Giải phương trình:
a. (3x+5)3 + (2x-7)3 - (5x-2)3 = 0
b. (x2 + x - 2)3 + (x2 +5x +6)3 - 8(x2 + 3x +2)3 = 0
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6 tháng 3 2021
1) Ta có: \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}
25 tháng 3 2020
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
23 tháng 2 2022
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
LP
2
5 tháng 3 2022
1.\(\left(x+2\right)\left(2x-3\right)=x^2-4\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3\right)-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3-x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
2.\(x^2+3x+2=0\)
\(\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
3.\(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
4.\(x^3+x^2-12x=0\)
\(\Leftrightarrow x\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x\left(x+4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=3\end{matrix}\right.\)
5 tháng 3 2022
a: \(\Leftrightarrow\left(x+2\right)\left(2x-3\right)-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3-x+2\right)=0\)
=>(x+2)(x-1)=0
=>x=-2 hoặc x=1
b: =>(x+1)(x+2)=0
=>x=-1 hoặc x=-2
c: =>(2x+3)(x+1)=0
=>x=-1 hoặc x=-3/2
d: =>x(x+4)(x-3)=0
hay \(x\in\left\{0;-4;3\right\}\)
N
giải các Phương trình sau
a) (5x+3)(x2+1)(x-1)=0
b) (4x-1)(x-3)-(x-3)(5x+2)=0
c) (x+6)(3x-1)+x2-36 =0
1
23 tháng 2 2023
a: =>(5x+3)(x-1)=0
=>x=1 hoặc x=-3/5
b: =>(x-3)(4x-1-5x-2)=0
=>(x-3)(-x-3)=0
=>x=-3 hoặc x=3
c: =>(x+6)(3x-1+x-6)=0
=>(x+6)(4x-7)=0
=>x=7/4 hoặc x=-6
a) \(\left(3x+5\right)^3+\left(2x-7\right)^3-\left(5x-2\right)^3=0\)
\(\Leftrightarrow\left[\left(3x+5\right)+\left(2x-7\right)\right]\left[\left(3x+5\right)^2-\left(3x+5\right)\left(2x-7\right)+\left(2x-7\right)^2\right]-\left(5x-2\right)^3=0\)
\(\Leftrightarrow\left(5x-2\right)\left[9x^2+30x+25-\left(6x^2-21x+10x-35\right)+4x^2-28x+49\right]-\left(5x-2\right)^3=0\)
\(\Leftrightarrow\left(5x-2\right)\left(7x^2+13x+109\right)-\left(5x-2\right)^3=0\)
\(\Leftrightarrow\left(5x-2\right)\left[7x^2+13x+109-\left(5x-2\right)^2\right]=0\)
\(\Leftrightarrow\left(5x-2\right)\left(7x^2+13x+109-25x^2+20x-4\right)=0\)
\(\Leftrightarrow\left(5x-2\right)\left(-18x^2+33x+105\right)=0\)
\(\Leftrightarrow-3\left(5x-2\right)\left(6x^2-11x-35\right)=0\)
\(\Leftrightarrow-3\left(5x-2\right)\left(6x^2+10x-21x-35\right)=0\)
\(\Leftrightarrow-3\left(5x-2\right)\left[2x\left(3x+5\right)-7\left(3x+5\right)\right]=0\)
\(\Leftrightarrow-3\left(5x-2\right)\left(2x-7\right)\left(3x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-2=0\\2x-7=0\\3x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{7}{2}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
a: Đặt \(3x+5=a;2x-7=b\)
=>a+b=3x+5+2x-7=5x-2
Phương trình ban đầu sẽ trở thành:
\(a^3+b^3-\left(a+b\right)^3=0\)
=>\(\left(a+b\right)^3-3ab\left(a+b\right)-\left(a+b\right)^3=0\)
=>-3ab(a+b)=0
=>ab(a+b)=0
=>(3x+5)(2x-7)(5x-2)=0
=>\(\left[{}\begin{matrix}3x+5=0\\2x-7=0\\5x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{7}{2}\\x=\dfrac{2}{5}\end{matrix}\right.\)
b: \(\left(x^2+x-2\right)^3+\left(x^2+5x+6\right)^3-8\left(x^2+3x+2\right)^3=0\)
=>\(\left(x^2+x-2\right)^3+\left(x^2+5x+6\right)^3-\left(2x^2+6x+4\right)^3=0\)(2)
Đặt \(x^2+x-2=c;x^2+5x+6=d\)
=>\(c+d=2x^2+6x+4\)
Phương trình (2) sẽ trở thành:
\(c^3+d^3-\left(c+d\right)^3=0\)
=>\(\left(c+d\right)^3-3cd\left(c+d\right)-\left(c+d\right)^3=0\)
=>-3cd(c+d)=0
=>cd(c+d)=0
=>\(\left(x^2+x-2\right)\left(x^2+5x+6\right)\left(2x^2+6x+4\right)=0\)
=>\(\left(x+2\right)\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+2\right)\left(x+1\right)=0\)
=>\(\left(x+2\right)^3\cdot\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x=-2\\x=1\\x=-1\\x=-3\end{matrix}\right.\)