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: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

\(=\dfrac{\left(x-4\right)\cdot\left(x+4\right)}{x}\cdot\dfrac{x}{\left(x-4\right)^2}=\dfrac{x+4}{x-4}\)

https://www.youtube.com/watch?v=cFZDEMTQQCs

j thế bạn

a) \(x-2y-\sqrt{x^2-4xy+4y^2}\)

\(=x-2y-\sqrt{\left(x-2y\right)^2}\)

\(=x-2y-\left|x-2y\right|\)

TH1: \(x-2y--\left(x-2y\right)\)

\(=x-2y+x-2y\)

\(=2x-4y\)

TH2: \(x-2y-\left(x-2y\right)\)

\(=x-2y-x+2y\)

\(=0\)

b) \(x^2+\sqrt{x^4-8x^2+16}\)

\(=x^2+\sqrt{\left(x^2-4\right)^2}\)

\(=x^2+\left|x^2-4\right|\)

TH1: 

\(x^2+-\left(x^2-4\right)\)

\(=x^2-x^2+4\)

\(=4\)

TH2: 

\(x^2+\left(x^2-4\right)\)

\(=x^2+x^2-4\)

\(=2x^2-4\)

c) \(2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\) (x>5)

\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}\)

\(=2x-1-\sqrt{x-5}\)

d) \(\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}\) (\(x>\sqrt{2}\))

\(=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}\)

\(=\sqrt{x^2-2}\)

e) \(\sqrt{\left(x^2-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}\)

\(=\left|x^2-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}\)

\(=\left|x^2-4\right|+\sqrt{\dfrac{\left(x-4\right)^2}{\left(x-4\right)^2}}\)

\(=\left|x^2-4\right|+1\)

TH1: 

\(x^2-4+1\)

\(=x^2-3\)

TH2:

\(-\left(x^2-4\right)+1\)

\(=-x^2+4+1\)

\(=-x^2+5\)

a: \(A=x-2y-\sqrt{x^2-4xy+4y^2}\)

=x-2y-|x-2y|

Khi x>=2y thì A=x-2y-x+2y=0

Khi x<2y thì A=x-2y+x-2y=2x-4y

b: \(B=x^2+\sqrt{x^4-8x^2+16}\)

\(=x^2+\left|x^2-4\right|\)

TH1: x>=2 hoặc x<=-2

B=x^2+x^2-4=2x^2-4

TH2: -2<=x<=2

B=x^2+4-x^2=4

c: \(C=2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\)

\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}=2x-1-\sqrt{x-5}\)

d: \(D=\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}=\sqrt{x^2-2}\)

B

Chọn B

Giúp tui với mấy bạn ơi

Đặt A=3(22 +1)(24+1)(28+1)(216+1)

=(4-1)(2^2+1)(2^4+1)(28+1)(2^16+1)

=[(2^2-1)(2^2+1)](2^4+1)(2^8+1)(2^16+1)

=(2^4-1)(2^4+1)(2^8+1)(2^16+1)

=(2^8-1)(2^8+1)(2^16+1)

=(2^16-1)

Theo mình ý a bn làm đc

\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+16}=\dfrac{\left(x+16\right)\left(\sqrt{x}+2\right)}{\left(x-16\right)\left(\sqrt{x}+16\right)}\)