Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)
\(b.\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =x^3+1-\left(x^3-1\right)\\ =x^3+1-x^3+1=2\)
\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)
\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)
\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)
\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)
\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)
\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
a) x2 - 2x - 4y2 - 4y
= (x2 - 4y2) - (2x + 4y)
= (x + 2y)(x - 2y) - 2(x + 2y)
= (x + 2y)(x - 2y - 2)
= (x + 2y)[x - 2(y + 1)]
b) x4 + 2x3 - 4x - 4
= (x4 - 4) + ( 2x3 - 4x)
= (x2 - 2)(x2 + 2) + 2x(x2 - 2)
= (x2 - 2)(x2 + 2 + 2x)
c) x3 + 2x2y - x -2y
= (x3 - x) + (2x2y - 2y)
= x(x2 - 1) + 2y(x2 - 1)
= (x + 2y)(x2 - 1)
a)(3x-1)2+2(3x-1)(2x+1)2(2x+1)=48x^4+56x^3+21x^2-12x-1 cái này tra google
b)(x2+1)(x-3)-(x-3)(x2+3x+9)=(x2+1)(x-3)-(x-3)(x+3)2=(x-3)[(x2+1)-(x+3)2 ]
c)(2x+3)2+(2x+5)2-2(2x+3)(2x+5)=(2x+3)2+(2x+5)2-(2x+3)(2x+5)-(2x+3)(2x+5)=(2x+3)(2x+3-2x+5)+(2x+5)(2x+5-2x+3)
=8(2x+3)+8(2x+5)=8(2x+3+2x+5)
=8(4x+8)
d)(x-3)(x+3)-(x-3)2 =(x-3)(x+3)-(x-3)(x-3)=(x-3)(x+3-x-3)=0
e)(2x+1)2+2(4x2-1)+(2x-1)2 =(2x+1)2+2[(2x)2 -1]+(2x-1)2 =(2x+1)(2x+1+2x-1)+(2x-1)(2x+1+2x-1)=4x(2x+1)+4x(2x-1)
=4x(2x+1+2x-1)=16x2
f)(x2-1)(x+2)-(x-2)(x2+2x+4)= (x2-1)(x+2)-(x-2)(x+2)2 =(x2-1)(x+2)-(x2-22)(x+2)=(x+2)(x2-1-x2-22) mình đoán câu f khai triển ra thế này nhưng kq không giống nhau nên chắc bạn phải tự làm rồi
1,(2x + 3 ) \(^{^{ }2}\)=\(\left(2x\right)^2+2.2x.3+3^2\)
=\(4x^2+12x+9\)
2, ( 3x + 2y )\(^2=\left(3x\right)^2+2.3x.2y+\left(2y\right)^2\)
=\(9x^2+12xy+4y^2\)
3,(3a -1 )\(^2=\left(3a\right)^2-2.3a.1+1^2\)
\(=9a^2-6a+1\)
4, (a - 2 )\(^2=a^2-2.a.2+2^2\)
=\(a^2-4a+4\)
5, ( 1 - 5a )\(^2=1^2-2.1.5a+\left(5a\right)^2\)
=\(1-10a+25a\)
6, ( x - 4 )\(^3=x^3-3x^24+3x4^2-4^3\)
=\(x^3-12x^2+48x-64\)
Sửa lại phần 2 nha
`(2x+1/2)^3= (2x)^3 + 3. (2x)^2 . 1/2 + 3.2x.(1/2)^2 + (1/2)^3 = 8x^3 + 6x^2 + 3/2 x + 1/8`
x2 - y2 + 4y Simplify ———————————— y2 + ysqt4
STEP2:Pulling out terms
2.1 Pull out factors :
y2 + ysqt4 = y • (y + sqt4)
Trying to factor a multi variable polynomial :
2.2 Factoring x2 - y2 + 4y
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Equation at the end of step2:
((((x3)-(3•(x2)))+2x)+1) x (x2-y2+4y) (((————————————————————————•d)•x)•((d•—)•y))+—————————— (((x2)-4x)+4) d y•(y+sqt4)
STEP3:
x Simplify — d
Equation at the end of step3:
((((x3)-(3•(x2)))+2x)+1) x (x2-y2+4y) (((————————————————————————•d)•x)•((d•—)•y))+—————————— (((x2)-4x)+4) d y•(y+sqt4)
STEP 4 :
Equation at the end of step4:
((((x3)-3x2)+2x)+1) (x2-y2+4y) (((———————————————————•d)•x)•xy)+—————————— (x2-4x+4) y•(y+sqt4)
STEP5:
x3 - 3x2 + 2x + 1 Simplify ————————————————— x2 - 4x + 4
Checking for a perfect cube :
5.1 x3 - 3x2 + 2x + 1 is not a perfect cube
Trying to factor by pulling out :
5.2 Factoring: x3 - 3x2 + 2x + 1
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 2x + 1
Group 2: x3 - 3x2
Pull out from each group separately :
Group 1: (2x + 1) • (1)
Group 2: (x - 3) • (x2)