3X x 3X = 81
tìm X
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CN
0
15 tháng 4 2020
R(x) = 2x2 + 3x - 1
- M(x) = -x3 + x2
x3 + x2 + 3x - 1
Vậy R(x) - M(x) = x3 + x2 + 3x - 1
DD
2
ND
Nguyễn Đức Trí
VIP
3 tháng 8 2023
\(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(B=x^3-9x^2+27x-27-\left(x^3-3x^2+9x+3x^2-9x+27\right)+\left(9x^2-1\right)\)
\(B=x^3-9x^2+27x-27-\left(x^3+27\right)+9x^2-1\)
\(B=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(B=27x-55\)
26 tháng 3 2020
a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2
= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25
= 36
b) (3x^2 - y)^2
= 9x^4 - 6x^2y + y^2
c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)
= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4
= 9x^2 + 54
d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2
= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x
= x^3 - 16x^2 + 25x
e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)
= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2
= x^3 + 2x^2 - 2x - 12
f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2
= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4
= x^6 + 2x^4 + 2x^2 + 124
MH
2
16 tháng 6 2017
\(x^{11}+3x^{10}+x^9+3x^8+x^7-3x^6-17x^5+3x^4+x^3+3x^2+x+3=0\)
\(\Leftrightarrow\left(x^{11}+2x^{10}+4x^9+6x^8+9x^7+6x^6+4x^5+2x^4+x^3\right)+\left(x^{10}+2x^9+4x^8+6x^7+9x^6+6x^5+4x^4+2x^3+x^2\right)-\left(5x^9+10x^8+20x^7+30x^6+45x^5+30x^4+20x^3+10x^2+5x\right)+\left(3x^8+6x^7+12x^6+18x^5+27x^4+18x^3+12x^2+6x+3\right)=0\)
\(\Leftrightarrow x^3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+x^2\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)-5\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x^3+x^2-5x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
Dễ thấy: \(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1>0\forall x\)
Nên \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
3 tháng 2 2021
a) Ta có: \(x^2-3x+7=1+2x\)
\(\Leftrightarrow x^2-3x+7-1-2x=0\)
\(\Leftrightarrow x^2-3x-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy: S={3;2}
b) Ta có: \(x^2-3x-10=0\)
\(\Leftrightarrow x^2-5x+2x-10=0\)
\(\Leftrightarrow x\left(x-5\right)+2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy: S={5;-2}
c) Ta có: \(x^2-3x+4=2\left(x-1\right)\)
\(\Leftrightarrow x^2-3x+4=2x-2\)
\(\Leftrightarrow x^2-3x+4-2x+2=0\)
\(\Leftrightarrow x^2-3x-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy: S={3;2}
d) Ta có: \(\left(x+1\right)\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=5\end{matrix}\right.\)
Vậy: S={-1;2;5}
e) Ta có: \(2x^2+3x+1=0\)
\(\Leftrightarrow2x^2+2x+x+1=0\)
\(\Leftrightarrow2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{-1}{2}\right\}\)
f) Ta có: \(4x^2-3x=2x-1\)
\(\Leftrightarrow4x^2-3x-2x+1=0\)
\(\Leftrightarrow4x^2-5x+1=0\)
\(\Leftrightarrow4x^2-4x-x+1=0\)
\(\Leftrightarrow4x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{1;\dfrac{1}{4}\right\}\)
3X=9
X=3
x=3
tick mk nha bạn