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a. \(8x\left(x-2007\right)-2x+4034=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy x=2017 hoặc x=1/4
b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)
\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy x=0 hoặc x=-4
c.\(4-x=2\left(x-4\right)^2\)
\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)
\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy x=4 hoặc x=7/2
d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)
\(\Rightarrow\left(x-2\right)\left(x^2+3\right)=0\)
Nxet: (x2+3)>0 với mọi x
=> x-2=0 <=>x=2
Vậy x=2
a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0
4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0
4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0
4\(x^2\) - 8029\(x\) + 2017 = 0
4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))2) - (\(\dfrac{8029}{4}\))2 + 2017 = 0
4.(\(x\) + \(\dfrac{8029}{8}\))2 = (\(\dfrac{8029}{4}\))2 - 2017
\(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)
a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\)
\(\Leftrightarrow x^3+9x+2=x^3+8\)
\(\Leftrightarrow x^3+9x=x^3+8-2\)
\(\Leftrightarrow x^3+9x=x^3+6\)
\(\Leftrightarrow x^3+9x=x^3+6x-x^3\)
\(\Leftrightarrow\frac{2}{3}\)
b) \(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-4=8x-16\)
\(\Leftrightarrow x^4-4=8x-16+16\)
\(\Leftrightarrow x^2+12=8x\)
\(\Leftrightarrow x^2+12=8x-8x\)
\(\Leftrightarrow x^2-8x+12=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)
Bài 3:
a) \(A=\left(2xy^2\right)\left(x^3-2xy+2y^2\right)\)
\(A=2xy^2\cdot x^3-2xy^2\cdot2xy+2xy^2\cdot2y^2\)
\(A=2x^4y^2-4x^2y^3+4xy^4\)
b) \(B=\left(x^2+y^2-z^2\right)\left(x^2+y^2+z^2\right)\)
\(B=x^2\cdot x^2+x^2\cdot y^2+x^2\cdot z^2+x^2\cdot y^2+y^2\cdot y^2+y^2\cdot z^2-x^2\cdot z^2-y^2\cdot z^2-z^2\cdot z^2\)
\(B=x^4+x^2y^2+x^2z^2+x^2y^2+y^4+y^2z^2-x^2z^2-y^2z^2-z^4\)
\(B=x^4+\left(x^2y^2+x^2y^2\right)+\left(x^2z^2-x^2z^2\right)+y^4+\left(y^2z^2-y^2z^2\right)-z^4\)
\(B=x^4+y^4-z^4+2x^2y^2\)
c) \(C=-\dfrac{1}{4}xy\left(4x^2y^2-x^2y-\dfrac{4}{5}\right)\)
\(C=-\dfrac{1}{4}xy\cdot4x^2y^2+\dfrac{1}{4}xy\cdot x^2y+\dfrac{1}{4}xy\cdot\dfrac{4}{5}\)
\(C=-x^3y^3+\dfrac{1}{4}x^3y^2+\dfrac{1}{5}xy\)
d) \(D=\left(x-y\right)^4\)
\(D=\left[\left(x-y\right)^2\right]^2\)
\(D=\left(x^2-2xy+y^2\right)^2\)
\(D=\left(x^2-2xy+y^2\right)\left(x^2-2xy+y^2\right)\)
\(D=x^4-2x^3y+x^2y^2-2x^3y+4x^2y^2-2xy^3+x^2y^2-2xy^3+y^4\)
\(D=x^4+6x^2y^2+y^4\)
4/
a/ \(A=\dfrac{7y^5z^2-14y^3z^4+2,1y^4z^3}{-7y^3z^2}=\dfrac{7y^5z^2}{-7y^3z^2}+\dfrac{-14y^3z^4}{-7y^3z^2}+\dfrac{2,1y^4z^3}{-7y^3z^2}=-y^2+2z^2-0,3yz\)
b/ \(A=\dfrac{9x^3y+3xy^3-6x^2y^2}{-3xy}=\dfrac{9x^3y}{-3xy}+\dfrac{3xy^3}{-3xy}+\dfrac{-6x^2y^2}{-3xy}=-3x^2-y^2+2xy\)
a: Xét tứ giác AEMF có
\(\widehat{MEA}=\widehat{MFA}=\widehat{FME}=90^0\)
Do đó: AEMF là hình chữ nhật
a)Tứ giác AEMF có :
\(\widehat{MEA}=\widehat{MFA}=\widehat{FME}=90^0\)
=>AEMF là hình chữ nhật
