giải pt
2(x-3)(x2 +1)+15x-5x2=0
Những câu hỏi liên quan
2(x-3)(x2+1)+15x-5x2=0
\(\Leftrightarrow\left(x-3\right)\left(2x^2+2\right)+5x\left(3-x\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\)
=>(x-3)(x-2)(2x-1)=0
=>x=3 hoặc x=2 hoặc x=1/2
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\(2\left(x-3\right)\left(x^2+1\right)+15x-5x^2=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2+2\right)-5x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left[\left(2x^2-4x\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left[2x\left(x-2\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
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\(2\left(x-3\right)\left(x^2+1\right)+15x-5x^2=0\)
\(\Leftrightarrow2x^3+2x-6x^2-6+15x-5x^2=0\)
\(\Leftrightarrow2x^3-11x^2+17x-6=0\)
\(\Leftrightarrow2x^3-4x^2-7x^2+14x+3x-6=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-7x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-7x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-6x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-1\right)-3\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{2;\dfrac{1}{2};3\right\}\)
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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 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.
Giải các phương trình sau:a)
2
x
+
1
2
−
2
x
−
1
2
;
b)
x
2
−
3
x
2
+
5
x...
Đọc tiếp
Giải các phương trình sau:
a) 2 x + 1 2 − 2 x − 1 = 2 ;
b) x 2 − 3 x 2 + 5 x 2 − 3 x + 6 = 0 ;
c) x 2 − x − 1 x 2 − x − 2 = 0 .
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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Giải các bất phương trình sau:a)
2
x
−
7
11
−
4
x
;
b)
x
−
2
2
−
x
2
−
8
x
+
3
≥
0
;
c)
2
3
−
3...
Đọc tiếp
Giải các bất phương trình sau:
a) 2 x − 7 > 11 − 4 x ; b) x − 2 2 − x 2 − 8 x + 3 ≥ 0 ;
c) 2 3 − 3 x − 6 2 > 1 + 3 x 6 ; d) x − 5 x + 1 + 4 x + 3 < − 5 x 2
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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Bài 1. Giải các phương trình sau:
a) x2 - 6x + 5 = 0 b) 2x2 + 4x – 8 = 0
c) 4y2 – 4y + 1 = 0 d) 5x2 - x + 2 = 0
\(a,x^2-6x+5=0\\ \Rightarrow\left(x^2-5x\right)-\left(x-5\right)=0\\ \Rightarrow x\left(x-5\right)-\left(x-5\right)=0\\ \Rightarrow\left(x-1\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
\(b,2x^2+4x-8=0\\ \Rightarrow x^2+2x-4=0\\ \Rightarrow\left(x^2+2x+1\right)-5=0\\ \Rightarrow\left(x+1\right)^2-\sqrt{5^2}=0\\ \Rightarrow\left(x+1+\sqrt{5}\right)\left(x+1-\sqrt{5}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1-\sqrt{5}\\x=-1+\sqrt{5}\end{matrix}\right.\)
\(c,4y^2-4y+1=0\\ \Rightarrow\left(2y-1\right)^2=0\\ \Rightarrow2y-1=0\\ \Rightarrow y=\dfrac{1}{2}\)
\(d,5x^2-x+2=0\)
Ta có:\(\Delta=\left(-1\right)^2-4.5.2=1-40=-39\)
Vì \(\Delta< 0\Rightarrow\) pt vô nghiệm
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