Giải các phương trình sau:
a, |x-2|=3
b, |x+1|=|2x+3|
c, |3x|=x+7
d, |-4x|= -2x+11
e, |3-2x|=3x-7
f, |3x|- x - 4=0
g, 9-|-5x|+2x=0
h, |x-9|=2x+5
i, |6-x|=2x-3
k, |3x-1|=4x+1
A) 2x³+6x²=x²+3x
B) (2x+5)²=(x+2)²
C) x²-5x+6=0
D) (2x-7)²-6(2x-7)(x-3)=0
E) (x-2)(x+1)=x²-4
G) 2x(2x-3)=(3-2x)(2-5x)
H) (1-x)(5x+3)=(3x-7)(x-1)
F) (x+6)(3x-1)+x+6=0
I) (4x-1)(x-3)=(x-3)(5x+2)
K) (x+4)(5x+9)-x-4=0
H) (x+3)(x-5)+(x+3)(3x-4)=0
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
B> <2X+5>2-<X+2>2=0
<2X+5-X-2><2X+X+2>=0
<X+3><3X+7>=0
X+3=0 HOẶC 3X+7=0
X=-3 HOẶC X=-7/3
C>X2-5X+6=0
X2-4X+4-X+2=0
<X-2>2-<X-2>=0
<X-2.><X-3>=0
X-2=0 HOẶC X-3=0
X=2 HOẶC X=3
D> <2X-7><2X-7-6<X-3>>=0
<2X-7><-4X+11>=0
2X-7=0 HOẶC -4X+11=0
X=7/2 HOẶC X=11/4
E><X-2><X+1>=X2-4
<X-2><X+1>-<X2-4>=0
<X-2><X+1>-<X-2><X+2>=0
-X+2=0
X=2
CÒN NHIÊU TỰ LÀM ĐI MỆT WA
Help me
A) 2x³+6x²=x²+3x
B) (2x+5)²=(x+2)²
C) x²-5x+6=0
D) (2x-7)²-6(2x-7)(x-3)=0
E) (x-2)(x+1)=x²-4
G) 2x(2x-3)=(3-2x)(2-5x)
H) (1-x)(5x+3)=(3x-7)(x-1)
F) (x+6)(3x-1)+x+6=0
I) (4x-1)(x-3)=(x-3)(5x+2)
K) (x+4)(5x+9)-x-4=0
H) (x+3)(x-5)+(x+3)(3x-4)=0
A) 2x³+6x²=x²+3x
B) (2x+5)²=(x+2)²
C) x²-5x+6=0
D) (2x-7)²-6(2x-7)(x-3)=0
E) (x-2)(x+1)=x²-4
G) 2x(2x-3)=(3-2x)(2-5x)
H) (1-x)(5x+3)=(3x-7)(x-1)
F) (x+6)(3x-1)+x+6=0
I) (4x-1)(x-3)=(x-3)(5x+2)
K) (x+4)(5x+9)-x-4=0
H) (x+3)(x-5)+(x+3)(3x-4)=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) 2x³+6x²=x²+3x
B) (2x+5)²=(x+2)²
C) x²-5x+6=0
D) (2x-7)²-6(2x-7)(x-3)=0
E) (x-2)(x+1)=x²-4
G) 2x(2x-3)=(3-2x)(2-5x)
H) (1-x)(5x+3)=(3x-7)(x-1)
F) (x+6)(3x-1)+x+6=0
I) (4x-1)(x-3)=(x-3)(5x+2)
K) (x+4)(5x+9)-x-4=0
H) (x+3)(x-5)+(x+3)(3x-4)=0
M) (2x+3)(-x+7)=0
Bài 1 : giải những các phương trình sau A. X² - 2x - 3 = 0 B. X² - 3x = 0 C. X² - 4x - 5 = 0 D. 5x² + 2x - 7 = 0 E. 2x² - 8 = 0 G. 3x² -7x + 1 = 0 H. X² - 4x + 1 = 0
a: =>(x-3)(x+1)=0
=>x=3 hoặc x=-1
b: =>x(x-3)=0
=>x=0 hoặc x=3
c: =>(x-5)(x+1)=0
=>x=5 hoặc x=-1
d: =>5x^2+7x-5x-7=0
=>(5x+7)(x-1)=0
=>x=1 hoặc x=-7/5
e: =>x^2-4=0
=>x=2 hoặc x=-4
h: =>x^2-4x+4-3=0
=>(x-2)^2=3
=>\(x=2\pm\sqrt{3}\)
Giải phương trình
a. 2x+6=0
b.4x+20=0
c. 2(x+1)=5x - 7
d. 2x-3=0
e.3x-1=2x_5
f.15-7x=9-3x
g. x-3=18
h. 2x+1=15-5x
i.3x-2=2x+5
k.-4x+8=0
l. 2x+3=0
m. 4x+5=3x
a) \(2x+6=0\Leftrightarrow2x=-6\Leftrightarrow x=-3
\)
b)\(4x+20=0\Leftrightarrow4x=-20\Leftrightarrow x=-5\)
c)\(2\left(x+1\right)=5x+7\Leftrightarrow2x+1=5x+7\Leftrightarrow2x-5x=7-1\Leftrightarrow-3x=6\Leftrightarrow x=-2\)
d)\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
e)\(3x-1=2x-5\Leftrightarrow3x-2x=-5+1\Leftrightarrow x=-4\)
f)\(15-7x=9-3x\Leftrightarrow-7x+3x=9-15\Leftrightarrow-4x=-6\Leftrightarrow x=\frac{3}{2}
\)
g)\(x-3=18\Leftrightarrow x=18+3\Leftrightarrow x=21\)
h)\(2x+1=15-5x\Leftrightarrow2x+5x=15-1\Leftrightarrow7x=14\Leftrightarrow x=2\)
i)\(3x-2=2x+5\Leftrightarrow3x-2x=5+2\Leftrightarrow x=7\)
k)\(-4x+8=0\Leftrightarrow-4x=-8\Leftrightarrow x=2\)
l)\(2x+3=0\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)
m)\(4x+5=3x\Leftrightarrow4x-3x=-5\Leftrightarrow x=-5\)
a) 5-4x=3x-9
b) (x-4)(3x+9)=0
c) (x/(x+4))+(12/(x-4))=(4x+48)/(x*x-16)
d) 4-2x=7-x
e) (x+4)(8-4x)=0
f) (x/(x+5))+(11/(x-5))=(x+55)/(x*x-25)
g) ((3x+2)/2) - ((3x+1)/6)= (5/3) +2x
h) 2x-(3-5x)=4(x+3)
i) 3x -6+x=9-x
k) 2t-3+5t= 4t+12
m.n giúp mik ạ ...
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 \(S=\left\{5\right\}\)
Giải các phương trình sau: a) 5x+9 = 2x b) (x+1).(4x-3)= (2x+5)(x+1) c) x/x-2 +x/x+2 = 4x/ x²-4 d) 11x-9= 5x+3 e) (2x+3)(3x-4) =0
c) \(\dfrac{x}{x-2}+\dfrac{x}{x+2}=\dfrac{4x}{x^2-4}.ĐKXĐ:x\ne2;-2\)
<=>\(\dfrac{x\left(x+2\right)}{x^2-4}+\dfrac{x\left(x-2\right)}{x^2-4}=\dfrac{4x}{x^2-4}\)
<=>x2+2x+x2-2x=4x
<=>2x2-4x=0
<=>2x(x-2)=0
<=>\(\left[{}\begin{matrix}2x=0< =>x=0\\x-2=0< =>x=2\left(loại\right)\end{matrix}\right.\)
Vậy pt trên có nghiệm là S={0}
d) 11x-9=5x+3
<=>11x-5x=9+3
<=>6x=12
<=>x=2
Vậy pt trên có nghiệm là S={2}
e) (2x+3)(3x-4) =0
<=> \(\left[{}\begin{matrix}2x+3=0< =>x=\dfrac{-3}{2}\\3x-4=0< =>x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy pt trên có tập nghiệm là S={\(\dfrac{-3}{2};\dfrac{4}{3}\)}
a) 5x+9 =2x
<=> 5x-2x=9
<=> 3x=9
<=> x=3
Vậy pt trên có nghiệm là S={3}
b) (x+1)(4x-3)=(2x+5)(x+1)
<=> (x+1)(4x-3)-(2x+5)(x+1)=0
<=>(x+1)(2x-8)=0
<=>\(\left[{}\begin{matrix}x+1=0< =>x=-1\\2x-8=0< =>2x=8< =>x=4\end{matrix}\right.\)
Vậy pt trên có tập nghiệm là S={-1;4}
c)
<=>
<=>x2+2x+x2-2x=4x
<=>2x2-4x=0
<=>2x(x-2)=0
<=>
Vậy pt trên có nghiệm là S={0}
d) 11x-9=5x+3
<=>11x-5x=9+3
<=>6x=12
<=>x=2
Vậy pt trên có nghiệm là S={2}
e) (2x+3)(3x-4) =0
<=>
Vậy pt trên có tập nghiệm là S={}
Giải các phương trình sau:
a.|3x-2|=2x+7 b.|-4,5x+5|=6+2,5x e.|3x-9|=2x+5
c.|5x-4|=3x+8 d.|7-4x|=2x+11 f.|6-5x|=2x-3
g.|3x-1|=3x+8 i.|x+5|=2x-2 m.|3x-1|=3x+1