Giúp mình với: (a2+4b2)2-16a2b2
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NT
2
24 tháng 10 2021
1: \(a^2-4b^2-2a-4b\)
\(=\left(a-2b\right)\left(a+2b\right)-2\left(a+2b\right)\)
\(=\left(a+2b\right)\left(a-2b-2\right)\)
2: \(x^3+2x^2-2x-1\)
\(=\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+3x+1\right)\)
NT
2
TG
10 tháng 8 2021
\(\dfrac{1}{a-2b}.\sqrt{b^2\left(a^2-4ab+4b^2\right)}=\dfrac{1}{a-2b}.b.\left|a-2b\right|=\dfrac{1}{a-2b}.b.\left(2b-a\right)=-b\)
10 tháng 8 2021
\(\dfrac{1}{a-2b}\cdot\sqrt{b^2\cdot\left(a^2-4ab+b^2\right)}\)
\(=\dfrac{1\cdot\left(a-2b\right)}{a-2b}\cdot b\)
=b
19 tháng 12 2021
a: \(\Leftrightarrow a^2-4a+4+b^2-6b+9+c^2-2c+1>=0\)
\(\Leftrightarrow\left(a-2\right)^2+\left(b-3\right)^2+\left(c-1\right)^2>=0\)
Dấu '=' xảy ra (a,b,c)=(2;3;1)
C
2
7 tháng 8 2021
a) Ta có: \(M=-x^2-4x+20\)
\(=-\left(x^2+4x-20\right)\)
\(=-\left(x^2+4x+4-24\right)\)
\(=-\left(x+2\right)^2+24\le24\forall x\)
Dấu '=' xảy ra khi x=-2
NN
2
5 tháng 8 2023
\(\left(a-1\right)^2\ge0\Rightarrow a^2+1-2a\ge0\Rightarrow a^2+1\ge2a\left(1\right)\)
\(\left(2b-3\right)^2\ge0\Rightarrow4b^2+9-12b\ge0\Rightarrow4b^2+9\ge12b\left(2\right)\)
\(\left(c\sqrt[]{3}-\sqrt[]{3}\right)^2\ge0\Rightarrow3c^2+3-6c\ge0\Rightarrow3c^2+3\ge6c\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow a^2+1+4b^2+9+3c^2+3\ge2a+12b+6c\)
\(\Rightarrow a^2+4b^2+3c^2+1+9+3\ge2a+12b+6c\)
\(\Rightarrow a^2+4b^2+3c^2+13\ge2a+12b+6c\)
\(\Rightarrow a^2+4b^2+3c^2\ge2a+12b+6c-13\)
mà \(2a+12b+6c-13>2a+12b+6c-14\)
\(\Rightarrow a^2+4b^2+3c^2>2a+12b+6c-14\)
\(\Rightarrow dpcm\)
5 tháng 8 2023
⇔(a−1)2+(2b−3)2+3(c−1)2+1>0 (luôn đúng)
⇒⇒ BĐT ban đầu đúng
19 tháng 12 2023
a) (a - 2b)x(a + 2b)
b) x2-(y-3)2
=> (x-y+3)(x+y-3)
c) (2a + b - a)(2a + b + a)
=> (a+b)(3a+b)
d) (4(x - 1))2 - (5(x + y))2
⇔ (4x - 4 - 5x - 5y)(4x - 4 + 5x + 5y)
⇔ -(x + 5y + 4)(9x + 5y + -4)
e) (x + 5)2
f) (5x - 2y)2
h) (x - 5)(x2 + 5x + 25)
k) (x + 5)3
AH
Akai Haruma
Giáo viên
4 tháng 7 2021
Lời giải:
$(a+2b-c)(a+2b+c)-(a^2+4b^2-c^2)=(a+2b)^2-c^2-a^2-4b^2+c^2$
$=(a+2b)^2-a^2-4b^2$
$=a^2+4ab+4b^2-a^2-4b^2=4ab$
J
0
PA
1
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 2 2022
Đặt \(P=a+b+c\)
\(P^2=\left(a+b+c\right)^2=\left(1.a+\dfrac{1}{2}.2b+\dfrac{1}{3}.3c\right)^2\le\left(1^2+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2\right)\left(a^2+4b^2+9c^2\right)\)
\(\Rightarrow P^2\le\dfrac{49}{36}\left(a^2+4b^2+9c^2\right)=\dfrac{49}{36}\)
\(\Rightarrow-\dfrac{7}{6}\le P\le\dfrac{7}{6}\)
\(P_{min}=-\dfrac{7}{6}\) khi \(\left(a;b;c\right)=\left(-\dfrac{6}{7};-\dfrac{3}{14};-\dfrac{2}{21}\right)\)
\(P_{max}=\dfrac{7}{6}\) khi \(\left(a;b;c\right)=\left(\dfrac{6}{7};\dfrac{3}{14};\dfrac{2}{21}\right)\)
T
4 tháng 9 2023
\(a,a^2-2a-4b^2-4b\)
\(=\left(a^2-4b^2\right)-\left(2a+4b\right)\)
\(=\left(a-2b\right)\left(a+2b\right)-2\left(a+2b\right)\)
\(=\left(a+2b\right)\left(a-2b-2\right)\)
\(b,x^3-2x^2+4x-8\)
\(=x^2\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4\right)\)
\(c,x^3+36x-12x^2\)
\(=x^3-6x^2-6x^2+36x\)
\(=x^2\left(x-6\right)-6x\left(x-6\right)\)
\(=\left(x-6\right)\left(x^2-6x\right)\)
\(=x\left(x-6\right)^2\)
\(d,5a^2+3\left(a+b\right)^2-5b^2\)
\(=\left(5a^2-5b^2\right)+3\left(a+b\right)^2\)
\(=5\left(a^2-b^2\right)+3\left(a+b\right)^2\)
\(=5\left(a-b\right)\left(a+b\right)+3\left(a+b\right)^2\)
\(=\left(a+b\right)\left[5\left(a-b\right)+3\left(a+b\right)\right]\)
\(=\left(a+b\right)\left(5a-5b+3a+3b\right)\)
\(=\left(a+b\right)\left(8a-2b\right)\)
\(=2\left(a+b\right)\left(4a-b\right)\)
\(e,x^3-3x^2+3x-1-y^3\)
\(=\left(x^3-3x^2+3x-1\right)-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)
\(=\left(x-y-1\right)\left(x^2-2x+1+xy-y+y^2\right)\)
\(=\left(x-y-1\right)\left(x^2+y^2-xy-y+1\right)\)
#Urushi☕
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
4 tháng 9 2023
\(c.\\ x^3+36x-12x^2\\ =x\left(x^2-12x+36\right)\\ =x.\left(x^2-2.x.6+6^2\right)\\ =x.\left(x-6\right)^2\\ ---\\ d.\\ 5a^2+3\left(a+b\right)^2-5b^2\\ =\left(5a^2-5b^2\right)+3\left(a+b\right)^2\\ =5.\left(a^2-b^2\right)+3.\left(a+b\right)\left(a+b\right)\\ =5\left(a+b\right)\left(a-b\right)+3\left(a+b\right)\left(a+b\right)\\ =\left(a+b\right)\left(5a-5b+3a+3b\right)\\ =\left(a+b\right)\left(8a-2b\right)\\ =2\left(a+b\right)\left(4a-b\right)\)
\(e.\\ x^3-3x^2+3x-1-y^3\\ =\left(x-1\right)^3-y^3\\ =\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right).y+y^2\right]\\ =\left(x-y-1\right).\left[\left(x^2-2x+1\right)+y\left(x+y-1\right)\right]\)
\(\left(a^2+4b^2\right)^2-16a^2b^2\)
\(=\left(a^2+4b^2\right)^2-\left(4ab\right)^2\)
\(=\left(a^2+4b^2-4ab\right)\left(a^2+4b^2+4ab\right)\)
\(=\left(a-2b\right)^2\left(a+2b\right)^2\)
\(\left(a^2+4b^2\right)^2-16a^2b^2=\left(a^2+4b^2\right)-\left(4ab\right)^2=\left(a^2-4ab+4b^2\right)\left(a^2+4ab+4b^2\right)=\left(a-2b\right)^2\left(a+2b\right)^2\)