(4x2+2x-18)2-(4x2+2x)2 =M.(4x2+2x-9)
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DL
1
TA
5 tháng 8 2021
a, `(8x^3-4x^2): 4x -(4x^2-5x) : 2x + (2x)^2`
`=4x (2x^2-x) : 4x - 2x(2x-5/2 ) :2x + 4x^2`
`=2x^2-x-2x+5/2+4x^2`
`=6x^2-3x+5/2`
b, `(3x^3-x^2y) :x^2 -(xy^2+x^2y) :xy + 2x(x+1)`
`=x^2 (3x-y) :x^2 -xy(y+x) + (2x^2+2x)`
`=3x-y-y-x+2x^2+2x`
`=2x^2+4x-2y`
CM
7 tháng 10 2017
Ta có
4 x 2 + 2 x - 18 2 - 4 x 2 + 2 x 2 = 4 x 2 + 2 x - 18 + 4 x 2 + 2 x 4 x 2 + 2 x - 18 - 4 x 2 - 2 x = 8 x 2 + 4 x - 18 - 18 = 2 4 x 2 + 2 x - 9 - 18 = - 36 4 x 2 + 2 x - 9 ⇒ m = - 36
Đáp án cần chọn là: C
HV
2
11 tháng 10 2021
1: Ta có: \(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)
14 tháng 2 2022
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
14 tháng 2 2022
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
2 tháng 10 2021
\(M=343-8x^3-64+8x^3=279\\ N=8x^3-1-1+8x^3=16x^3=16\cdot1000=16000\)
CM
3 tháng 10 2018
Ta có:
8 x 3 - 1 = 2 x 3 - 1 3 = 2 x - 1 . 4 x 2 + 2 x + 1 = - 1 - 2 x . 4 x 2 + 2 x + 1
Do đó, ( 8 x 3 - 1 : 1 - 2 x = - 4 x 2 + 2 x + 1 = - 4 x 2 - 2 x - 1
Chọn B. - 4 x 2 - 2 x - 1
17 tháng 9 2021
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
18 tháng 9 2021
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
\(\Leftrightarrow M\cdot\left(4x^2+2x-9\right)=\left(4x^2+2x-18-4x^2-2x\right)\left(4x^2+2x-18+4x^2+2x\right)\)
\(\Leftrightarrow M\cdot\left(4x^2+2x-9\right)=-18\cdot\left(8x^2+4x-18\right)\)
\(\Leftrightarrow M=-18\cdot2=-36\)