Giải các phương trình sau:
a. (x−1)(5x+3)=(3x−8)(x−1)(x−1)(5x+3)=(3x−8)(x−1)
b. 3x(25x+15)−35(5x+3)=0
a) (x-1)(5x+3)=(3x-8)(x-1)
b) 3x(25x+15)-35(5x+3)=0
a) (x - 1)(5x + 3) = (3x - 8)(x - 1)
\(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Leftrightarrow x-1=0\Rightarrow x=1\)
và\(2x+11=0\Rightarrow x=\frac{-11}{2}\)
GPT sau:
a) ( x-1)(5x+3)= (3x - 8 )(x-1)
b) 3x ( 25x + 15 )- 35 ( 5x+3) = 0
c) (2-3x ) ( x-11)=(3x-2)(2- 5x)
Giups mk vs thank cacs bn
b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)
a)(x-1)(5x+3)=(3x-8)(x-1)
\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0
\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)
a) Ta có: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\Leftrightarrow5x^2+3x-5x-3=3x^2-3x-8x+8\)
\(\Leftrightarrow5x^2-2x-3=3x^2-11x+8\)
\(\Leftrightarrow5x^2-2x-3-3x^2+11x-8=0\)
\(\Leftrightarrow2x^2+9x-11=0\)
\(\Leftrightarrow2x^2+11x-2x-11=0\)
\(\Leftrightarrow x\left(2x+11\right)-\left(2x+11\right)=0\)
\(\Leftrightarrow\left(2x+11\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+11=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-11\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=1\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{11}{2};1\right\}\)
b) Ta có: \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow3x\cdot5\cdot\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(5x+3\right)\left(15x-35\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\15x-35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\15x=35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) Ta có: \(\left(2-3x\right)\left(x-11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\Leftrightarrow2x-22-3x^2+33x=6x-15x^2-4+10x\)
\(\Leftrightarrow-3x^2+35x-22=-15x^2+16x-4\)
\(\Leftrightarrow-3x^2+35x-22+15x^2-16x+4=0\)
\(\Leftrightarrow12x^2+19x-18=0\)
\(\Leftrightarrow12x^2+27x-8x-18=0\)
\(\Leftrightarrow3x\left(4x+9\right)-2\left(4x+9\right)=0\)
\(\Leftrightarrow\left(4x+9\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+9=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-9\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{9}{4};\dfrac{2}{3}\right\}\)
giải các phương trình sau:
a.3(x-2)-10=5(2x + 1)
b.3x + 2=8 -2(x-7)
c.2x-(2+5x)= 4(x + 3)
d.5-(x +8)=3x + 3(x-9)
e.3x - 18 + x= 12-(5x + 3)
a. 3(x-2)-10=5(2x + 1)
<=> 3x - 6 - 10 = 10x + 5
<=> 3x - 10x = 5 + 6 + 10
<=> -7x = 21
<=> x = -3
b. 3x + 2=8 -2(x-7)
<=> 3x + 2 = 8 - 2x + 14
<=> 3x + 2x = 8 + 14 - 2
<=> 5x = 20
<=> x = 4
c. 2x-(2+5x)= 4(x + 3)
<=> 2x - 2 - 5x = 4x + 12
<=> 2x - 5x - 4x = 12 + 2
<=> -7x = 14
<=> x = -2
d. 5-(x +8)=3x + 3(x-9)
<=> 5 - x - 8 = 3x + 3x - 27
<=> -x - 3x - 3x = -27 + 8 - 5
<=> -7x = -24
<=> x = 24/7
e. 3x - 18 + x= 12-(5x + 3)
<=> 3x - 18 + x = 12 - 5x - 3
<=> 3x + x - 5x = 12 - 3 + 18
<=> -x = 27
<=> x = - 27
a. 3(x-2)-10=5(2x + 1)
<=> 3x - 6 - 10 = 10x + 5
<=> 3x - 10x = 5 + 6 + 10
<=> -7x = 21
<=> x = -3
b. 3x + 2=8 -2(x-7)
<=> 3x + 2 = 8 - 2x + 14
<=> 3x + 2x = 8 + 14 - 2
<=> 5x = 20
<=> x = 4
c. 2x-(2+5x)= 4(x + 3)
<=> 2x - 2 - 5x = 4x + 12
<=> 2x - 5x - 4x = 12 + 2
<=> -7x = 14
<=> x = -2
d. 5-(x +8)=3x + 3(x-9)
<=> 5 - x - 8 = 3x + 3x - 27
<=> -x - 3x - 3x = -27 + 8 - 5
<=> -7x = -24
<=> x = 24/7
e. 3x - 18 + x= 12-(5x + 3)
<=> 3x - 18 + x = 12 - 5x - 3
<=> 3x + x - 5x = 12 - 3 + 18
<=> -x = 27
<=> x = - 27
Mọi người giúp mình giải 2 bài phương trình tích này nhá!!!
a/ (x-1)(5x+3) = (3x-8)(x-1)
b/ 3x(25x+15) - 35(5x+3)=0
a) (x - 1)(5x + 3) = (3x - 8)(x - 1)
⇔ (x - 1)(5x + 3) - (3x - 8)(x - 1) = 0
⇔ (x - 1)(5x + 3 - 3x + 8) = 0
⇔ (x - 1)(2x + 11) = 0
⇔\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-11}{2}\end{matrix}\right.\)
Vậy S = {1; \(\frac{-11}{2}\)}
b) 3x(25x + 15) - 35(5x + 3) = 0
⇔ 15x(5x + 3) - 35(5x + 3) = 0
⇔ 5(3x - 7)(5x + 3) = 0
⇔ \(\left[{}\begin{matrix}3x-7=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=\frac{-3}{5}\end{matrix}\right.\)
Vậy S = {\(\frac{7}{3};\frac{-3}{5}\)}
a/ \(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{11}{2}\end{matrix}\right.\)
b/ \(3x.5\left(5x+3\right)-5.7\left(5x+3\right)=0\)
\(\Leftrightarrow5\left(3x-7\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-7=0\\5x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=-\frac{3}{5}\end{matrix}\right.\)
a. (x – 1)(5x + 3) = (3x – 8)(x – 1)
b. 3x(25x + 15) – 35(5x + 3) = 0
c. (2 – 3x)(x + 11) = (3x – 2)(2 – 5x)
d. (2x2 + 1)(4x – 3) = (2x2 + 1)(x – 12)
e. (2x – 1)2 + (2 – x)(2x – 1) = 0
f. (x + 2)(3 – 4x) = x2 + 4x + 4
\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)
\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\left(5x+3\right).5\left(3x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)
\(c,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\left(3x-2\right)\left(2-5x\right)+\left(3x-2\right)\left(x+11\right)=0\)
\(\left(3x-2\right)\left(2-5x+x+11\right)=0\)
\(\left(3x-2\right)\left(13-4x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=2\\4x=13\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}}\)
còn đâu tự lm lười :_#
giải các phương trình sau:
a.|2-5x|=|2x-3|
b.|2-5x|-2|x+1|=3x+5
c.|3x-1|+2|x-1|=|5x-3|
a: =>|5x-2|=|2x-3|
=>5x-2=2x-3 hoặc 5x-2=-2x+3
=>3x=-1 hoặc 7x=5
=>x=5/7 hoặc x=-1/3
b: =>|5x-2|-|2x+2|=3x+5
TH1 x<-1
PT sẽ là 2-5x+2x+2=3x+5
=>-3x+4=3x+5
=>-6x=1
=>x=-1/6(loại)
TH2: -1<=x<2/5
Pt sẽ là 2-5x-2x-2=3x+5
=>-7x=3x+5
=>-4x=5
=>x=-5/4(loại)
Th3: x>=2/5
PT sẽ là 5x-2-2x-2=3x+5
=>3x-4=3x+5
=>0x=9(loại)
Bài 1: Giải các phương trình sau:
a) 3x – 15 = 0 b) 4x + 20 = 0 c) -5x – 20 = 0 d) 3x + 1 = 7x – 11
e) 3 + 2x = 2(x + 1) g
a: 3x-15=0
nên 3x=15
hay x=5
b: 4x+20=0
nên 4x=-20
hay x=-5
c: -5x-20=0
nên -5x=20
hay x=-4
Giải các phương trình sau:
a) \(5x - 30 = 0\);
b) \(4 - 3x = 11\);
c) \(3x + x + 20 = 0\);
d) \(\dfrac{1}{3}x + \dfrac{1}{2} = x + 2\).
a) \(5x - 30 = 0\)
\(5x = 0 + 30\)
\(5x = 30\)
\(x = 30:5\)
\(x = 6\)
Vậy phương trình có nghiệm \(x = 6\).
b) \(4 - 3x = 11\)
\( - 3x = 11 - 4\)
\( - 3x = 7\)
\(x = \left( { 7} \right):\left( { - 3} \right)\)
\(x = \dfrac{-7}{3}\)
Vậy phương trình có nghiệm \(x = \dfrac{7}{3}\).
c) \(3x + x + 20 = 0\)
\(4x + 20 = 0\)
\(4x = 0 - 20\)
\(4x = - 20\)
\(x = \left( { - 20} \right):4\)
\(x = - 5\)
Vậy phương trình có nghiệm \(x = - 5\).
d) \(\dfrac{1}{3}x + \dfrac{1}{2} = x + 2\)
\(\dfrac{1}{3}x - x = 2 - \dfrac{1}{2}\)
\(\dfrac{{ - 2}}{3}x = \dfrac{3}{2}\)
\(x = \dfrac{3}{2}:\left( {\dfrac{{ - 2}}{3}} \right)\)
\(x = \dfrac{{ - 9}}{4}\)
Vậy phương trình có nghiệm \(x = \dfrac{{ - 9}}{4}\).
Giải các bất phương trình sau và biểu diễn tập nghiệm trên trục số:
a.3x-5 >15-x b.3(x-2).(x+2)<3x^2+x
c.(2x+1)^2+(1-x).3x<hoặc=(x+2)^2
d.5x-20/3 - 2x^2+x/2 > x.(1-3x)/3 -5x/4
e.4-2x <hoặc= 3x-6
f.(x+4).(5x-1)>5x^2+16x+2
g)x.(2x-1)-8<5-2x(1-x)
h)3x-1/4 - 3.(x-2)/8 - 1>5-3x/2
a: 3x-5>15-x
=>4x>20
hay x>5
b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
=>3x2+x>3x2-12
=>x>-12