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\(3x^2-5x+2+3x^2+5x=\left(3x^2+3x^2\right)+\left(-5x+5x\right)+2=6x^2+2\)
a) \(3x^2-6xy=3x\left(x-2y\right)\)
b) \(x^3-6x^2+9x=x\left(x^2-6x+9\right)=x\left(x-3\right)^2\)
c) \(=x\left(x-2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x-3\right)\)
d) \(=2x\left(3x-5\right)-3\left(3x-5\right)=\left(3x-5\right)\left(2x-3\right)\)
\(a,=3x\left(x-2y\right)\\ b,=x\left(x-3\right)^2\\ c,Sửa:x^2-2xy-3x+6y=x\left(x-2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x-3\right)\\ d,=\left(3x-5\right)\left(2x-3\right)\)
( 36x +12x5 - 8x4 + 10x3 - 6x2 + 2x - 1) : (x4 + 4x3 - 3x2 + 2x - 1)
=12x-56
Dư 120x3-98x2+112x+55
(Mình tính ra nháp r hí hí)
ĐKXĐ của phân thức x ≠ 1.
Ta có:
Vậy min A = 2 khi và chỉ khi x - 2 = 0 ⇔ x =2
1)
a) \(=15x^3-20x^2+10x\)
b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)
2)
a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)
\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))
b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)
(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ
Giúp em vs ạ . Em cảm ơn
a: \(C=x^3-3x^2+3x+2023\)
\(C=x^3-3x^2+3x-1+2024\)
\(=\left(x-1\right)^3+2024\)
Khi x=101 thì \(C=\left(101-1\right)^3+2024\)
\(=100^3+2024\)
\(=1000000+2024=1002024\)
b: \(D=x^3-6x^2+12x-100\)
\(=x^3-6x^2+12x-8-92\)
\(=\left(x-2\right)^3-92\)
Khi x=-98 thì \(D=\left(-98-2\right)^3-92\)
\(=-100^3-92\)
\(=-1000000-92=-1000092\)
a) \(\Rightarrow\left(x-1\right)^3=0\Rightarrow x=1\)
b) \(\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\Rightarrow\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)(do \(\left\{{}\begin{matrix}x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\\x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\end{matrix}\right.\))
c) \(\Rightarrow4x\left(x^2-9\right)=0\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(x-2\right)^3=0\Rightarrow x=2\)
a) \(x^3-3x^2+3x-1=0\Rightarrow\left(x-1\right)^3=0\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
b) \(x^6-1=0\Rightarrow\left(x^3\right)^2-1=0\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^3-1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(4x^3-36x=0\Rightarrow4x\left(x^2-36\right)=0\Rightarrow4x\left(x-6\right)\left(x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-6=0\\x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)
d) \(x^3-6x^2+12x-8=0\) (đề bài như vậy mới làm đc, nếu là +8 thì mình xin bó tay nhé)
\(\Rightarrow x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3=0\)
\(\Rightarrow\left(x-2\right)^3=0\Rightarrow x-2=0\Rightarrow x=2\)