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MC
21 tháng 10 2018
mk chỉ phân tích thôi bạn tự chia nha!
a, \(16x^4-81=(4x^2)^2-9^2=(4x^2-9)(4x^2+9)\)
\(=[(2x)^2-3^2](4x^2+9)\)
\(=(2x+3)(2x-3)(4x^2+9)\)
b, \(x^3-3x^2+3x-1=(x-1)^3\)
\(x^2-2x+1=(x-1)^2\)
c, \(18x^5+9x^4+3x^3+6x^2+3x+1=(18x^5+9x^4+3x^3)+(6x^2+3x+1)\)
\(=(6x^2+3x+1)(3x^3+1)\)
câu c bạn đánh sai 1 dấu phép toán kìa!!!!
HN
2
27 tháng 10 2021
\(a,\Leftrightarrow x\left(x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)=0\\ \Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
27 tháng 10 2021
a) \(\Leftrightarrow x\left(x+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)
b) \(\Leftrightarrow x\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
c) \(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) \(\Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)
1 tháng 2
a: \(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)
\(=\left(x-8\right)\left(x^2-x-2\right)\)
\(=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
b: \(x^3-x^2-x-2\)
\(=x^3-2x^2+x^2-2x+x-2\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x-2\right)\cdot\left(x^2+x+1\right)\)
c: \(x^3+x^2-x+2\)
\(=x^3+2x^2-x^2-2x+x+2\)
\(=x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-x+1\right)\)
d: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
e: Sửa đề: \(x^3-7x-6\)
\(=x^3-x-6x-6\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)
f: \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
g: \(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
h: \(\left(x^2-3\right)^2+16\)
\(=x^4-6x^2+9+16\)
\(=x^4-6x^2+25\)
\(=x^4+10x^2+25-16x^2\)
\(=\left(x^2+5\right)^2-\left(4x\right)^2\)
\(=\left(x^2+5+4x\right)\left(x^2+5-4x\right)\)
3 tháng 3 2020
a) (2x + 5)(x - 3) = (x - 4)(3 - x)
<=> (2x + 5)(x - 3) + (x - 3)(x - 4) = 0
<=> (2x + 5 + x - 4)(x - 3) = 0
<=> (3x + 1)(x - 3) = 0
<=> \(\left[{}\begin{matrix}3x+1=0\\x-3=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\frac{1}{3}\\x=3\end{matrix}\right.\)
Vậy S = {-1/3; 3}
b) 18x2(x + 4) - 12(x2 + 4x) = 0
<=> 18x2(x + 4) - 12x(x + 4) = 0
<=> 6x(x + 4)(3x - 2) = 0
<=> \(\left[{}\begin{matrix}x=0\\x+4=0\\3x-2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=-4\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy S = {0; -2; 2/3}
HD
0
NT
0
HT
2
NK
2
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 12 2021
a.
\(x^2-16x=0\)
\(\Leftrightarrow x\left(x-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)
b.
\(9x^2+6x+4y^2-8y+5=0\)
\(\Leftrightarrow\left(9x^2+6x+1\right)+\left(4y^2-8y+4\right)=0\)
\(\Leftrightarrow\left(3x+1\right)^2+\left(2y-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\2x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
8 tháng 12 2021
\(a,x^2-16x=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(b,9x^2+6x+4y^2-8y+5=0\)
\(\Leftrightarrow\left(9x^2+6x+1\right)+4\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)^2+4\left(y-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x+1\right)^2=0\\4\left(y-1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=1\end{matrix}\right.\)
\(9x^5-18x^4-16x+32=0\)
\(\left(9x^5-18x^4\right)-\left(16x-32\right)=0\)
\(9x^4\left(x-2\right)-16\left(x-2\right)=0\)
\(\left(x-2\right)\left(9x^4-16\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\9x^4-16=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\9x^4=16\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x^4=\frac{16}{9}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\\left(x^2\right)^2=\left(\frac{\pm4}{3}\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\pm\sqrt{\frac{4}{3}}\end{cases}}\)
Vậy,..........
(=)(9x5-18x4)-(16x-32)=0
(=)2x4(x-2)-16(x-2)=0
(=)(2x4-16)(x-2)=0
(=)2x4-16=0 hoặc x-2=0
2x4-16=0
(=)2x4=16
(=)x4=8
x-2=0
(=)x=2
vậy x=2 hoặc x=bấm máy giùm nha