Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Leftrightarrow x^2-6x+9-x^2+4=1\)
=>-6x=-12
hay x=2
\(2021x\left(x-2020\right)-x+2020=0\)
\(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)
\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
Ta có: \(2021x\left(x-2020\right)-x+2020=0\)
\(\Leftrightarrow\left(x-2020\right)\left(2021x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
Ta có: \(\left(2x+3\right)^2-4\left(x-1\right)^2=17\)
\(\Leftrightarrow4x^2+12x+9-4x^2+8x-4=17\)
\(\Leftrightarrow20x=12\)
hay \(x=\dfrac{3}{5}\)
Ta có:
\(F\left(1\right)=\left(1-1+1\right)^{1994}+\left(1+1-1\right)^{1994}-2=0\)
\(\Rightarrow\)x=1 là 1 nghiệm của phương trình F(x)=0=> F(x) chia hết cho x-1
Đa thức chia có bậc 2 nên đa thức dư có bậc không vượt quá 1.
Gọi đa thức dư là : x + a, có :
\(F\left(x\right)=\left(x^2-1\right)Q\left(x\right)+x+a\)
F(x) chia hết cho x-1=> F(1)=0<=>a+1=0<=>a=-1
\(a,(x-2)^2-25=0\\\Leftrightarrow (x-2)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
\(---\)
\(b,4x(x-2)+x-2=0\\\Leftrightarrow4x(x-2)+(x-2)=0\\\Leftrightarrow(x-2)(4x+1)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(---\)
\(c,4x(x-2)-x(3+4x)(?)\)
\(d,(2x-5)^2-3x(5-2x)=0\\\Leftrightarrow(2x-5)^2+3x(2x-5)=0\\\Leftrightarrow(2x-5)(2x-5+3x)=0\\\Leftrightarrow(2x-5)(5x-5)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\5x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=1\end{matrix}\right.\)
\(---\)
\(e,x^2-25-(x+5)=0(sửa.đề)\\\Leftrightarrow(x^2-5^2)-(x+5)=0\\\Leftrightarrow (x-5)(x+5)-(x+5)=0\\\Leftrightarrow(x+5)(x-5-1)=0\\\Leftrightarrow(x+5)(x-6)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
\(---\)
\(f,5x(x-3)-x+3=0\\\Leftrightarrow5x(x-3)-(x-3)=0\\\Leftrightarrow(x-3)(5x-1)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
\(Toru\)
\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)
a) \(\Leftrightarrow x^3-8-x^3+9x=1\)
\(\Leftrightarrow9x=9\Leftrightarrow x=1\)
b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)
\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)
Ta có: \(\dfrac{x-2}{x-1}< =0\)
nên x-1>0 và x-2<=0
=>1<x<=2