\(\dfrac{x^2-x}{x+3}\)_\(\dfrac{x^2}{x-3}\)=\(\dfrac{7x^2-3x}{9-x^2}\)
Giải Phương Trình
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Y
1
TA
28 tháng 5 2021
ĐK: ` x \ne \pm 3`
`(x^2-x)/(x+3)-(x^2)/(x-3)=(7x^2-3x)/(9-x^2)`
`<=> (x^2-x)(x-3)-x^2 (x+3) = -(7x^2-3x)`
`<=> −7x^2+3x=-7x^2+3x`
`<=> 0x=0 forall x`
Vậy `S=RR \\ {+-3}`.
TL
2
NH
26 tháng 2 2023
\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{7x+5}{x^2-9}\left(x\ne3;x\ne-3\right)\\ < =>\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{7x+5}{\left(x-3\right)\left(x+3\right)}\)
suy ra:
`2(x+3)+3(x-3)=7x+5`
`<=>2x+6+3x-9=7x+5`
`<=>2x+3x-7x=5-6+9`
`<=> -2x=8`
`<=> x=-4(tm)`
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2023
ĐKXĐ: \(x\ne\pm3\)
\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{7x+5}{x^2-9}\)
\(\Leftrightarrow\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{7x+5}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2\left(x+3\right)+3\left(x-3\right)=7x+5\)
\(\Leftrightarrow2x+6+3x-9=7x+5\)
\(\Leftrightarrow2x=-8\)
\(\Leftrightarrow x=-4\) (thỏa)
Vậy pt có nghiệm \(x=-4\)
19 tháng 2 2022
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
5 tháng 2 2022
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)
3 tháng 2 2022
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
19 tháng 8 2021
1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)
\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(3x+9+4x-12=3x-7\)
\(\Leftrightarrow4x=-7+12-9=-4\)
hay \(x=-1\left(nhận\right)\)
2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(3x+12-4x+16=3x-4\)
\(\Leftrightarrow28-4x=-4\)
\(\Leftrightarrow4x=32\)
hay \(x=8\left(tm\right)\)
19 tháng 8 2021
3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
Suy ra: \(5x^2-12+3x+3=5x^2-5x\)
\(\Leftrightarrow3x-9+5x=0\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(nhận\right)\)
HP
9 tháng 5 2022
a. \(x-\dfrac{x+2}{3}< 3x+\dfrac{x}{2}+5\)
\(\Leftrightarrow\dfrac{6x}{6}-\dfrac{2\left(x+2\right)}{6}< \dfrac{18x}{6}+\dfrac{3x}{6}+\dfrac{30}{6}\)
\(\Rightarrow6x-2x-4-18x-3x-30< 0\)
\(\Leftrightarrow-17x< 34\)
\(\Leftrightarrow x>-2\)
b. \(\dfrac{x}{2}+\dfrac{1-x}{3}>0\)
\(\Leftrightarrow3x+2-2x>0\)
\(\Leftrightarrow x>-2\)
c. \(\left(x-9\right)^2-x\left(x+9\right)< 0\)
\(\Leftrightarrow x^2-18x+81-x^2-9x< 0\)
\(\Leftrightarrow-27x< -81\)
\(\Leftrightarrow x>3\)
\(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\\ \Leftrightarrow\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=-\dfrac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}\\ đkxđ:\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\\ \Leftrightarrow\dfrac{\left(x^2-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{x^3-3x^2-x^2+3x-x^3-3x^2+7x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{0}{\left(x-3\right)\left(x+3\right)}=0\\ \Rightarrow0=0\left(luon.dung\right)\)