Với \(2x-1\ge0\Leftrightarrow x\ge\frac{1}{2}\)

\(\Rightarrow D=2\left(2x-1\right)-3\left(2x+3\right)\)

\(\Rightarrow D=4x-2-6x-9\)

\(\Rightarrow D=-2x-11\)

Với \(2x+3< 0\Rightarrow x< -\frac{3}{2}\)

\(\Rightarrow D=2\left(1-2x\right)-3\left(2x+3\right)\)

\(\Rightarrow D=2-4x-6x-9\)

\(\Rightarrow D=-10x-7\)

Thanks nha!

(2x - 1)(2x + 1) + 4x(1 - x)

= 4x² - 1 + 4x - 4x²

= 4x - 1

a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012

2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013

3M=2^0+2^2013

M=(2^0+2^2013)÷3

Vậy.......

b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012

3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013

4N=3-3^2013

N=(3-3^2013)÷4

Vậy........

K tao nhé ko lên lớp tao đánh m😈😈😈

Bt dễ thế mà ko làm dc😂😂😂😂😂

phần c đâu

a) \(=x^2+2x-2x=x^2\)

b) \(=4-x^2+x^2=4\)

c) \(=x^2-x^3+x^3+27=x^2+27\)

d) \(=4x^2+4xy+y^2+4x^2-8x^2-4xy=y^2\)

\(ĐKXĐ:x\ge4\)

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

\(=\sqrt{\left(\sqrt{x-4}\right)^2+2.2\sqrt{x-4}+2^2}-\sqrt{x-4}\)

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

\(=\sqrt{x-4}+2-\sqrt{x-4}=2\)( vì \(x\ge4\)nên \(\sqrt{x-4}\ge0\))

\(\dfrac{2sin8a-sin16a}{2sin8a+sin16a}=\dfrac{2sin8a-2sin8a.cos8a}{2sin8a+2sin8a.cos8a}=\dfrac{2sin8a\left(1-cos8a\right)}{2sin8a\left(1+cos8a\right)}=\dfrac{1-cos8a}{1+cos8a}=\dfrac{1-\left(1-2sin^24a\right)}{1+\left(1-2sin^24a\right)}=\dfrac{2sin^24a}{2-2sin^24a}=\dfrac{sin^24a}{1-sin^24a}=\dfrac{sin^24a}{cot^24a}=tan^24a\)

\(=\dfrac{2sin8a-2sin8a.cos8a}{2sin8a+2sin8a.cos8a}=\dfrac{2sin8a\left(1-cos8a\right)}{2sin8a\left(1+cos8a\right)}=\dfrac{1-cos8a}{1+cos8a}\)

\(=\dfrac{1-\left(1-2sin^24a\right)}{1+\left(2cos^24a-1\right)}=\dfrac{2sin^24a}{2cos^24a}=tan^24a\)

\(B=9x^4-\left(2x+1\right)^2-\left(9x^4+6x^2+1\right)\\ =9x^4-4x^2-4x-1-9x^4-6x^2-1\\ =-10x^2-4x-2\)

\(B=\left(3x^2+1-2x\right)\left(3x^2+1+2x\right)-\left(3x^2+1\right)^2\\ B=\left(3x^2+1\right)^2-4x^2-\left(3x^2+1\right)^2=-4x^2\)

a: Ta có: \(M=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

\(=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)

\(=a+\sqrt{a}-2\sqrt{a}-1+1\)

\(=a-\sqrt{a}\)

c: \(P=4\left(x-3\right)-3\left|x+3\right|\)

Trường hợp 1: x>=-3

\(P=4x-12-3x-9=x-21\)

Trường hợp 2: x<-3

P=4x-12+3x+9=7x-3