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a) \(4x^2-12x+9\)
\(=\left(2x\right)^2-2.2.3+3^2\)
\(=\left(2x-3\right)^2\)
b) \(4x^2+4x+1\)
\(=\left(2x\right)^2+2.2x.1+1^2\)
\(=\left(2x+1\right)^2\)
c) \(1+12x+36x^2\)
\(=1^2+2.6x+\left(6x\right)^2\)
\(=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2\)
\(=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2\)
\(=\left(3x-4y\right)^2\)
e) Viết = công thức trực quan hộ mình
f) \(-x^2+10x-25\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2.5x+5^2\right)\)
\(=-\left(x-5\right)^2\)
Bài 1 : Tìm x,biết :
a, x2(x + 5) - 9x = 45
⇔ x2(x + 5) - 9x - 45 = 0
⇔ x2(x + 5) - 9(x + 5) = 0
⇔ (x + 5)(x2 - 9) = 0
⇔ (x + 5)(x - 3)(x + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\\x=-3\end{matrix}\right.\)
Vậy x ={-5; 3; -3}
b, 9(5 - x) + x2 - 10x = -25
⇔ 45 - 9x + x2 - 10x + 25 = 0
⇔ x2 - 19x + 70 = 0
⇔ x2 - 14x - 5x + 70 = 0
⇔ (x2 - 5x) - (14x - 70) = 0
⇔ x(x - 5) - 14(x - 5) = 0
⇔ (x - 5)(x - 14) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)
Vậy x ={5; 14}
a, x2( x+5 ) - 9x = 45
x3 + 5x2 - 9x - 45 = 0
x2( x+5 ) - 9( x+5) = 0
(x2 - 9)(x + 5) = 0
(x + 3)(x - 3)(x + 5) = 0
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-5\end{matrix}\right.\)
b, 9( 5-x ) + x2 -10x = -25
45 - 9x + x2 - 10x + 25 = 0
x2 - 19x + 70 = 0
x2 - 14x - 5x + 70 = 0
x( x-14 ) - 5( x-14) = 0
(x - 5)(x - 14) = 0
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)
a) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
b) \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left[3\left(x+y-1\right)\right]^2-\left[2\left(2x+3y+1\right)\right]^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3+4x+6y+2\right)\left(3x+3y-3-4x-6y-2\right)\)
\(=\left(7x+9y-1\right)\left(-x-3y-5\right)\)
c) \(-4x^2+12xy-9y^2+25\)
\(=-\left(2x\right)^2+2.2x.3y-\left(3y\right)^2+5^2\)
\(=-\left[\left(2x\right)^2-2.2x.3y+\left(3y\right)^2-5^2\right]\)
\(=-\left[\left(2x-3y\right)^2-5^2\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-4m\left(m-n\right)-n^2\)
\(=\left(x-y\right)^2-4m\left(m-n\right)-n^2\)
\(=\left(x-y-n\right)\left(x-y+n\right)-4m\left(m-n\right)\)
a, Ta có : \(-x^2+2x-1-3\)
\(=-\left(x-1\right)^2-3\)
Ta thấy : \(\left(x-1\right)^2\ge0\forall x\)
=> \(-\left(x-1\right)^2-3\le-3\forall x\)
Vậy Max = -3 <=> x = 1 .
b, Ta có : \(-x^2-4x-4+4\)
\(=-\left(x+2\right)^2+4\)
Ta thấy : \(\left(x+2\right)^2\ge0\forall x\)
=> \(-\left(x+2\right)^2+4\le4\forall x\)
Vậy Max = 4 <=> x = -2 .
c, Ta có : \(-9x^2+24x-16-2\)
\(=-9\left(x^2-\frac{2.4x}{3}+\frac{16}{9}\right)-2\)
\(=-9\left(x-\frac{4}{3}\right)^2-2\)
Ta thấy : \(\left(x-\frac{4}{3}\right)^2\ge0\forall x\)
=> \(-9\left(x-\frac{4}{3}\right)^2-2\le-2\forall x\)
Vậy Max = -2 <=> x = \(\frac{4}{3}\) .
d, Ta có : \(-x^2+4x-4+3\)
\(=-\left(x-2\right)^2+3\)
Ta thấy : \(\left(x-2\right)^2\ge0\forall x\)
=> \(-\left(x-2\right)^2+3\le3\forall x\)
Vậy Max = 3 <=> x = 2 .
e, Ta có : \(-x^2+2x-1-4y^2-4y-1+7\)
\(=-\left(x-1\right)^2-4\left(y^2+y+\frac{1}{4}\right)+7\)
\(=-\left(x-1\right)^2-4\left(y+\frac{1}{2}\right)^2+7\)
Ta thấy : \(\left\{{}\begin{matrix}\left(x-1\right)^2\\\left(y+\frac{1}{2}\right)^2\end{matrix}\right.\) \(\ge0\forall xy\)
=> \(\left\{{}\begin{matrix}-\left(x-1\right)^2\\-4\left(y+\frac{1}{2}\right)^2\end{matrix}\right.\) \(\le0\forall xy\)
=> \(=-\left(x-1\right)^2-4\left(y+\frac{1}{2}\right)^2\le0\forall xy\)
=> \(=-\left(x-1\right)^2-4\left(y+\frac{1}{2}\right)^2+7\le7\forall xy\)
Vậy Max = 7 <=> \(\left\{{}\begin{matrix}x=1\\y=-\frac{1}{2}\end{matrix}\right.\)
a, ( x2 + x )2 - 14 ( x2 + x ) + 24
= (x2 + x)2 - 2(x2 + x) -12(x2 + x) + 24
= (x2 + x).(x2 + x -2) - 12(x2 + x -2)
= (x2 + x -2).(x2 + x -12)
= (x2 + 2x - x - 2).(x2 + 4x - 3x - 12)
=[x.(x+2)-(x+2)].[x.(x+4)-3(x+4)]
= (x+2).(x-1).(x+4).(x-3)
= x4 + 2x3 - 13x2 - 14x + 24
b, ( x2 + x )2 + 4x2 + 4x - 12
= x4 + 2x3 + x2 + 4x2 + 4x -12
= x4 + 2x3 + 5x2 + 4x -12
c, x4 + 2x3 + 5x2 + 4x - 12
= x4 - x3 + 3x3 - 3x2 + 8x2 - 8x +12x -12
= x3(x-1) + 3x2(x-1) + 8x(x-1) + 12(x-1)
= (x-1) . (x3 + 3x2 + 8x +12)
= (x-1) . ( x3 +2x2 + x2 + 2x + 6x +12)
= (x-1). [x2(x+2) + x(x+2) + 6(x+2)]
= (x-1).(x+2).(x2 + x+ 6)
a, \(x^2-10x=-25\)
=> \(x^2-10x+25=0\)
=> \(\left(x-5\right)^2=0\)
=> x - 5 = 0
=> x = 5
b, \(4x^2-4x=-1\)
=> \(4x^2-4x+1=0\)
=> \(\left(2x-1\right)^2=0\)
=> 2x - 1 = 0
=> 2x = 1
=> x = \(\frac{1}{2}\)