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Mk xin lỗi nha, câu c sai đề
c) (x+6)4 + (x+8)4 = 272
2: \(\Leftrightarrow\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)=2\left(x-1\right)\left(x+1\right)\)
=>x^2-3x-4+x^2+3x-4=2x^2-2
=>2x^2-8=2x^2-2(loại)
3: \(\Leftrightarrow\left(x^2-x\right)\left(x-3\right)+x^2\left(x+3\right)=-7x^2+3x\)
=>x^3-3x^2-x^2+3x+x^3+3x^2+7x^2-3x=0
=>2x^3+6x^2=0
=>2x^2(x+3)=0
=>x=0(nhận) hoặc x=-3(loại)
a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
b) \(\dfrac{3}{4}xy+\dfrac{3}{4}x^2y-\dfrac{3}{4}xy^2\Leftrightarrow\dfrac{3}{4}xy+\dfrac{3}{4}xy\left(x-y\right)\Leftrightarrow\dfrac{3}{4}xy\left(x-y+1\right)\)
c) \(x\left(x-2\right)+y\left(2-x\right)\Leftrightarrow x\left(x-2\right)-y\left(x-2\right)=\left(x-y\right)\left(x-2\right)\)
d) \(x\left(3-2x\right)+6-4x\Leftrightarrow x\left(3-2x\right)+2\left(3-2x\right)\Leftrightarrow\left(x+2\right)\left(3-2x\right)\)
a) x3 - 5x2 + 8x - 4
= x3 - x2 - 4x2 + 4x + 4x - 4
= x2( x - 1) - 4x( x - 1) + 4( x - 1)
= ( x - 1)( x- 2)2
a.
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow\left(x^2+5x\right)-\left(2x+10\right)=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b.
\(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=1\end{matrix}\right.\)
bài 2:
ĐKXĐ: x khác -1
\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\)
\(\Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x+1}\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow0x=-1\)
\(\Leftrightarrow x\in\varnothing\)
Suy ra pt vô nghiệm
b.
ĐKXĐ: x khác \(\dfrac{3}{2}\)
\(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow\dfrac{x^2+4x+4}{2x-3}-\dfrac{2x-3}{2x-3}=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\) ( loại)
a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
a)\(\dfrac{3}{x^2+5x+4}+\dfrac{2}{x^2+10x+24}=\dfrac{4}{3}+\dfrac{9}{x^2+3x-18}\left(đkxđ:x\ne-1;-4;-6;3\right)\)
\(\Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{4}{3}+\dfrac{9}{\left(x+6\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{4}{3}+\dfrac{1}{x-3}-\dfrac{1}{x+6}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{4}{3}+\dfrac{1}{x-3}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x-3}=\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{-4}{\left(x+1\right)\left(x-3\right)}=\dfrac{4}{3}\)
\(\Leftrightarrow\left(x+1\right)\left(3-x\right)=3\)
\(\Leftrightarrow2x-x^2+3=3\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\left(tm\right)\)
b)\(x^2-y^2+2x-4y-10=0\)
\(\Leftrightarrow x^2+2x+1-y^2-4y-4-7=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)
\(\Leftrightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)
Mà x,yEN*=>x-y-1<x+y+3
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y-1=1\\x+y+3=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-y-1=-7\\x+y+3=-1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
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