Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 5(x+3)-2x(3+x)=0
(x+3)(5-2x)=0
\(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
b) 4x(x-2017)-x+2017=0
4x(x-2017)-(x-2017)=0
(x-2017)(4x-1)=0
\(\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
c) (x+1)2 = x2 + 1
x2+2x+1-x2-1=0
2x=0
Pt có vô số nghiệm
a) \(5\left(x+3\right)-2x\left(3+x\right)=0\) (1)
\(\Leftrightarrow5\left(x+3\right)-2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{-3;\dfrac{5}{2}\right\}\)
b) \(4x\left(x-2017\right)-x+2017=0\)
cách làm hơi khó, cho đáp án thôi nhé: \(x=2017;x=\dfrac{1}{4}\)
c) \(\left(x+1\right)^2=x^2+1\) (3)
\(\Leftrightarrow x^2+2x+1=x^2+1\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{0\right\}\)
2)
a) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0 ; x=-1 ; x=1
b) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
1)
a) \(\left(x-2\right)\left(x^2+3x+4\right)\)
\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)
\(\Leftrightarrow x^3+x^2-2x-8\)
b) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
c) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)
\(=17x^2+5x-6-6x^3\)
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
a,\(3x\left(x-1\right)+x-1=0\)
\(\Rightarrow3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(3x+1\right).\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
c,\(\left(2x-1\right)^2-25=0\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
a/ \(x^2-2x=-1\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\Rightarrow x=1\)
Vậy..............
b/ \(x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\Rightarrow x=-1\)
Vậy.......
c/ \(4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)
\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)=3x^2\)
\(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)
\(\Leftrightarrow-4x=0\Rightarrow x=0\)
Vậy...................
d/ \(x\left(x-2017\right)-x^2\left(2017-x\right)=0\)
\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)
\(\Leftrightarrow x^3-2016x^2-2017x=0\)
\(\Leftrightarrow x^3+x^2-2017x^2-2017x=0\)
\(\Leftrightarrow x\left(x^2+x\right)-2017\left(x^2+x\right)=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x-2017\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\\x-2017=0\Rightarrow x=2017\end{matrix}\right.\)
Vậy pt có 3 nghiệm là.....(tự ghi ra)
\(a,x^2-2x=-1\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(b,x^2+2x+1=0\)
\(\Leftrightarrow\left(x+2\right)^2=0\)
\(\Rightarrow x+2=0\Rightarrow x=-2\)
\(c,4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)
\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)-3x^2=0\) \(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)
\(\Leftrightarrow-4x=0\Rightarrow x=0\)
\(d,x\left(x-2017\right)-x^2\left(2017-x\right)=0\)
\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)
\(\Leftrightarrow x^3+x^2-2017x-2017=0\)
\(\Leftrightarrow x^2\left(x+1\right)-2017\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2017\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-2107=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x^2=2017\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\\left[{}\begin{matrix}x=\sqrt{2017}\\x=-\sqrt{2017}\end{matrix}\right.\end{matrix}\right.\)