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\(\left(a-x-y\right)^3-\left(a+x-y\right)^3\)
\(=\left[\left(a-x-y\right)-\left(a+x-y\right)\right]\left[\left(a-x-y\right)^2+\left(a-x-y\right)\left(a+x-y\right)+\left(a+x-y\right)^2\right]\)
\(=-2x.\left[a^2+x^2+y^2-2ax+2xy-2ay+\left(a-y\right)^2-x^2+a^2+x^2+y^2+2ax-2xy-2ay\right]\)
\(=-2x\left[a^2+x^2+y^2-2ax+2xy-2ay+a^2-2ay+y^2-x^2+a^2+x^2+y^2+2ax-2xy-2ay\right]\)
\(=-2x\left(3a^2+x^2+3y^2-4ay\right)\)
a: \(\left(x-1\right)^3+27\)
\(=\left(x-1+3\right)\left(x^2-2x+1+3x-3+3\right)\)
\(=\left(x+2\right)\left(x^2+x+1\right)\)
b: \(\left(x-2\right)^3-8\)
\(=\left(x-2-2\right)\left(x^2-4x+4+2x-4+4\right)\)
\(=\left(x-4\right)\left(x^2-2x+4\right)\)
a. 12xy2 - 8x2y = 4xy . (3y - 2x)
b. 3x + 3y - x2 - xy = (3x + 3y) - (x2 + xy) = 3 . (x + y) - x . (x + y) = (x + y)(3 - x)
1) \(x^2-2x+5+y^2-4y=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)
Để PT bằng 0 thì:
\(\left(x-1\right)^2=0\)và \(\left(y-2\right)^2=0\)
\(\Rightarrow x=1\)và \(y=2\)
2) \(y^2+2y+5-12x+9x^2=0\)
\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)
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..............<Giải thích như câu đầu>......................
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\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)
\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)
3) \(x^2+20+9y^2+8x-12y=0\)
\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)
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...............<Giải thích như câu đầu>..............
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\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)
\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)
1) \(x^2-2x+5+y^2-4y=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)
Để PT bằng 0 thì:
\(\left(x-1\right)^2=0\)và \(\left(y-2\right)^2=0\)
\(\Rightarrow x=1\)và \(y=2\)
2) \(y^2+2y+5-12x+9x^2=0\)
\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)
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..............<Giải thích như câu đầu>......................
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\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)
\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)
3) \(x^2+20+9y^2+8x-12y=0\)
\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)
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...............<Giải thích như câu đầu>..............
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\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)
\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)
Cái này chia trường hợp thoy bn
mk lm 1 ý thoy ý còn lại tương tự nha
\(\left|x+2\right|+\left|2x-3\right|=5\) (*)
Với x \(\ge\)1.5 ta có
(*) <=> x+2+2x-3=5
<=> 3x=6 <=>x=2(TM)
Với x \(\le-2\)ta có
(*) <=> -x-2+3-2x=5
<=>x=-4/3 (loại)
Với \(-2< x< 1.5\) ta có
x+2+3-2x=5
<=>-x=0
<=>x=0 TM
vây x=0 hoặc x= 2
Bài 1 :
1) 4x2 - y2 = ( 2x + y ) ( 2x - y )
2) 9x2 - 4y2 = ( 3x - 2y ) ( 3x + 2y )
3) 4x2 + y2 + 4xy = ( 2x + y )2
Bài 2:
1) 2x2 + 8x = 0
=> 2x ( x + 4 ) = 0
=> \(\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
2) 3 ( x - 4 ) + x2 - 4x = 0
=> 3 ( x - 4 ) + x ( x - 4 ) = 0
=> ( x - 4 ) ( 3 + x ) = 0
=> \(\orbr{\begin{cases}x-4=0\\3+x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)
3) 3 ( x - 2 ) = x2 - 2x
=> 3 ( x - 2 ) - x2 + 2x = 0
=> 3 ( x - 2 ) - x ( x - 2 ) = 0
=> ( x - 2 ) ( 3 - x ) = 0
=> \(\orbr{\begin{cases}x-2=0\\3-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
4) x ( x - 2 ) - 6 ( 2 - x ) = 0
=> x ( x - 2 ) + 6 ( x - 2 ) = 0
=> ( x - 2 ) ( x + 6 ) = 0
=> \(\orbr{\begin{cases}x-2=0\\x+6=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
5) 2x ( x + 5 ) = x2 + 5x
=> 2x ( x + 5 ) - x2 - 5x = 0
=> 2x ( x + 5 ) - x ( x + 5 ) = 0
=> ( x + 5 ) ( 2x - x ) = 0
=> \(\orbr{\begin{cases}x+5=0\\2x-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-5\\x=0\end{cases}}\)
6 ) ( x - 2 )2 - x ( x + 3 ) = 9
=> x2 - 4x + 4 - x2 - 3x = 9
=> - 7x + 4 = 9
=> - 7x = 5
=> x = \(-\frac{5}{7}\)
\(1,4x^2-y^2=\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
\(2,9x^2-4y^2=\left(3x\right)^2-\left(2y\right)^2=\left(3x-2y\right)\left(3x+2y\right)\)
\(3,4x^2+y^2+4xy=\left(2x\right)^2+2.2x.y+y^2=\left(2x+y\right)^2\)
\(1,2x^2+8x=0\Rightarrow2x\left(x+4\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
\(2,3\left(x-4\right)+x^2-4x=0\)
\(\Rightarrow3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Rightarrow\left(3+x\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3+x=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(3,3\left(x-2\right)=x^2-2x\)
\(\Rightarrow3\left(x-2\right)-x^2+2x=0\)
\(\Rightarrow3\left(x-2\right)-x\left(x-2\right)=0\)
\(\Rightarrow\left(3-x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
\(4,x\left(x-2\right)-6\left(2-x\right)=0\)
\(\Rightarrow x\left(x-2\right)+6\left(x-2\right)=0\)
\(\Rightarrow\left(x+6\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+6=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-6\\x=2\end{cases}}\)
2x(5 - x) + (x - 1)(2x + 3) = 8
=> 10x - 2x2 + 2x2 + 3x - 2x - 3 = 8
=> 11x - 3 = 8
=> 11x = 8 + 3
=> 11x = 11
=> x = 11 : 11
=> x = 1