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\(\dfrac{15-2x}{4}-\dfrac{x+1}{3}+\dfrac{6x-1}{2}=\dfrac{x-3}{6}\)
\(\Leftrightarrow45-6x-4x-4+36x-6=2x-12\) (quy đồng và khử mẫu)
\(\Leftrightarrow24x=23\)
\(\Leftrightarrow x=\dfrac{23}{24}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`(x+1)(x+3)=2x^2-2`
`<=>x^2+x+3x+3=2x^2-2`
`<=>x^2-4x-5=0`
`<=>x^2-5x+x-5=0`
`<=>x(x-5)+(x-5)=0`
`<=>(x-5)(x+1)=0`
`<=>` $\left[ \begin{array}{l}x=5\\x=-1\end{array} \right.$
Vậy `S={5,-1}`
Ta có: \(\left(x+1\right)\left(x+3\right)=2x^2-2\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2x^2+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x+3-2\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3-2x+2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
Vậy: S={-3;5}
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\(10x^2-5x\cdot\left(2x+3\right)=15\)
\(\Rightarrow10x^2-10x^2-15x-15=0\)
\(\Rightarrow-15x-15=0\)
\(\Rightarrow-15\cdot\left(x+1\right)=0\)
Vì -15< 0
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1:
a: =>3x=6
=>x=2
b: =>4x=16
=>x=4
c: =>4x-6=9-x
=>5x=15
=>x=3
d: =>7x-12=x+6
=>6x=18
=>x=3
2:
a: =>2x<=-8
=>x<=-4
b: =>x+5<0
=>x<-5
c: =>2x>8
=>x>4
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài `1:`
`h)(3/4x-1)(5/3x+2)=0`
`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`
______________
Bài `2:`
`b)3x-15=2x(x-5)`
`<=>3(x-5)-2x(x-5)=0`
`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`
`d)x(x+6)-7x-42=0`
`<=>x(x+6)-7(x+6)=0`
`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`
`f)x^3-2x^2-(x-2)=0`
`<=>x^2(x-2)-(x-2)=0`
`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`
`h)(3x-1)(6x+1)=(x+7)(3x-1)`
`<=>18x^2+3x-6x-1=3x^2-x+21x-7`
`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`
`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`
`j)(2x-5)^2-(x+2)^2=0`
`<=>(2x-5-x-2)(2x-5+x+2)=0`
`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`
`w)x^2-x-12=0`
`<=>x^2-4x+3x-12=0`
`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`
`m)(1-x)(5x+3)=(3x-7)(x-1)`
`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`
`<=>(1-x)(5x+3+3x-7)=0`
`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`
`p)(2x-1)^2-4=0`
`<=>(2x-1-2)(2x-1+2)=0`
`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`
`r)(2x-1)^2=49`
`<=>(2x-1-7)(2x-1+7)=0`
`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`
`t)(5x-3)^2-(4x-7)^2=0`
`<=>(5x-3-4x+7)(5x-3+4x-7)=0`
`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`
`u)x^2-10x+16=0`
`<=>x^2-8x-2x+16=0`
`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`
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1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)
ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow-7x+3=-4x-15\)
\(\Leftrightarrow-7x+4x=-15-3\)
\(\Leftrightarrow-3x=-18\)
\(\Leftrightarrow x=6\)( tmđk )
Vậy x = 6 là nghiệm của phương trình
2) 2x + 3 < 6 - ( 3 - 4x )
<=> 2x + 3 < 6 - 3 + 4x
<=> 2x - 4x < 6 - 3 - 3
<=> -2x < 0
<=> x > 0
Vậy nghiệm của bất phương trình là x > 0
![](https://rs.olm.vn/images/avt/0.png?1311)
do mũ chẵn nên (15-2x)^4=(2x-15)^4
ta có x-7+x-8=2x-15
đặt x-7=a,x-8=b thì 2x-15=a+b
ta có a^4+b^4=(a+b)^4
a^4+b^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4
suy ra 4a^3b+6a^2b^2+4ab^3=0
2ab(2a^2+3ab+2b^2)=0
suy ra 2ab=0 hoặc 2a^2+3ab+2b^2=0
ta có 2a^2+3ab+2b^2=0
=2a^2+3ab+9/8b^2+2b^2-9/8b^2
=2(a^2+3/2ab+9/16b^2)+7/8b^2
=2(a+3/4b)^2+7/8b^2>=0
dấu = xảy ra khi a=0,b=0
vậy x-7=0 và x-8=0 TH này ko xảy ra do ko đồng nhất nghiệm
TH 2ab=0
suy ra a=0 hoặc b=0 hoặc cả a và b = 0
như ta đã ns ở trên thì TH cả a và b =0 ko thỏa mãn
vậy a=0 hoặc b=0
x-7=0 hoặc x-8=0
x=7 hoặc x=8
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a: \(\Leftrightarrow\dfrac{3}{x-2}=\dfrac{2x-1}{x-2}-\dfrac{x\left(x-2\right)}{x-2}\)
=>3=2x-1-x^2+2x
=>3=-x^2+4x-1
=>x^2-4x+1+3=0
=>x^2-4x+4=0
=>x=2(loại)
b: =>(x+2)(2x-4)=x(2x+3)
=>2x^2-4x+4x-8=2x^2+3x
=>3x=-8
=>x=-8/3(nhận)
\(\left(x+15\right):\dfrac{3}{4}=2x-30\)
\(\Leftrightarrow\left(x+15\right)\cdot\dfrac{4}{3}=2x-30\)
\(\Leftrightarrow\dfrac{4}{3}x+20=2x-30\)
\(\Leftrightarrow2x-\dfrac{4}{3}x=20+30\)
\(\Leftrightarrow\dfrac{2}{3}x=50\)
\(\Leftrightarrow x=50\cdot\dfrac{3}{2}\)
\(\Leftrightarrow x=75\)
Vậy: ...