Thu gọn đa thức:
A = x(x-2) +(x-1)(x+3) - (x2 +5)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TB
1
H9
HT.Phong (9A5)
CTVHS
17 tháng 8 2023
a) \(\left(x+1\right)\left(x-1\right)\)
\(=x^2-1^2\)
\(=x^2-1\)
b) \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)\)
\(=\left(x^2\right)^2-1^2\)
\(=x^4-1\)
c) \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)\left(x^2+1\right)-x^8\)
\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)-x^8\)
\(=\left(x^4-1\right)\left(x^4+1\right)-x^8\)
\(=\left(x^4\right)^2-1-x^8\)
\(=x^8-1-x^8\)
\(=-1\)
20 tháng 11 2021
\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)
15 tháng 10 2023
2:
a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)
b: \(2\left(x-1\right)+x^2-x\)
\(=2\left(x-1\right)+x\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
c: \(3x^2+14x-5\)
\(=3x^2+15x-x-5\)
\(=3x\left(x+5\right)-\left(x+5\right)=\left(x+5\right)\left(3x-1\right)\)
3:
a: \(2x\left(x-1\right)-2x^2=4\)
=>\(2x^2-2x-2x^2=4\)
=>-2x=4
=>x=-2
b: \(x\left(x-3\right)-\left(x+2\right)\left(x-1\right)=5\)
=>\(x^2-3x-\left(x^2+x-2\right)=5\)
=>\(x^2-3x-x^2-x+2=5\)
=>-4x=3
=>x=-3/4
c: \(4x^2-25+\left(2x+5\right)^2=0\)
=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)
=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)
=>4x(2x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
29 tháng 8 2021
a: Ta có: \(\left(x-2\right)^2-\left(2x-1\right)^2+\left(3x-1\right)\left(x-5\right)\)
\(=x^2-4x+4-4x^2+4x-1+3x^2-15x-x+5\)
\(=-16x+8\)
b: Ta có: \(\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
=27x-55
ML
2
18 tháng 5 2021
a) \(\left(x-1\right)\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
b) \(x+1x^2+1=x^2+x+1=x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)với mọi x.
=> Pt vô nghiệm.
c) \(x^2+4x=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
P/s: Check lại đề ý b nhé.
18 tháng 5 2021
a) Ta có:(x-1)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy: S={1;-5}
b) Ta có: \(x^2+x+1=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)(Vô lý)
Vậy: \(S=\varnothing\)
c) Ta có: \(x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy: S={0;-4}
A=x.(x-2)+(x-1).(x+3)-(x2+5)
A=x2-2x+x2+3x-x-3-x2+5
A=x2+2
ban kia lam dung roi do
k tui nha
thanks