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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
Ta có: \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2-2^2=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)
Ta có : \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=2\\2x-1=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{1}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,x2-25-(x+5) b,mình quên mất rồi.Đợi tí nhé
(x2-25)-(x+5)=0
(x2-52)+(x-5)=0
(x-5)(x+5)+(x-5)=0
(x-5)(x+5+1)=0
x-5=0 hoặc x+5+1=0
x=0+5 hoặc x=0-5-1
x=5 hoặc x=-6
Vậy x=5 và x=-6
Giải bpt
A) (x^2+1)×(4x-2)≫0(lớn hơn hoặc =0)
B) (x-2)×x^2>0
Mog mn giúp ạ
E cần gấp
Thak mn
![](https://rs.olm.vn/images/avt/0.png?1311)
I 2x-3 I = I x+1 I
2x-3 = x+1
x+1 - 2x+3=0
x (1-2) +1+3=0
-1x +4 =0
-1x = 0-4
-1x =-4
x = -4 : -1
x =4
Trả lời:
\(\left|2x-3\right|=\left|x+1\right|\)
\(\Rightarrow2x-3=x+1\) hoặc \(2x-3=-\left(x+1\right)\)
TH1: \(2x-3=x+1\)
\(2x-x=1+3\)
\(x=4\)
TH2: \(2x-3=-\left(x+1\right)\)
\(2x-3=-x-1\)
\(2x+x=-1+3\)
\(3x=2\)
\(x=\frac{2}{3}\)
Vậy \(x=4;x=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (-2) . ( x+7 ) + (-5) = 7
<=>(-2).(x+7)=7+5
<=>x+7=12:(-2)
<=>x+7=-6
<=>x=(-6)-7
<=>x=-13
Vậy x=-13
b)(x+4) : (-7) = 14
<=>x+4=14 x (-7)
<=>x+4=-98
<=>x=-98-4
<=>x=-102
Vậy x= -102
c) 72 : ( x+5) - 4 = -12
<=>72:(x+5)=(-12)+4
<=>x+5=72:(-8)
<=>x+5=-9
<=>x=-9-5
<=>x=-14
Vậy x= -14
d) (x+3) : (-6 ) + 12 = 8
<=>(x+3) :(-6)=8-12
<=>x+3=(-4)x(-6)
<=>x+3=24
<=>x=24-3
<=>x=21
Vậy x= 21
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{x+1}{29}+\dfrac{x+3}{27}=\dfrac{x-3}{33}+\dfrac{x-7}{37}\)
\(\Leftrightarrow\dfrac{x+30}{29}+\dfrac{x+30}{27}-\dfrac{x+30}{33}-\dfrac{x+30}{37}=0\)
\(\Leftrightarrow x+30=0\)
hay x=-30
![](https://rs.olm.vn/images/avt/0.png?1311)
7x+2/5x+7=7x-1/5x+1=>37/5x+7=34/5x+1=>37/5x-34/5x=1-7=>3/5x=-6=>x=-6:3/5=-10 vay x=-10 nho ****
(2x-123)-(x+27)=0
2x-123-x-27=0
x-150=0
x=150