Giải các phương trình sau: 3x - 11 = 0.
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(4x-9=0\) \(\Leftrightarrow4x=9\) \(\Leftrightarrow x=\dfrac{9}{4}\)
Vậy \(x=\dfrac{9}{4}\)
b) \(-2x+50=0\) \(\Leftrightarrow2x=50\) \(\Leftrightarrow x=25\)
Vậy \(x=25\)
c) \(3x+11=0\) \(\Leftrightarrow3x=-11\) \(\Leftrightarrow x=-\dfrac{11}{3}\)
Vậy \(x=-\dfrac{11}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)
\(\Leftrightarrow6-8x-10+2x-5=0\)
\(\Leftrightarrow-6x+11=0\)
\(\Leftrightarrow-6x=-11\)
hay \(x=\dfrac{11}{6}\)
b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)
\(\Leftrightarrow6-12x-11+3x-1=0\)
\(\Leftrightarrow-9x-6=0\)
\(\Leftrightarrow-9x=6\)
hay \(x=-\dfrac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, -2x>15 x>-15/2 c, th1 x+2>0 vs x+3 <0 suy ra x>-2 vs x<-3 . th2 x+2<0,x+3>0 suy ra x<-2 ,x>-3
b, 112-x2>0
x2<112 x<11
a) \(3x-8>5x+7\)
\(\Leftrightarrow-8>5x+7-3x\)
\(\Leftrightarrow-8>2x+7\)
\(\Leftrightarrow-8-7>2x\)
\(\Leftrightarrow-15>2x\)
\(\Leftrightarrow-\frac{15}{2}>x\)
\(\Rightarrow x< -\frac{15}{2}\)
b) \(\left(11-x\right)\left(11+x\right)>0\)
\(\Leftrightarrow x=\pm11\)
\(\Rightarrow-11< x< 11\)
c) \(\left(x+2\right)\left(x+3\right)< 0\)
\(\Leftrightarrow x=-2;-3\)
\(\Rightarrow-3< x< -2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(2 – 3x)(x + 11) = (3x – 2)(2 – 5x)
⇔ (2 – 3x)(x + 11) – (3x – 2)(2 – 5x) = 0
⇔ (2 – 3x)(x + 11) + (2 – 3x)(2 – 5x) = 0
⇔ (2 – 3x)[(x + 11) + (2 – 5x)] = 0
⇔ (2 – 3x)(x + 11 + 2 – 5x) = 0
⇔ (2 – 3x)(13 – 4x) = 0 ⇔ 2 – 3x = 0 hoặc 13 – 4x = 0
2 – 3x = 0 ⇔ x = 2/3
13 – 4x = 0 ⇔ x = 13/4
Vậy phương trình có nghiệm x = 2/3 hoặc x = 13/4
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 7x+35=0
=>7x=-35
=>x=-5
b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
=>8-x-8(x-7)=1
=>8-x-8x+56=1
=>-9x+64=1
=>-9x=-63
hay x=7(loại)
a, \(7x=-35\Leftrightarrow x=-5\)
b, đk : x khác 7
\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)
vậy pt vô nghiệm
2, thiếu đề
![](https://rs.olm.vn/images/avt/0.png?1311)
4) Ta có: \(\dfrac{2x-5}{5}-\dfrac{x+3}{3}=\dfrac{2-3x}{2}-x-2\)
\(\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}-\dfrac{10\left(x+3\right)}{30}=\dfrac{15\left(2-3x\right)}{30}-\dfrac{30\left(x+2\right)}{30}\)
\(\Leftrightarrow12x-30-10x-30=30-45x-30x-60\)
\(\Leftrightarrow-22x-60=-75x-30\)
\(\Leftrightarrow-22x+75x=-30+60\)
\(\Leftrightarrow53x=30\)
\(\Leftrightarrow x=\dfrac{30}{53}\)
Vậy: \(S=\left\{\dfrac{30}{53}\right\}\)
5) Ta có: \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=5\)
\(\Leftrightarrow\dfrac{2\left(5x-3\right)}{12}-\dfrac{3\left(7x-1\right)}{12}=\dfrac{60}{12}\)
\(\Leftrightarrow10x-6-21x+3=60\)
\(\Leftrightarrow-11x-3=60\)
\(\Leftrightarrow-11x=63\)
\(\Leftrightarrow x=-\dfrac{63}{11}\)
Vậy: \(S=\left\{-\dfrac{63}{11}\right\}\)
`9,x^3+x^2-2=0`
`x^3-x^2+2x^2-2=0`
`<=>x^2(x-1)+2(x-1)(x+1)=0`
`<=>(x-1)(x^2+2x+2)=0`
`<=>x=1`
`14,x^2-2x+1=0`
`<=>(x-1)^2=0`
`<=>x-1=0`
`<=>x=1`
`15,x^3+3x^2+3x+1=0`
`<=>(x+1)^3=0`
`<=>x+1=0`
`<=>x=-1`
3x - 11 = 0 ⇔ 3x = 11 ⇔
. Vậy phương trình có một nghiệm
.