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23: \(=\left(2a-b\right)^2-\left(2a-2b\right)^2\)
\(=\left(2a-b-2a+2b\right)\left(2a-b+2a-2b\right)\)
\(=b\left(4a-3b\right)\)
24: \(=\left(3a+3b\right)^2-\left(2a-4b\right)^2\)
\(=\left(3a+3b-2a+4b\right)\left(3a+3b+2a-4b\right)\)
\(=\left(a+7b\right)\left(5a-b\right)\)
25: \(=\left(4a-2b\right)^2-\left(4a-4b\right)^2\)
\(=\left(4a-2b-4a+4b\right)\left(4a-2b+4a-4b\right)\)
\(=2b\left(8a-6b\right)\)
=4b(4a-3b)
a) Ta có: \(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+105+15\)
\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+120\)
\(=\left(x^2+8x\right)^2+12\left(x^2+8x\right)+10\left(x^2+8x\right)+120\)
\(=\left(x^2+8x\right)\left(x^2+8x+12\right)+10\left(x^2+8x+12\right)\)
\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)
b) Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-2-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x\right)^2+3\left(12x^2+11x\right)-2\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x\right)\left(12x^2+11x+3\right)-2\left(12x^2+11x+3\right)\)
\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
c) Ta có: \(\left(x^2+2x\right)^2+9x^2+18x+20\)
\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)
\(=\left(x^2+2x\right)^2+5\left(x^2+2x\right)+4\left(x^2+2x\right)+20\)
\(=\left(x^2+2x\right)\left(x^2+2x+5\right)+4\left(x^2+2x+5\right)\)
\(=\left(x^2+2x+5\right)\left(x^2+2x+4\right)\)
a: \(=12x^{n+2}+4x^2-8x^{n+2}\)
\(=4x^{n+2}+4x^2\)
b: \(=2x^{2n}+4x^ny^n+2y^{2n}-4x^ny^n-2y^{2n}\)
\(=2x^{2n}\)
c: \(=\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)
\(=x^{6n}-y^{6n}\)
d: \(=4^n\cdot4-3\cdot4^n=4^n\)
= 28x2n-1 + 4xn+1 - 20xn - 28x2n-1 + 20x2 = 4xn+1
đúng 100%
Ta có:
\(N=\left(2x+1\right)^2-2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(N=\left(2x+1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x-1\right)^2\)
\(N=\left(2x+1-2x+1\right)^2=2^2=4\)
Vậy với mọi giá trị của x thì biểu thức \(N=4\)
Chúc bạn học tốt!!!
a: \(\left(\dfrac{1}{\left(2x-y\right)^2}+\dfrac{2}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{1}{\left(2x+y\right)^2}\right)\cdot\dfrac{\left(2x+y\right)^2}{16x}\)
\(=\dfrac{4x^2+4xy+y^2+2\left(4x^2-y^2\right)+4x^2-4xy+y^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}\cdot\dfrac{\left(2x+y\right)^2}{16x}\)
\(=\dfrac{8x^2+2y^2+8x^2-2y^2}{\left(2x-y\right)^2}\cdot\dfrac{1}{16x}\)
\(=\dfrac{16x^2}{16x}\cdot\dfrac{1}{\left(2x-y\right)^2}=\dfrac{x}{\left(2x-y\right)^2}\)
b: \(\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)
\(=\dfrac{2x+4-4}{\left(x+2\right)^2}:\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\)
\(=\dfrac{2x}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{-x}=\dfrac{-2\left(x-2\right)}{x+2}\)
câu này post hồi học lớp 8 = )) giờ tốt nghiệp c3 thì có người trả lời :'))
khbiet nên cười hay khóc đây
4xn( 8xn-1 - 1 ) - 2xn+1( 16xn-2 - 1 )
= 32x2n-1 - 4xn - 32x2n-1 + 2xn+1
= 2xn+1 - 4xn
= 2xn( x - 2 )