rút gọn : a) 5.(2x-1)^2+4(x-1)(x+3)-2(5-3x)^2
b) (2a^2+2a+1)(2a^2-2a+1)=(2a^2+1)^2
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26 tháng 7 2021
a) A=(4-5x)2-(3+5x)2=(4-5x-3-5x)(4-5x+3+5x)=(-25x+1)1=-25x+1
26 tháng 7 2021
B=(3x-1)(1+3x)-(3x+1)2=9x2-1-(3x+1)2=9x2-1-(9x2+6x+1)=9x2-1-9x2-6x-1=-6x-2=-2(3x+1)
18 tháng 10 2021
a) \(=6a-3+15-5a=a+12\)
b) \(=25x-12x+4+35-14x=-x+39\)
d) \(=2ab+8a^2-b^2-4ab+2ab-6a^2=2a^2-b^2\)
e) \(=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4=-x^5+2x+1\)
f) \(=6y^3-3y^2+y-y+y^2-y^3-y^2+y=5y^3-3y^2+y\)
PL
18 tháng 10 2021
a) 3( 2a -1) +5( 3-a)
= 3. 2a -3.1 +5. 3- 5.a
= 6a -3+ 15-5a
=(6a -5a )+ (-3+ 15)
b) 25x - 4(3x - 1) +7(5 - 2x)
= 25x -4.3x + 4.1 + 7.5 - 7.2
=25x - 12x + 4 +35 - 14x
= (25x-12x-14x)+(4+35)
= -x=39
c) -12x3 -x1-2x-18x2
= -36x-x-2x-36x
= -75x
d) (2a-b)(b+4a)+2a(b-3a)
= 2ab+2a4a-bb-b4a+2ab-2a3b
= 2ab+8a2-b2-4ab+2ab-6a2
=(2ab-4ab+2ab)+(8a2-6a2)-b2
= 2a2-b2
e) (x+1)(2+x-x2+x3-x4)
= (x+1)(2-2x)
= x2-x2x+1.2-1.2x
=(2x-2x)-2x2+2
= -2x2+2
24 tháng 8 2021
Bài 2:
a) \(\left(x+5\right)^2=x^2+10x+25\)
b) \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)
c) \(\left(2u+3v\right)^2=4u^2+12uv+9v^2\)
d) \(\left(-\dfrac{1}{8}a+\dfrac{2}{3}bc\right)^2=\dfrac{1}{64}a^2-\dfrac{1}{6}abc+\dfrac{4}{9}b^2c^2\)
e) \(\left(\dfrac{x}{y}-\dfrac{1}{z}\right)^2=\dfrac{x^2}{y^2}-\dfrac{2x}{yz}+\dfrac{1}{z^2}\)
f) \(\left(\dfrac{mn}{4}-\dfrac{x}{6}\right)\left(\dfrac{mn}{4}+\dfrac{x}{6}\right)=\dfrac{m^2n^2}{16}-\dfrac{x^2}{36}\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
Bài 1:
$M=(2a+b)^2-(b-2a)^2=[(2a+b)-(b-2a)][(2a+b)+(b-2a)]$
$=4a.2b=8ab$
$N=(3a+1)^2+2a(1-2b)+(2b-1)^2$
$=(9a^2+6a+1)+2a-4ab+(4b^2-4b+1)$
$=9a^2+8a+4b^2-4b-4ab+2$
$A=(m-n)^2+4mn=m^2-2mn+n^2+4mn$
$=m^2+2mn+n^2=(m+n)^2$
16 tháng 7 2020
Bài 1:
a) Ta có: \(\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=4x^2-4x+1+4\left(x^2+2x-3\right)-2\left(25-30x+9x^2\right)\)
\(=4x^2-4x+1+4x^2+8x-12-50+60x-18x^2\)
\(=-10x^2+64x-61\)
b) Ta có: \(\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a^2+1\right)^2\)
\(=\left(2a^2+1\right)^2-\left(2a\right)^2-\left(2a^2+1\right)^2\)
\(=-4a^2\)
c) Ta có: \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)
\(=\left(9x-1+1-5x\right)^2\)
\(=\left(4x\right)^2=16x^2\)
d)
Sửa đề: \(\left(x^2+5x-1\right)^2+2\left(5x-1\right)\left(x^2+5x-1\right)+\left(5x-1\right)^2\)
Ta có: \(\left(x^2+5x-1\right)^2+2\left(5x-1\right)\left(x^2+5x-1\right)+\left(5x-1\right)^2\)
\(=\left(x^2+5x-1+5x-1\right)^2\)
\(=\left(x^2+10x-2\right)^2\)
\(=x^4+100x^2+4+20x^3-40x-4x^2\)
\(=x^4+20x^3+96x^2-40x+4\)
e) Ta có: \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=x\left(x^2-1\right)-\left(x^3+1\right)\)
\(=x^3-x-x^3-1\)
=-x-1
f) Ta có: \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^3-16x-x^4+1\)