2(x-3)(x2+1)+15x-5x2=0
Những câu hỏi liên quan
Tìm đa thức M biết:a)
x
3
-
5
x
2
+x - 5 (x - 5).M;b) (
x
2
- 4x - 3).M 2
x
4
- 13
x
3
+ 14
x
2
+ 15x.
Đọc tiếp
Tìm đa thức M biết:
a) x 3 - 5 x 2 +x - 5 = (x - 5).M;
b) ( x 2 - 4x - 3).M = 2 x 4 - 13 x 3 + 14 x 2 + 15x.
2 tháng 12 2021 lúc 10:11
a) x2 – x;
b) 5x2(x – 2y) – 15x(x – 2y);
c) 3(x – y) – 5x(y – x).
\(a,x^2-x=x\left(x-1\right)\\ b,5x^2\left(x-2y\right)-15x\left(x-2y\right)=\left(5x^2-15x\right)\left(x-2y\right)=5x\left(x-3\right)\left(x-2y\right)\\ c,3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(5x+3\right)\left(x-y\right)\)
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bài 3 phân tích đa thức sau thành nhân tử
a 4x2 -16 + (3x +12) (4-2x)
b x3 + X2Y -15x -15y
c 3(x+8) -x2 -8x
d x3 -3x2 + 1 -3x
e 5x2 -5y2 -20x + 20y
kkk =0)
a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)
\(=\left(2x-4\right)\left(2x+4\right)-3\left(x+4\right)\left(2x-4\right)\)
\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)
\(=-\left(2x-4\right)\left(x+8\right)\)
b) \(x^3+x^2y-15x-15y\)
\(=x^2\left(x+y\right)-15\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-15\right)\)
c) \(3\left(x+8\right)-x^2-8x\)
\(=3\left(x+8\right)-x\left(x+8\right)\)
\(=\left(x+8\right)\left(3-x\right)\)
d) \(x^3-3x^2+1-3x\)
\(=x^3+1-3x^2-3x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d) \(5x^2-5y^2-20x+20y\)
\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y-4\right)\)
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Giải hộ e bài này với ai 👍
Câu 1 : a, 4x2 -3x-1=0 / d, 4x4-5x2+1=0
b, x2 - (1+căn 5)x + căn 5= 0 / e,x2 +3=|4x| / f, 2x + 5cănx +3 =0 / g, (x2 +x +1 ).(x2+x+2)=2 / h, x4-5x2+4=0
c, x4 + x2 -20=0 / k, x phần x2-1 -- 1 phần 2(x+1) = 1phan 2
tìm x
5x2 - 15x = 0
5x2 - 15x = 0
5x(x-3)=0
suy ra 2 trường hợp
x=0
x-3=0=>x=3
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5x2-15x=0
5x(x-3) =0
TH1: 5x=0 TH2: x-3=0
=>x=0 => x=3
Vậy x thuộc {0;3}
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1. (x3 – 3x2 + x – 3) : (x – 3) 2. (2x4 – 5x2 + x3 – 3 – 3x) : (x2 – 3) 3. (x – y – z)5 : (x – y – z)3 4. (x2 + 2x + x2 – 4) : (x + 2) 5. (2x3 + 5x2 – 2x + 3) : (2x2 – x + 1) 6. (2x3 – 5x2 + 6x – 15) : (2x – 5)
1: \(=x^2+1\)
3: \(=\left(x-y-z\right)^2\)
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Tìm x, biết.
a) x+ 5x2 = 0 b)(x+3)2+(4+x)(4-x)=10
c) 5x( x – 1) = x - 1 d) x2 -2x -3 = 0
\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)
\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
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5x2 -15x = 0
3 (x+5)-2x(x+5)=0
\(5x^2-15x=0\Leftrightarrow5x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\\ 3\left(x+5\right)-2x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
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1) \(5x^2-15x=0\)
\(\Rightarrow5x\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
2) \(3\left(x+5\right)-2x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(3-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
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\(5x^2-15x=0\)
\(\Rightarrow x.\left(5x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\5x=15\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy x ∈ {0 ; 3}
\(3.\left(x+5\right)-2x.\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right).\left(3-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\3-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\2x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy x ∈ \(\left\{-5;\dfrac{3}{2}\right\}\)
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Cứu với ạ
Làm tính chia
1) (x3 – 3x2 + x – 3) : (x – 3) 2) (2x4 – 5x2 + x3 – 3 – 3x) : (x2 – 3)
3) (x – y – z)5 : (x – y – z)3 4) (x2 + 2x + x2 – 4) : (x + 2)
| 5) (2x3 + 5x2 – 2x + 3) : (2x2 – x + 1) | 6) (2x3 – 5x2 + 6x – 15):(2x – 5) |