Giải phương trình:
a) 4x-16=3x(x-4)
b) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
Giải phương trình:
a) 4x-16=3x(x-4)
b) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
a) \(4x-16=3x\left(x-4\right)\)
\(4\left(x-4\right)=3x\left(x-4\right)\)
\(3x\left(x-4\right)-4\left(x-4\right)=0\)
\(\left(x-4\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\left(đk:x\ne0,2\right)\)
\(\dfrac{x\left(x+2\right)-\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
\(x^2+2x-x+2=2\)
\(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
a) \(4x-16=3x\left(x-4\right)\)
\(\Leftrightarrow\) \(4\left(x-4\right)=3x\left(x-4\right)\)
\(\Leftrightarrow\) \(4\left(x-4\right)=3x\left(x-4\right)=0\)
\(\Leftrightarrow\left(4-3x\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-3x=0\\x-4=0\end{matrix}\right.\)
TH1 \(\Leftrightarrow3x=0+4\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=4\div3\)
\(\Leftrightarrow x=\dfrac{4}{3}\)
TH2 \(x-4=0\)
\(\Leftrightarrow\) \(x=0+4\)
\(\Leftrightarrow x=4\)
\(\Leftrightarrow x=\left[{}\begin{matrix}\dfrac{4}{3}\\4\end{matrix}\right.\)
b) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\) \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{1\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\) \(x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x+2=2\)
\(\Leftrightarrow x^2+x=2-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(x=0\)
\(\Leftrightarrow\) \(x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Leftrightarrow x=-1\)
\(\Leftrightarrow x=\left[{}\begin{matrix}0\\-1\end{matrix}\right.\)
Làm bài 21 phần e và f
\(\left(2x-5\right)^2\)- \(\left(x+2\right)^2\) = 0
\(x^2-x-\left(3x-3\right)=0\)
\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow4x^2-20x+25-\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow4x^2-20x+25-x^2-4x-4=0\)
\(\Leftrightarrow3x^2-24x+21=0\)
\(\Leftrightarrow3x^2-21x-3x+21=0\)
\(\Leftrightarrow3x\left(x-7\right)-3\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
\(x^2-x-\left(3x-3\right)=0\)
\(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(\dfrac{40}{20}\)\(-\dfrac{5(3x+7)}{20}\)\(+\dfrac{4(x+17)}{20}\)=\(\dfrac{0}{20}\)
<=>40-5(3x+7)+4(x+17)=0
<=>40-15x-35+4x+68=0
<=>-15x+4x=-40+35-68
<=>-11x=-73
<=>x=-73:(-11)=\(\dfrac{73}{11}\)
Vậy S={\(\dfrac{73}{11}\)}
(x^2-x)(x^2+3x+2) = 6 Giải phyowng trình
<=>x(x-1)(x+1)(x+2)=6
<=>(x^2+x)(x^2+x-2)=6
đặt x^2 + x = t
<=>t(t-2)=6
<=>t^2 -2t + 1 = 7
<=>(t-1)^2 = 7
<=> (x^2 + x)^2=7
<=>x^2 + x = căn 7<=> (x+1/2)^2 = căn7 + 5/4
=> x+1/2 = căn(căn7+5/4) = 1,973765769
<=>x=1,473765769
TH2:x+1/2 = -căn(căn7 + 5/4) = -1,973765769
<=>x=-2.473765769
mặt khác (x+1/2)^2=-căn7 + 5/4
=>x+1/2 = căn(-căn7 +5/4) => vô lí
TH4:x+1/2 = -căn(-căn7 + 5/4) => vô lí
=> TH 3 4 loại
vậy phương trình có tập nghiệm S = 1,473765769;-2.473765769
2x-3/4-4x-5/3=5-x/6
giúp mk vs
Cho phương trình (m-3)x+m^2 -9=0, m là tham số. a)Giải phương trình khi m=3 b)Tìm m để:+phương trình có nghiệm duy nhất +có vô số nghiệm
giúp hộ mk vs.Mk cảm trước nha
a: Khi m=3 thì pt sẽ là 0x+0=0(luôn đúng)
b: Để phương trình có nghiệm duy nhất thì m-3<>0
hay m<>3
Để phương trình có vô số nghiệm thì m-3=0
hay m=3
1: =>(x+4)(x-3)=0
=>x=-4 hoặc x=2
2: =>(x-6)(x+1)-2(x+1)=0
=>(x+1)(x-8)=0
=>x=-1 hoặc x=8
3: =>x2-2x-3=0
=>(x-3)(x+1)=0
=>x=3 hoặc x=-1
\(1,x^2+x-12=0\\ \Leftrightarrow\left(x^2+4x\right)-\left(3x+12\right)=0\\ \Leftrightarrow x\left(x+4\right)-3\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ 2,\left(x-6\right)\left(x+1\right)=2\left(x+1\right)\\ \Leftrightarrow\left(x-6\right)\left(x+1\right)-2\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x-6-2\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=6\end{matrix}\right.\)
\(3,x^2-2x+1=4\\ \Leftrightarrow\left(x-1\right)^2=4\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-2\\x-1=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ 4,\left(x-3\right)\left(x+3\right)=16\\ \Leftrightarrow x^2-9-16=0\\ \Leftrightarrow x^2-25=0\\ \Leftrightarrow x=\pm5\\ 5,x^2-x-6=0\\ \Leftrightarrow\left(x^2-3x\right)+\left(2x-6\right)=0\\ \Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
\(6,\left(2x-1\right).\dfrac{5x-10}{10}+\left(2x-1\right).\dfrac{3x+12}{6}-\left(2x-1\right)x=0\\ \Leftrightarrow\left(2x-1\right).\dfrac{x-2}{2}+\left(2x-1\right).\dfrac{x+4}{2}-\left(2x-1\right)x=0\\ \Leftrightarrow\left(2x-1\right)\left(\dfrac{x-2}{2}+\dfrac{x+4}{2}-x\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(\dfrac{x-2}{2}+\dfrac{x+4}{2}-\dfrac{x}{2}\right)=0\\ \Leftrightarrow\left(2x-1\right).\dfrac{x-2+x+4-x}{2}=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
a: \(\Leftrightarrow\left(x-1+2x+3\right)\left[\left(x^2-2x+1\right)-\left(x-1\right)\left(2x+3\right)+\left(2x+3\right)^2\right]-\left(3x+2\right)\left(9x^2-6x+4\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x^2-2x+1-2x^2-3x+2x+3+4x^2+12x+9-9x^2+6x-4\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(-6x^2+15x+9\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(-2x^2+5x+3\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(-2x^2+6x-x+3\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-3\right)\left(-2x-1\right)=0\)
hay \(x\in\left\{-\dfrac{2}{3};3;-\dfrac{1}{2}\right\}\)
b: \(\Leftrightarrow\left(x^2-4x\right)^2+4\left(x^2-4x\right)-2\left(x^2-4x\right)-8=0\)
\(\Leftrightarrow\left(x^2-4x\right)\cdot\left(x^2-4x+4\right)-2\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2\cdot\left[\left(x-2\right)^2-6\right]=0\)
hay \(x\in\left\{2;\sqrt{6}+2;-\sqrt{6}+2\right\}\)
a) (2x +1)(3 – x)(4 - 2x) = 0 b)2x(x – 3) + 5(x – 3) = 0
c) (x2 – 4) – (x – 2)(3 – 2x) = 0 d) x2 – 5x + 6 = 0
e) (2x + 5)2 = (x + 2)2 f) 2x3 + 6x2 = x2 + 3x
a: (2x+1)(3-x)(4-2x)=0
=>(2x+1)(x-3)(x-2)=0
hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)
b: 2x(x-3)+5(x-3)=0
=>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
c: =>(x-2)(x+2)+(x-2)(2x-3)=0
=>(x-2)(x+2+2x-3)=0
=>(x-2)(3x-1)=0
=>x=2 hoặc x=1/3
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
e: =>(2x+5+x+2)(2x+5-x-2)=0
=>(3x+7)(x+3)=0
=>x=-7/3 hoặc x=-3
f: \(\Leftrightarrow2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)
giải phương trình
a: \(\Leftrightarrow2x^3-5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2-5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2-6x+x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(2x+1\right)=0\)
hay \(x\in\left\{0;3;-\dfrac{1}{2}\right\}\)
b: \(\Leftrightarrow\left(3x-1\right)\left(x^2+x\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+x-7x+10\right)=0\)
=>3x-1=0
hay x=1/3