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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
a: \(=\dfrac{4x^2+4x+1-4x^2+4x-1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x\cdot5}{4x\left(2x+1\right)}=\dfrac{10}{2x+1}\)
b: \(=\left(\dfrac{1}{x^2+1}+\dfrac{x-2}{x+1}\right):\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{x+1+x^3+x-2x^2-2}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x^3-2x^2+2x-1}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x\left(x^2-x+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{1}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}\)
\(=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
Mình giải từ cuối lên , mình giải dần -)
n, <=> x(2x-1)-3(2x-1)=0
<=> (x-3)(2x-1)=0
<=> x= 3 hoặc x= 1/2
m, <=> (x+2)(x2-3x+5)-x2(x+2)=0
<=> (x+2)(x2-3x+5-x2)=0
<=> (x+2)(5-3x)=0
=> x= -2 hoặc5/3