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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.
\(\frac{x^2}{4}+x+3=\frac{x^2}{4}+x+1+2=\left(\frac{x}{2}+1\right)^2+2>0;\forall x\)
b.
\(A=-3x^2+2x-5=-3\left(x^2-2.\frac{1}{3}x+\frac{1}{9}\right)-\frac{14}{3}=-3\left(x-\frac{1}{3}\right)^2-\frac{14}{3}\le-\frac{14}{3}\)
\(A_{max}=-\frac{14}{3}\) khi \(x=\frac{1}{3}\)
c.
Đề thiếu (để ý 2 số hạng cuối)
\(A=x^4-2x^3+x^2+3x^2-6x+3-1\)
\(=\left(x^2-x\right)^2+3\left(x-1\right)^2-1\ge-1\)
\(A_{min}=-1\) khi \(x=1\)
d.
\(27x^2-\frac{9}{2}x+\frac{3}{16}=3\left(9x^2-\frac{3}{2}x+\frac{1}{16}\right)=3\left(3x-\frac{1}{4}\right)^2\)
e.
\(=\left[\left(b+c\right)+a\right]^2+\left[\left(b+c\right)-a\right]^2+\left[a-\left(b-c\right)\right]^2+\left[a+\left(b-c\right)\right]^2\)
\(=2\left(b+c\right)^2+2a^2+2a^2+2\left(b-c\right)^2\)
\(=4a^2+2b^2+4bc+2c^2+2b^2-4bc+2c^2\)
\(=4\left(a^2+b^2+c^2\right)\)
f.
\(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=\left(a^2c^2+b^2d^2+2ac.bd\right)+\left(a^2d^2+b^2c^2-2ad.bc\right)\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
Câu a : \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
Câu b : \(9m^2+n^2-6mn=\left(3m-n\right)^2\)
Câu c : \(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
Câu d : \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
\(a,4x^2+4xy+y^2=\left(2x\right)^2+4xy+y^2=\left(2x+y\right)^2\)
\(b,9m^2+n^2-6mn=\left(3m\right)^2-6mn+n^2=\left(3m-n\right)^2\)
\(c,16a^2+25b^2+40ab=\left(4a\right)^2+40ab+\left(5b\right)^2=\left(4a+5b\right)^2\)
@Yukru ơi! giúp câu D với!
Chúc bạn học tốt!
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
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) \(x^2+8xy+16y^2=\left(x+4y\right)^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
d) \(x^2-2x+4=x^2-2.x.4+4^2\)
\(=\left(x-4\right)^2\)
e) \(25x^2+4y^2-20xy=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
^...^ 
^_^ Bài làm có gì ko hiểu bạn cứ hỏi nhé ^_^
mạng của mk bị lỗi bạn xem cái phần cuối cùng nhé xl bạn nhiều vì mạng của mk bị lỗi 
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. \(x^2-2x-24=0\)
\(\Leftrightarrow\left(x^2-6x\right)+\left(4x-24\right)=0\)
\(\Leftrightarrow x\left(x-6\right)+4\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)
Bài 1:
b) \(16x^2-8x+1=\left(4x-1\right)^2\)
c) \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)
\(=\left[\left(x+3\right)\left(x+6\right)\right]\left[\left(x+4\right)\left(x+5\right)\right]+1\)
\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)
Đật \(x^2+9x+19=t\) , pt trở thành
\(\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2=\left(x^2+9x+19\right)^2\)
d) \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)
\(=\left(x^2+2x+1\right)+2\left(x+1\right)\left(y+1\right)+\left(y^2+2y+1\right)\)
\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+1+y+1\right)^2=\left(x+y+2\right)^2\)
e) \(x^2-2x\left(y+2\right)+y^2+4y+4\)
\(=x^2-2x\left(y+2\right)+\left(y+2\right)^2\)
\(=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\)
a)_ Sai đề
N = (x2 - 4x - 5)(x2 - 4x - 19) + 49
Đặt x2 - 4x - 5 = t, ta có:
t(t - 14) + 49
t2 - 14t + 49
= (t - 7)2
= (x2 - 4x - 12)2
= (x2 - 6x + 2x - 12)2
= [x(x - 6) + 2(x - 6)]2
= [(x + 2)(x - 6)]2
[(x + 2)(x - 6)]2 lớn hơn hoặc bằng 0
Vậy Min N = 0 khi x = - 2 hoặc x = 6.
T = x2 - 6x + y2 - 2y + 12
= x2 - 2 . x . 3 + 9 + y2 - 2 . y . 1 + 1 + 2
= (x - 3)2 + (y - 1)2 + 2
(x - 3)2 lớn hơn hoặc bằng 0
(y - 1) lớn hơn hoặc bằng 0
(x - 3)2 + (y - 1)2 + 2 lớn hơn hoặc bằng 2
Vậy Min T = 2 khi x = 3 và y = 1.
Chúc bạn học tốt ^^
a. \(9x^2+25-12xy+5y^2-10y\)
\(=\left(9x^2-12xy+4y^2\right)+\left(25+y^2-10y\right)\)
\(=9\left(x^2-\frac{4xy}{3}+\frac{4y^2}{9}\right)+\left(5-y\right)^2\)
\(=9\left(x-\frac{2y}{3}\right)^2+\left(5-y\right)^2\)
\(a,=\left(x+\dfrac{5}{2}\right)^2\\ b,=\left(2x+3y\right)^2\\ c,=a^2+b^2+c^2+2ab-2bc-2ac\\ d,=\left(4x-1\right)^2\\ e,=a^2+b^2+c^2+2ab+2bc+2ac\\ f,=a^2+b^2+c^2-2ab+2bc-2ac\)