-6-x+1=2x+4\)
\(\Leftrightarrow2x-5=2x+4\left(vôlý\right)\)
2: Ta có: \(\dfrac{x-5}{2x-3}-\dfrac{x}{2x+3}=\dfrac{1-6x}{4x^2-9}\)
Suy ra: \(\left(x-5\right)\left(2x+3\right)-x\left(2x-3\right)=1-6x\)
\(\Leftrightarrow2x^2-7x-15-2x^2+6x+6x-1=0\)
\(\Leftrightarrow5x=16\)
hay \(x=\dfrac{16}{5}\)
1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)
\(\Leftrightarrow-3x-12-3+5x-x+4=0\)
\(\Leftrightarrow x=11\left(nhận\right)\)
2. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Leftrightarrow \frac{3(x-2)-(x-1)}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)
\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)
Vậy pt vô nghiệm
`8(x-3)(x+1)=8x^2 +11`
`<=>8(x^2 +x-3x-3)-8x^2 -11=0`
`<=>8x^2 +8x-24x-24-8x^2 -11=0`
`<=>-16x-35=0`
`<=>-16x=35`
`<=>x=-35/16`
\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(x\ne0;x\ne2\right)\\ < =>\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)
suy ra
`x^2 +2x-2=x-2`
`<=>x^2 +2x-x-2+2=0`
`<=>x^2 +x=0`
`<=>x(x+1)=0`
\(< =>\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ < =>x=-1\)
\(a,8\left(x-3\right)\left(x+1\right)=8x^2+11\\ \Leftrightarrow\left(8x-24\right)\left(x+1\right)=8x^2+11\\ \Leftrightarrow8x^2-24x+8x-24-8x^2-11=0\\ \Leftrightarrow-16x-35=0\\ \Leftrightarrow x=\dfrac{-35}{16}\)
Vậy \(x=-\dfrac{35}{16}\)
\(b,đkxđ:x\ne2;x\ne0\)
\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}-\dfrac{1}{x}=0\\ \Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=0\\ \Leftrightarrow x^2+2x-2-x+2=0\\ \Leftrightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)
Vậy \(x=-1\)
@ducminh
1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)
=>-4x<12
hay x>-3
2: \(\Leftrightarrow6+2x+2>2x-1-12\)
=>8>-13(đúng)
4: \(\dfrac{2x+1}{x-3}\le2\)
\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)
=>x-3<0
hay x<3
6: =>(x+4)(x-1)<=0
=>-4<=x<=1
\(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
\(\Leftrightarrow\dfrac{20\left(2x-1\right)}{60}+\dfrac{15\left(3x-2\right)}{60}=\dfrac{12\left(4x-3\right)}{60}\)
`<=> 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`
a) \(\dfrac{2x+1}{x-2}=3\Rightarrow2x+1=3x-6\Rightarrow x=7\)
b) \(\dfrac{2x-3}{x+1}=\dfrac{1}{2}\Rightarrow4x-6=x+1\Rightarrow3x=7\Rightarrow x=\dfrac{7}{3}\)
a) \(\dfrac{2x+1}{x-2}=3\)
dkxd : x ≠ 2
MTC : x - 2
Quy đồng mẫu thức :
⇒ \(\dfrac{2x+1}{x-2}=\dfrac{3\left(x-2\right)}{x-2}\)
Suy ra : 2x + 1 = 3(x - 2)
\(\) \(\Leftrightarrow\) 2x + 1 = 3x - 6
\(\Leftrightarrow\) 2x + 1 - 3x + 6 = 0
\(\Leftrightarrow\) -1x + 7 = 0
\(\Leftrightarrow\) -1x = -7
\(\Leftrightarrow\) x = \(\dfrac{-7}{-1}=7\)
Vậy S = \(\left\{7\right\}\)
b) \(\dfrac{2x-3}{x+1}=\dfrac{1}{2}\)
dkxd : x ≠ -1
MTC : 2(x + 1)
Quy đồng mẫu thức :
⇒ \(\dfrac{2\left(2x-3\right)}{2\left(x+1\right)}=\dfrac{1\left(x+1\right)}{2\left(x+1\right)}\)
Suy ra : 2(2x - 3) = x + 1
\(\Leftrightarrow\) 4x - 6 - x - 1 = 0
\(\Leftrightarrow\) 3x - 7 = 0
\(\Leftrightarrow\) 3x = 7
\(\Leftrightarrow\) x = \(\dfrac{7}{3}\)
Vậy S = \(\left\{\dfrac{7}{3}\right\}\)
Chúc bạn học tốt