Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2-\left(x+3\right)\left(3x+1\right)=\)\(9\)
\(\Leftrightarrow x^2-9-\left(x+3\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x+3\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3-3x-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-2x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\-2x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\-2x=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-2\end{cases}}}\)
Vậy phương trình có tập nghiệm \(S=\left\{-3;-2\right\}\)
\(x^3+4x+5=0\)
\(\Leftrightarrow\left(x^3+1\right)+\left(4x+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)+4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=0\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\\left(x-\frac{1}{2}\right)^2=\frac{-19}{4}\left(vn\right)\end{cases}}\)(vn: vô nghiệm).\(\Leftrightarrow x=-1\)
Vậy phương trình có nghiệm duy nhất : \(x=-1\)
Hình bạn tự vẽ nha
a) Chứng minh AB//DG và AD//BF
Từ đó theo Ta lét ta có
ΔΔADE có AD//BF ; F∈∈AE;B∈∈DE
⇒⇒AEEK=DEBEAEEK=DEBE (1)
ΔΔDEG có DG//AB;A∈∈GE;B∈∈DE
⇒⇒EGAE=DEEBEGAE=DEEB (2)
Từ (1)(2) thì AEEK=EGAEAEEK=EGAE
⇒⇒AE2=EG.EK
a)(2x2+1)(3x3-2x2+3
= 6x5-4x4+6x2+3x3-2x2+3
= 6x5-4x4+3x3+4x2+3
b)(-3x+1)(4x4-x³+x)
= -12x5+3x4-3x2+4x4-x³+x
= -12x5+7x4-x3-3x2+x
a, \(9-x^2+2xy-y^2\)
\(=9-\left(x-y\right)^2\)
\(=\left(3-x+y\right)\left(3+x-y\right)\)
b, \(x^4-x^2+4x-4\)
\(=x^4-\left(x-2\right)^2\)
\(=\left(x^2-x+2\right)\left(x^2+x-2\right)\)
\(=\left(x^2-x+2\right)\left(x^2+2x-x-2\right)\)
\(=\left(x^2-x+2\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2-x+2\right)\)
c, \(x^3-2x^2y+xy^2\)
\(=x^3-x^2y-x^2y+xy^2\)
\(=x^2\left(x-y\right)-xy\left(x-y\right)\)
\(=\left(x^2-xy\right)\left(x-y\right)\)
\(=x\left(x-y\right)^2\)
d, \(1-x^2-2xz-z^2\)
\(=1-\left(x+z\right)^2\)
\(=\left(1-x-z\right)\left(1+x+z\right)\)
A)
\(9-x^2+2xy-y^2=3^2-\left(x-y\right)^2\\ =\left(x-y+3\right)\left(3-x+y\right)\)
B)
\(x^4-x^2+4x-4=\left(x^2\right)^2-\left(x-2\right)^2\\ =\left(x^2+x-2\right)\left(x^2-x+2\right)\\ =\left(x^2-x+2x-2\right)\left(x^2+x-2x-2\right)\\ =\left(x-1\right)\left(x+2\right)\left(x+1\right)\left(x-2\right)\)
C)
\(x^3-2x^2y+xy^2\\ =x\left(x^2-2xy+y^2\right)\\ =x\left(x-y\right)^2\)
D)
\(1-x^2-2xz-z^2\\ =1^2-\left(x+z\right)^2\\ =\left(1+x+z\right)\left(1-x-z\right)\)
a: \(=\dfrac{x^3-x^2-7x+3}{x-3}=\dfrac{x^3-3x^2+2x^2-6x-x+3}{x-3}=x^2+2x-1\)
b: \(=\dfrac{2x^4-4x^2-3x^3+6x+x^2-2}{x^2-2}=2x^2-3x+1\)
\(x^3-3x^2+2=x^3-2x^2-2x-\left(x^2-2x-2\right)\)
\(=x.\left(x^2-2x-2\right)-\left(x^2-2x-2\right)\)
\(=\left(x-1\right).\left(x^2-2x-2\right)\)
\(1,x^3-3x^2+2=0\)
\(x^3-x^2-2x^2+2=0\)
\(x^2\left(x-1\right)-2\left(x^2-1\right)=0\)
\(\left(x-1\right)\left(x^2-2x-2\right)=0\)