Tìm x biết:
\(A=\frac{x-2}{3x+2}\)
a, A=0
b, A<0
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PN
2
15 tháng 8 2021
a: Ta có: \(\left(x+1\right)^2-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b: Ta có: \(2\left(3x-2\right)^2=9x^2-4\)
\(\Leftrightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(6x-4-3x-2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
13 tháng 8 2021
a: Ta có: \(3x\left(3x-1\right)-\left(3x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow9x^2-3x-9x^2+1=0\)
\(\Leftrightarrow3x=1\)
hay \(x=\dfrac{1}{3}\)
b: Ta có: \(x^2-5x+25-5x=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
hay x=5
LP
18 tháng 8 2021
a)(2x-3)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)
Vậy x=3/2 hoặc x=-5
LT
18 tháng 8 2021
a) \(\left(2x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-5\right\}\)
b) \(3x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: \(S=\left\{2;\dfrac{7}{2}\right\}\)
c) \(5x\left(2x-3\right)-6x+9=0\)
\(\Leftrightarrow5x\left(2x-3\right)-3\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\5x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};\dfrac{3}{5}\right\}\)
DD
2
GH
27 tháng 7 2023
a
\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)
b
\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)
c
\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)
27 tháng 7 2023
a: =>(2x+15)(x^2+4)=0
=>2x+15=0
=>2x=-15
=>x=-15/2
b; =>(x-2)(5x-3)=0
=>x=2 hoặc x=3/5
c: =>(x+3)(2-x)=0
=>x=2 hoặc x=-3
14 tháng 11 2021
a) (x-4)(x+4)-x(x+2)=0
x2-16-x2-2x = 0
-16 - 2x = 0
2x = -16
x = -16/2
x = -8
b) 3x(x-2)-x+2=0
(3x-1)(x-2)=0
=> x ∈ {1/3 ; 2 }
c) 6x - 12x2 = 0
6x(1-2x) = 0
=> x ∈ {0; 1/2 }
d) mình thấy có vẻ hơi sai đề nên mình ko giải được, bạn thông cảm nha
2 tháng 11 2021
\(a,\Rightarrow3x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ c,Đề.sai\\ d,Sửa:\left(x-2\right)^2-16\left(5-2x\right)^2=0\\ \Rightarrow\left[x-2-4\left(5-2x\right)\right]\left[x-2+4\left(5-2x\right)\right]=0\\ \Rightarrow\left(x-2-20+8x\right)\left(x-2+20-8x\right)=0\\ \Rightarrow\left(9x-22\right)\left(18-7x\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{9}\\x=\dfrac{18}{7}\end{matrix}\right.\)
30 tháng 10 2021
a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)
b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
24 tháng 9 2021
b: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=4\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=4\)
\(\Leftrightarrow12x=12\)
hay x=2
d: Ta có: \(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Leftrightarrow3x^2-6x+3-3x^2+15x=1\)
\(\Leftrightarrow9x=-2\)
hay \(x=-\dfrac{2}{9}\)
2 tháng 11 2021
Bài 1:
a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)
\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)
b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)
e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
29 tháng 10 2023
a: \(\left(x+5\right)^2-\left(x-5\right)^2-2x+1=0\)
=>\(x^2+10x+25-\left(x^2-10x+25\right)-2x+1=0\)
=>\(x^2+8x+26-x^2+10x-25=0\)
=>18x+1=0
=>\(x=-\dfrac{1}{18}\)
b: \(\left(2x-7\right)^2-\left(x+3\right)^2=3x^2+6\)
=>\(4x^2-28x+49-\left(x^2+6x+9\right)-3x^2-6=0\)
=>\(x^2-28x+43-x^2-6x-9=0\)
=>34-34x=0
=>34x=34
=>x=1
c: \(\left(3x+2\right)^2-9\left(x-5\right)\left(x+5\right)=225-5x\)
=>\(9x^2+12x+4-9\left(x^2-25\right)-225+5x=0\)
=>\(9x^2+17x+4-225-9x^2+225=0\)
=>17x+4=0
=>x=-4/17
a, để A=0 thì \(\frac{x-2}{3x+2}=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
b, để A< 0 thì \(\frac{x-2}{3x+2}< 0\Leftrightarrow x-2< 3x+2\Leftrightarrow-2-2< 3x-x\Leftrightarrow-4< 2x\Leftrightarrow-2< x\Leftrightarrow x>-2\)