a,|9+x|=2x
b,|x+6|-9=2x
c,|2x-3|+x =21
d,|4+2x|=-4x
e,|5x-4|=|x+2|
f,|2x-6|+|x+3|=8
g,|x-2|+|x-5|=3
h,|x+5|+|x-3|=9
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a) \(\left|x+9\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=2x\\x+9=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
b) \(\left|5x\right|-3x=2\Leftrightarrow\left|5x\right|=3x+2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=3x+2\\-5x=3x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1}{4}\end{matrix}\right.\)
c) \(\left|x+6\right|-9=2x\Leftrightarrow\left|x+6\right|=2x+9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=2x+9\\-x-6=2x+9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
d) \(\left|2x-3\right|+x=21\Leftrightarrow\left|2x-3\right|=21-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=21-x\\2x-3=x-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-18\end{matrix}\right.\)
e) \(\left|2x+4\right|=-4x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=4x\\2x+4=-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{-2}{3}\end{matrix}\right.\)
i) \(\left|3x-1\right|+2=x\Leftrightarrow\left|3x-1\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=x-2\\3x-1=2-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{3}{4}\end{matrix}\right.\)
g) \(\left|x+15\right|+1=3x\Leftrightarrow\left|x+15\right|=3x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-1\\x+15=1-3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3,5\end{matrix}\right.\)
h) \(\left|2x-5\right|+x=2\Leftrightarrow\left|2x-5\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=2-x\\2x-5=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=3\end{matrix}\right.\)
a) |9+x|=2x
TH1: 9+x=2x
<=> 9=2x-x
<=> x=9
TH2: -9-x=2x
<=> -9=3x
<=> x=-3
b) |5x|-3x=2
TH1: 5x-3x=2
<=> 2x=2
<=> x=1
TH2: -5x-3x=2
<=> -8x=2
<=>x=-4
c) |x+6|-9=2x
TH1: x+6-9=2x
<=> -3=x
TH2: -x-6-9=2x
<=> -15=3x
<=>x=-5
d) |2x-3|+x=21
TH1: 2x-3+x=21
<=> 3x=24
<=> x=8
TH2: -2x+3+x=21
<=> -x=18
<=> x=-18
e,i,g,h tương tự
a) 5 - 4x = 3x - 9
\(\Leftrightarrow5-4x-3x+9=0\)
\(\Leftrightarrow14-7x=0\)
\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b) \(\left(x-4\right)\left(3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{-3;4\right\}\)
c) \(\dfrac{x}{x+4}+\dfrac{12}{x-4}=\dfrac{4x+48}{x\cdot x-16}\)(1)
ĐKXĐ: \(x\ne\pm4\)
\(\left(1\right)\Leftrightarrow\dfrac{x\left(x-4\right)+12\left(x+4\right)-4x-48}{\left(x+4\right)\left(x-4\right)}=0\)
\(\Leftrightarrow x^2-4x+12x+48-4x-48=0\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-4\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
d) \(4-2x=7-x\)
\(\Leftrightarrow4-2x-7+x=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
e) \(\left(x+4\right) \left(8-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\8-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-4;2\right\}\)
f) \(\dfrac{x}{x+5}+\dfrac{11}{x-5}=\dfrac{x+55}{x\cdot x-25}\left(2\right)\)
ĐKXĐ: \(x\ne\pm5\)
\(\left(2\right)\Leftrightarrow\dfrac{x\left(x-5\right)+11\left(x+5\right)-x-55}{\left(x+5\right)\left(x-5\right)}=0\)
\(\Leftrightarrow x^2-5x+11x+55-x-55=0\)
\(\Leftrightarrow x^2+5x=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-5\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
g) \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)-3x-1-10-12x}{6}=0\)
\(\Leftrightarrow9x+6-3x-1-10-12x=0\)
\(\Leftrightarrow-6x-5=0\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy \(S=\left\{-\dfrac{5}{6}\right\}\)
h) \(2x-\left(3-5x\right)=4\left(x+3\right)\)
\(\Leftrightarrow2x-3+5x-4x-12=0\)
\(\Leftrightarrow3x-15=0\)
\(\Leftrightarrow x=5\)
Vậy \(S=\left\{5\right\}\)
i) \(3x-6+x=9-x\)
\(\Leftrightarrow3x-6+x-9+x=0\)
\(\Leftrightarrow5x-15=0\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
k)\(2t-3+5t=4t+12\)
\(\Leftrightarrow2t-3+5t-4t-12=0\)
\(\Leftrightarrow3t-15=0\)
\(\Leftrightarrow t=5\)
Vậy \(S=\left\{5\right\}\)
c. x^2-5x +6 = 0
<=> x^2 - 5x = -6
<=> - 4x = -6
<=> x= -6/-4
Mình chỉ phân tích đa thức thành nhân tử thôi , phần còn lại bạn tự tính nha keo dài lắm
A) 2x2(x+3) - x(x+3) = 0 <=> x(x - 3)(2x-1)=0
B) (2x+5)2 - (x+2)2=0 <=> (x+3)(3x+7)=0
C) (x2-2x) - (3x-6)=0 <=> (x-2)(x-3)=0
D) (2x-7)(2x-7-6x+18)=0 <=> (2x-7)(-4x+11)=0
E) (x-2)(x+1) - (x-2)(x+2)=0 <=> (x-2)*(-1)=0 <=> x-2=0
G) (2x-3)(2x+2-5x)=0 <=> (2x-3)(-3x+2)=0
H) (1-x)(5x+3+3x-7)=0 <=> (1-x)(8x-4)=0
F) (x+6)*3x=0
I) (x-3)(4x-1-5x-2)=0 <=> (x-3)(-x-3)=0
K) (x+4)(5x+8)=0
H) (x+3)(4x-9)=0
c. x^2-5x+6=0
<=> x^2-5x=-6
<=> -4x=-6
<=> x=-6/-4
vậy tập nghiệm của pt là s={-6/-4}
a) | 5/4x -7/2| - | 5/8x + 3/5| = 0
|5/4x - 7/2| = | 5/8x + 3/5|
TH1: 5/4x - 7/2 = 5/8x + 3/5
=> 5/4x - 5/8x = 3/5 +7/2
5/8x = 41/10
x = 41/10:5/8
x = 164/25
TH2: 5/4x - 7/2 = -5/8x - 3/5
=> 5/4x + 5/8x = -3/5 +7/2
15/8x = 29/10
x = 29/10 : 15/8
x = 116/75
KL: x = 164/25 hoặc x = 116/75
các bài cn lại b lm tương tự nha! h lm dài lắm!
a,\(\left|9+x\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}9+x=2x\\9x+x=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}9=x\\9=-3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
Vậy...
Trường hợp 2 chưa chắc chắn lắm!!!
a) \(\left|9+x\right|=2x\)
Xét trường hợp 1:
\(9+x=2x\)
\(\Leftrightarrow9+x-2x=0\)
\(\Leftrightarrow9-x=0\)
\(\Leftrightarrow x=9\)
Xét trường hợp 2:
\(9+x=-2x\)
\(\Leftrightarrow9+x-\left(-2x\right)=0\)
\(\Leftrightarrow9+x+2x=0\)
\(\Leftrightarrow9+3x=0\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-9:3\)
\(\Leftrightarrow x=-3\)
Vậy x=9 hoặc x=-3
b) \(\left|x+6\right|-9=2x\)
\(\Leftrightarrow\left|x+6\right|=2x+9\)
Xét trường hợp 1:
\(x+6=2x+9\)
\(\Leftrightarrow x+6-\left(2x+9\right)=0\)
\(\Leftrightarrow x+6-2x-9=0\)
\(\Leftrightarrow-3-x=0\)
\(\Leftrightarrow x=-3\)
Xét trường hợp 2:
\(x+6=-\left(2x+9\right)\)
\(\Leftrightarrow x+6-\left[-\left(2x+9\right)\right]=0\)
\(\Leftrightarrow x+6+\left(2x+9\right)=0\)
\(\Leftrightarrow x+6+2x+9=0\)
\(\Leftrightarrow3x+15=0\)
\(\Leftrightarrow3x=-15\)
\(\Leftrightarrow x=-15:3\)
\(\Leftrightarrow x=-5\)
Vậy x=-3 hoặc x=-5