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DM
Doãn Minh Cường
Giáo viên
12 tháng 9 2021
\(x^2-5y+y^2-2xy+5x=\left(x^2-2xy+y^2\right)+\left(5x-5y\right)\)
MD
12 tháng 9 2021
a/ x2 – 5y + y2 -2xy + 5x = ( x2 - 2xy + y2 ) - 5( y - x ) = ( x - y )2 - 5( y - x ) = ( y - x )2 - 5( y - x ) = ( y - x )( y - x - 5 )
b/ 4x2 – 81(y – 2)2 = 4x2 - 92(y – 2)2= 4x2 – ( 9y – 18)2 = ( 2x -9y -18 )( 2x + 9y + 18 )
c/ x2z – y2z + 2yz – z = ( x2z + yz ) - ( y2z - yz ) - z = z( x2 + y ) - z( y2 - y ) -z = z( x2 + y - y2 +y - 1 ) = z( x2 + 2y - y2 - 1 ) \(=z[x^2-\left(y^2-2y+1\right)]=z[x^2-\left(y-1\right)^2=z\left(x-y+1\right)\left(x+y-1\right)\)
d/ x3 – 8y3 + x2 + 2xy + 4y2 = ( x3 – 8y3 ) + x2 + 2xy + 4y2 = ( x -2y )( x2 + 2xy + 4y2 ) + ( x2 + 2xy + 4y2 0 = ( x2 + 2xy + 4y2)( x -2y +1)
e/ 7x2 – 11x + 4 = 7x2 -7x -4x +4 = 7x( x-1 ) - 4( x - 1 ) = ( x - 1 )( 7x - 4 )
g/ 13x2 + 2xy – 15y2 = 13x2 - 13xy + 15xy - 15y2 = 13x( x - y ) + 15y( x - y ) = ( x - y )( 13x + 15y )
h/ x3 + 3x2 + 3x + 2 = x3 +2x2 + x2 +2x + x + 2 = x2( x + 2 ) + x( x + 2 ) + ( x + 2 ) = ( x + 2 )( x2 + x + 1 )
i/ x3 – 3x2 + 3x – 2 + xy – 2y = x3 - 2x2 - x2 + 2x + x - 2 +xy - 2y = x2( x - 2 ) - x( x - 2 ) + ( x - 2 ) + y( x - 2 ) = ( x - 2 )( x2 - x +1 + y )
DH
12 tháng 6 2017
a, \(C=5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(C=5.\left(4x^2-4x+1\right)+4\left(x^2+3x-x-3\right)-2.\left(25-75x+9x^2\right)\)
\(C=20x^2-20x+5+4x^2+8x-12-50+150x-18x^2\)
\(=\left(20x^2+4x^2-18x^2\right)+\left(-20x+8x+150x\right)+\left(5-12-50\right)\)
\(C=6x^2+138x-57\)
Chúc bạn học tốt!!! Cũng không chắc có đúng hay sai nữa do cồng kềnh quá !
DT
9 tháng 2 2021
(2x2+x-6)+3(2x2+x-3)-9=0
\(\Leftrightarrow\) 2x2 + x - 6 + 6x2 + 3x - 9 - 9 = 0
\(\Leftrightarrow\)2x2 + 6x2 + 3x + x = 6 + 9 + 9
\(\Leftrightarrow\)8x2 + 4x = 24
\(\Leftrightarrow\)8x2 + 4x - 24 = 0
\(\Leftrightarrow\)(x+2)(8x-12) = 0
\(\Leftrightarrow\)x + 2 = 0 hoặc 8x - 12 = 0
1) x + 2 = 0 \(\Leftrightarrow\)x = -2
2)8x - 12 = 0 \(\Leftrightarrow\)8x = 12 \(\Leftrightarrow\)x = \(\frac{12}{8}\)
Vậy Tập nghiệm của phương trình đã cho là S ={ -2 ; \(\frac{12}{8}\)}
XO
20 tháng 2 2021
A = \(\frac{x+9}{x^2+3x-4}+\frac{x+12}{x^2-5x+4}+\frac{x-5}{x^2-1}\)
\(=\frac{x+9}{\left(x-1\right)\left(x+4\right)}+\frac{x+12}{\left(x-1\right)\left(x-4\right)}+\frac{x-5}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{ }{ }\)
9 tháng 3 2022
\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)
\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)
\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)
\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)
\(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\)
\(=\frac{2x+1}{x-3}\)
b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)
thay \(x=-\frac{3}{2}\) vào P tâ đc: \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)
c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)
\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)
\(\Leftrightarrow4x+2=x^2-3x\)
\(\Leftrightarrow x^2-7x-2=0\)
\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)
bạn tự giải nốt nhé!!
d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)
\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
bạn tự làm nốt nhé
9 tháng 3 2022
a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\)
\(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\)
b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)
E = - 3\(x^2\) - \(x\) + 2
E = - 3.( \(x^2\) + 2.\(\dfrac{1}{6}\)\(x\) + \(\dfrac{1}{36}\)) + 2
E = -3.(\(x\) + \(\dfrac{1}{6}\))2 + \(\dfrac{25}{12}\)
Vì (\(x+\dfrac{1}{6}\))2 ≥ 0 ∀ \(x\) ⇒ -3.(\(x+\dfrac{1}{6}\))2 ≤ 0 ⇒ -3(\(x+\dfrac{1}{6}\))2 + \(\dfrac{25}{12}\) ≤ \(\dfrac{25}{12}\)
Emax = \(\dfrac{25}{12}\) ⇔ \(x\) = - \(\dfrac{1}{6}\)