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a) x2 + 2x +1
= (x + 1)2
b) 9x2 + y2 + 6xy
= (3x + y)2
c) 25a2 + 4b2 - 20ab
= (5a - 2b)2
d) x2 - x + 1/4
= (x - 1/2)2
a) =(3x+y)2
câu b) sao kì quá mik k hiểu bạn xem lại đề đi có viết thiếu dấu k
a) \(x^2+2x+1=\left(x+1\right)^2\)
b) \(9x^2+y^2+6xy=\left(3x+y\right)^2\)
c) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)
d) \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
bạn có thể giải thích bằng lời đẻ ôi hiểu rõ hơn có được hay ko
1a/ z2 - 6z + 5 - t2 - 4t = z2 - 2 . 3z + 32 - 4 - t2 - 4t = (z2 - 2 . 3z + 32) - (22 + 2 . 2t + t2) = (z - 3)2 - (2 + t)2
b/ x2 - 2xy + 2y2 + 2y2 + 1 = x2 - 2xy + y2 + y2 + 2y + 1 = (x2 - 2xy + y2) + (y2 + 2y + 1) = (x - y)2 + (y + 1)2
c/ 4x2 - 12x - y2 + 2y + 8 = (2x)2 - 12x - y2 + 2y + 32 - 1 = [ (2x)2 - 2 . 3 . 2x + 32 ] - (y2 - 2y + 1) = (2x - 3)2 - (y - 1)2
2a/ (x + y + 4)(x + y - 4) = x2 + xy - 4x + xy + y2 - 4y + 4x + 4y + 16 = x2 + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y2 + 16
= x2 + 2xy + y2 + 42 = (x + y)2 + 42
b/ (x - y + 6)(x + y - 6) = x2 + xy - 6x - xy - y2 + 6y + 6x + 6y - 36 = x2 + (xy - xy) + (-6x + 6x) + (6y + 6y) - y2 - 36
= x2 - y2 + 12y - 62 = x2 - (y2 - 12y + 62) = x2 - (y2 - 2 . 6y + 62) = x2 - (y - 6)2
c/ (y + 2z - 3)(y - 2z - 3) = y2 -2yz - 3y + 2yz - 4z2 - 6z - 3y + 6z + 9 = y2 + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z2 + 9
= y2 - 6y - 4z2 + 9 = (y2 - 6y + 9) - 4z2 = (y - 3)2 - (2z)2
d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x2 + 4y2 + 6yz - 2xy + 6yz + 9z2 - 3xz = (2xy - 2xy) + (3xz - 3xz) - x2 + (6yz + 6yz) + 9z2 + 4y2
= -x2 + 4y2 + 12yz + 9z2 = (4y2 + 12yz + 9z2) - x2 = [ (2y)2 + 2 . 2 . 3yz + (3z)2 ] - x2 = (2y + 3z)2 - x2
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 ^^
Bài 1 : \(a,\)\(16u^2v^4-8uv^2+1\)
\(=\left(4uv^2\right)^2-2.4uv^2.1+1^2\)
\(=\left(4uv^2-1\right)^2\)
\(b,\)\(4x^2-12x+4\)
\(\left(2x\right)^2-2.2x.3+3^2-5\)
\(=\left(2x-3\right)^2-\left(\sqrt{5}\right)^2\)
\(=\left(2x-3-\sqrt{5}\right)\left(2x-3+\sqrt{5}\right)\)
Bài 2 :
\(\left(x+1-2y\right)^2\)
\(=\left[\left(x-1\right)-2y\right]^2\)
\(=\left(x-1\right)^2-2\left(x-1\right).2y+\left(2y\right)^2\)
\(=x^2-2x+1-4xy+4y+4y^2\)
Bài 3 : ( Đề nhầm tí nha , coi lại nhé )
\(x^2+y^2=\left(x+y\right)^2-2xy\)
\(\Rightarrow x^2+y^2=x^2+2xy+y^2\)
\(\Rightarrow x^2+y^2=x^2+y^2\) ( luôn đúng với \(\forall x\))
\(\Rightarrow x^2+y^2=\left(x+y\right)^2-2xy\)\(\left(đpcm\right)\)
a) \(x^6-4=\left(x^3\right)^2-2^2=\left(x^3-2\right).\left(x^3+2\right)\)
b) \(-9x^2+1=1^2-\left(3x\right)^2=\left(1-3x\right).\left(1+3x\right)\)
c) \(x^{10}-9=\left(x^5\right)^2-3^2=\left(x^5-3\right).\left(x^5+3\right)\)
mk chỉ làm đk bài 1 thui ,thông cảm cho mk nha bạn
\(a;x^6-4=\left(x^3\right)^2-2^2=\left(x^3-2\right)\left(x^3+2\right)\)
\(b;-9x^2+1=1^2-3x^2=\left(1-3x\right).\left(1+3x\right)\)
\(c;x^{10}-9=\left(x^5\right)^2-3^2=\left(x^5-3\right).\left(x^5+3\right)\)
\(#LTH\)
Bài 1:
a) \(x^2\left(5x^3-x-6\right)\)
\(=x^2.5x^3-x^2.x-x^2.6\)
\(=5x^5-x^3-6x^2\)
b) \(\left(x^2-2xy+y^2\right)\left(x-y\right)\)
\(=x^2\left(x-y\right)-2xy\left(x-y\right)+y^2\left(x-y\right)\)
\(=x^3-x^2y-2x^2y+2xy^2+y^2x-y^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
Bài 2:
a) \(y^2+2y+1\)
\(=\left(y+1\right)^2\)
b) \(9x^2+y^2-6xy\)
\(=\left(3x\right)^2-2.3x.y+y^2\)
\(=\left(3x-y\right)^2\)
c) \(25a^2+4b^2+20ab\)
\(=\left(5a\right)^2+2.5a.2b+\left(2b\right)^2\)
\(=\left(5a+2b\right)^2\)
d) \(x^2-x+\dfrac{1}{4}\)
\(=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
d) x^2 - x + 1/4
= x^2 - 2.x + 1/4 + (1/2)^2
= ( x - 1/2)^2