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3 tháng 10 2021
a: Ta có: \(2\left(x-2\right)^3=2-x\)
\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
b: ta có: \(8x^3-72x=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
c: Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
NN
1
28 tháng 12 2021
\(1,3x-7=19\\ \Rightarrow3x=26\\ \Rightarrow x=\dfrac{26}{3}\\ 2,\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\\ 3,3x+\dfrac{2}{4}+1=5x-\dfrac{1}{3}\\ \Rightarrow5x-\dfrac{1}{3}-3x-\dfrac{2}{4}-1=0\\ \Rightarrow2x-\dfrac{11}{6}=0\\ \Rightarrow2x=\dfrac{11}{6}\\ \Rightarrow x=\dfrac{11}{12}\)
\(4,\dfrac{x}{15}+\dfrac{1}{2}-\dfrac{x}{50}=\dfrac{5}{6}\\ \Rightarrow\dfrac{x}{15}-\dfrac{x}{50}=\dfrac{5}{6}-\dfrac{1}{2}\\ \Rightarrow x\left(\dfrac{1}{15}-\dfrac{1}{50}\right)=\dfrac{1}{3}\\ \Rightarrow\dfrac{7}{150}x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{50}{7}\)
2 tháng 10 2021
Bài 1:
a: \(8x^3-2x=2x\left(4x^2-1\right)=2x\left(2x-1\right)\left(2x+1\right)\)
c: \(-5m^3\left(m+1\right)+m+1=\left(m+1\right)\left(-5m^3+1\right)\)
3 tháng 7 2021
a) Ta có: \(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow-\left(x-2\right)-2\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)\left[1+2\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
b) Ta có: \(8x^3-72x=0\)
\(\Leftrightarrow8x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy: S={0;3;-3}
c) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
d) Ta có: \(2x^3+3x^2+3+2x=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
hay \(x=-\dfrac{3}{2}\)
e) Ta có: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
Vậy: S={0;1;-2}
f) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
Vậy: S={0;2;12}
25 tháng 9 2021
1) \(\Leftrightarrow\left(x-4\right)\left(x+4\right)-x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4-x\right)=0\)
\(\Leftrightarrow\left(x-4\right)4=0\)
\(\Leftrightarrow x=4\)
2) \(\left(x+3\right)^2-\left(x-3\right)\left(x+5\right)=x^2+6x+9-x^2-2x+15=4x+24\)
3) \(2x^3+3x^2-2x+a=2x^2\left(x-2\right)+7x\left(x-2\right)+16\left(x-2\right)+32+a\)
Để \(2x^3+3x^2-2x+a⋮x-2\) thì \(32+a=0\Leftrightarrow a=-32\)
HP
25 tháng 9 2021
1.
x2 - 16 - x(x - 4) = 0
<=> (x2 - 42) - x(x - 4) = 0
<=> (x - 4)(x + 4) - x(x - 4) = 0
<=> (x + 4 - x)(x + 4) = 0
<=> 4(x + 4) = 0
<=> x + 4 = 0
<=> x = -4
2.
(x + 3)2 - (x - 3)(x + 5)
= x2 + 6x + 9 - (x2 + 5x - 3x - 15)
= x2 + 6x + 9 - x2 + 5x - 3x - 15
= x2 - x2 + 6x + 5x - 3x + 9 - 15
= 8x - 6
28 tháng 1 2022
a, \(A=2x^3-9x^5+3x^5-3x^2+7x^2-12=-6x^5+2x^3+4x^2-12\)
b, \(B=2x^4+x^2+2x-2x^3-2x^2+x^2-2x+1=2x^4-2x^3+1\)
c, \(C=2x^2+x-x^3-2x^2+x^3-x+3=3\)
2x3-3x2-2x+3=0 tìm x
\(=>x^2\left(2x-3\right)-\left(2x-3\right)=0\\ =>\left(2x-3\right)\left(x^2-1\right)=0\\ \)
TH1 TH2
2x - 3 = 0 x2 - 1 = 0
=> x = 3/2 => x = +1 và -1
HT