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a. x2 + 6x + 9 = (x + 3)2
b. 25 + 10x + x2 = (5 + x)2
c. x2 + 8x + 16 = (x + 4)2
d. x2 + 14x + 49 = (x + 7)2
e. 4x2 + 12x + 9 = (2x + 3)2
f. 9x2 + 12x + 4 = (3x + 2)2
h. 16x2 + 8 + 1 = (4x + 1)2
i. 4x2 + 12xy + 9y2 = (2x + 3y)2
k. 25x2 + 20xy + 4y2 = (5x + 2y)2
a) \(=\left(x+3\right)^2\)
b) \(=\left(x+5\right)^2\)
c) \(=\left(x+4\right)^2\)
d) \(=\left(x+7\right)^2\)
e) \(=\left(2x+3\right)^2\)
f) \(=\left(3x+2\right)^2\)
h) \(=\left(4x+1\right)^2\)
i) \(=\left(2x+3y\right)^2\)
k) \(=\left(5x+2y\right)^2\)
a) \(x^2-6x+9=x^2-2\cdot x\cdot3+3^2=\left(x-3\right)^2\)
b) \(4x^2-12xy+9y^2=\left(2x\right)^2-2\cdot2x\cdot3y+\left(3y\right)^2=\left(2x-3y\right)^2\)
c) \(4x^2-2x+1=\left(2x-1\right)^2\)
d) \(x^2+8xy+16y^2=\left(x+4y\right)^2\)
a. \(9x^2+25-12xy+5y^2-10y\)
\(=\left(9x^2-12xy+4y^2\right)+\left(25+y^2-10y\right)\)
\(=9\left(x^2-\frac{4xy}{3}+\frac{4y^2}{9}\right)+\left(5-y\right)^2\)
\(=9\left(x-\frac{2y}{3}\right)^2+\left(5-y\right)^2\)
a) \(x^2-6x+9=x^2-2.3.x+3^2=\left(x-3\right)^2\)
b)\(x^2+4x+4=x^2+2.2.x+2^2=\left(x+2\right)^2\)
c)\(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
d)\(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
e)\(x^2-8x+16=x^2-2.4.x+4^2=\left(x-4\right)^2\)
a) x2 -6x +9 = (x-3)2
b) x2+4x +4= (x+2)2
c) 4x2+4x+1= (2x+1)2
d) 4x2+12xy+9y2 = (2x+3y)2
e) x2-8x+16 = (x-4)2
Đây chính là hằng đẳng thức nhé bn....
a) 9x2 + 25 - 12xy + 5y2 - 10y
= ( 9x2 - 12xy + 4y2 ) + ( y2 - 10y + 25 )
= ( 3x - 2y )2 + ( y - 5 )2
b) 13x2 + 4x + 12xy + 4y2 + 1
= ( 9x2 + 12xy + 4y2 ) + ( 4x2 + 4x + 1 )
= ( 3x + 2y )2 + ( 2x + 1 )2
c) x2 + 20 + 9y2 + 8x - 12y
= ( x2 + 8x + 16 ) + ( 9y2 - 12y + 4 )
= ( x + 4 )2 + ( 3y - 2 )2
Bài 2: Tìm x
a) x2 - 6x + 5 = 0
<=> x2 - x - 5x + 5 = 0
<=> x(x - 1) - 5(x - 1) = 0
<=> (x - 1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Vậy x ={1; 5}
b) x2 - 2x - 24 = 0
<=> x2 + 4x - 6x - 24 = 0
<=> x(x + 4) - 6(x + 4) = 0
<=> (x + 4)(x - 6) = 0
<=> \(\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)
Vậy x ={-4; 6}
2.
a. \(x^2-6x+5=0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(5x-5\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b. \(x^2-2x-24=0\)
\(\Leftrightarrow\left(x^2-6x\right)+\left(4x-24\right)=0\)
\(\Leftrightarrow x\left(x-6\right)+4\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)
a)\(\left(3x\right)^2-2×3x+1^2=\left(3x-1\right)^2\)
b)\(\left(2x\right)^2+2×2x+1^2=\left(2x+1\right)^2\)
c)\(\left(2x\right)^2+2×2x×3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
d)\(-\left(4x^2-12xy+9y^2\right)=-\left[\left(2x\right)^2-2×2x×3y+\left(3x\right)^2\right]=-\left[\left(2x-3y\right)^2\right]\)