Rút gọn a)(2x-1)² - (4x+1)(x-2)
b)(x-3)³ - (x+1)(x²-x+1)
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16 tháng 8 2023
`@` `\text {Ans}`
`\downarrow`
`A= (2x - 3)^2 - (2x + 3)^2`
`= [(2x - 3) - (2x + 3)]*[(2x - 3) + (2x + 3)]`
`= (2x - 3 - 2x - 3) * (2x - 3 + 2x + 3)`
`= -6 * 4x`
`= -24x`
CT
1
HT
24 tháng 7 2016
\(\left(+\right)A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right).\)
\(A=8x^3-6x^2-18x+27-8x^3+2\)
\(A=6x^2-18x+29\)
\(\left(+\right)B=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x-1\right)\left(x+1\right)\)
\(B=x^3-3x^2+3x+1-x^3-3x^2-3x-1+6\left(x^2-1\right)\)
\(B=-6x^2+6x^2-6\)
\(B=-6\)
1 tháng 8 2021
A = \(\left(3x-1\right)^2+2\left(3x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
A = \(\left(3x-1+2x+1\right)^2\)
HN
1 tháng 8 2021
A)
<=>(3x)^2−2×3x+1+2(3x−1)(2x+1)+(2x+1)^2
<=>(3x)^2−2×3x+1+(6x−2)(2x+1)+(2x+1)^2
<=>(3x)^2−2×3x+1+12x^2+6x−4x−2+(2x+1)^2
<=>(3x)^2−2×3x+1+12x^2+6x−4x−2+(2x)^2+2×2x+1
<=>32x^2−2×3x+1+12x^2+6x−4x−2+(2x)^2+2×2x+1
<=>9x^2−2×3x+1+12x^2+6x−4x−2+(2x)^2+2×2x+1
<=>9x^2−2×3x+1+12x^2+6x−4x−2+2^2x^2+2×2x+1
<=>9x^2−2×3x+1+12x^2+6x−4x−2+4x^2+2×2x+1
<=>9x^2−6x+1+12x^2+6x−4x−2+4x^2+2×2x+1
<=>9x^2−6x+1+12x^2+6x−4x−2+4x^2+4x+1
<=>(9x^2+12x^2+4x^2)+(−6x+6x−4x+4x)+(1−2+1)
<=> 25x^2
B)
<=>2x(4x^2−6x+9)+3(4x^2−6x+9)+8(1−x)(1+x+x^2)
<=>8x^3−12x^2+18x+3(4x^2−6x+9)+8(1−x)(1+x+x^2)
<=>8x^3−12x^2+18x+12x^2−18x+27+8(1−x)(1+x+x^2)
<=>8x^3−12x^2+18x+12x^2−18x+27+(8−8x)(1+x+x^2)
<=>8x^3−12x^2+18x+12x^2−18x+27+8(1+x+x^2)−8x(1+x+x^2)
<=>8x^3−12x^2+18x+12x^2−18x+27+8+8x+8x^2−8x(1+x+x^2)
<=>8x^3−12x^2+18x+12x^2−18x+27+8+8x+8x^2−(8x+8x2+8x^3)
<=>8x^3−12x^2+18x+12x^2−18x+27+8+8x+8x^2−8x−8x^2−8x^3
<=>(8x^3−8x^3)+(−12x^2+12x^2+8x^2−8x^2)+(18x−18x+8x−8x)+(27+8)
<=> 35
DT
2
18 tháng 4 2020
Dùng hằng đẳng thức nhé !! :))) ..or nhân đa thức với đa thức nếu chưa học đến hằng đẳng thức -.- rồi cộng trừ với nhau .
VD : a) ( x+2).(x-2) - ( x-3 ).(x - 1 )
= x2 - 2x + 2x -4 - x2 + x + 3x - 1 ( đổi dấu vì trc là dấu trừ nhé )
= 4x - 5
ý b) tự làm nhé :3
DA
18 tháng 4 2020
b) -4x(2x+1)+(2x+1)^2+4x^2
=(2x)^2-2.2x(2x+1)+(2x+1)^2
=(2x-2x-1)^2
=(-1)^2
=1
T
24 tháng 10 2023
Bài 1.
a)
\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)
b)
\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)
Bài 2.
a)
\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)
b)
\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)
26 tháng 7 2023
a: A=x+3+|x-3|
=x+3+3-x(x<=3)
=6
b:\(B=\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
=x+2-x=2
c: \(C=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)
\(=\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)
22 tháng 10 2023
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
\(\left(2x-1\right)^2-\left(4x+1\right)\left(x-2\right)\\ =\left[\left(2x\right)^2-2\cdot2x\cdot1+1^2\right]-\left(4x^2-8x+x-2\right)\\ =4x^2-4x+1-4x^2+8x-x+2\\ =3x+3\)
\(\left(x-3\right)^3-\left(x+1\right)\left(x^2-x+1\right)\\ =\left(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\right)-\left(x^3-x^2+x+x^2-x+1\right)\\ =x^3-9x^2+27x-27-x^3+x^2-x-x^2+x-1\\ =-10x^2+27x-28\)