\(M=a+\dfrac{4a+2ab+2b+b^2+4a-2ab-2b+b^2-4a}{\left(2-b\right)\left(2+b\right)}\\ M=a+\dfrac{4a+2b^2}{\left(2-b\right)\left(2+b\right)}=\dfrac{4a-ab^2+4a+2b^2}{\left(2-b\right)\left(2+b\right)}\\ M=\dfrac{8a-ab^2+2b^2}{4-b^2}\)

Ta có \(8a-b^2\left(a-2\right)=8a-\dfrac{a^2\left(a-2\right)}{\left(a+1\right)^2}=\dfrac{8a^3+16a^2+8a-a^3+2a^2}{\left(a+1\right)^2}=\dfrac{7a^3+18a^2+8a}{\left(a+1\right)^2}\)

\(4-b^2=4-\dfrac{a^2}{\left(a+1\right)^2}=\dfrac{4a^2+8a+4-a^2}{\left(a+1\right)^2}=\dfrac{3a^2+8a+4}{\left(a+1\right)^2}\)

\(\Leftrightarrow M=\dfrac{7a^3+18a^2+8a}{3a^2+8a+4}=\dfrac{a\left(7a+4\right)\left(a+2\right)}{\left(3a+2\right)\left(a+2\right)}=\dfrac{a\left(7a+4\right)}{3a+2}\)

Với a > 0 và a ≠ 4 , ta có

T =   a −   2 a + 2   −   a +   2 a − 2   .   a   −   4 a   =    a −   2 2 −   a +   2 2   a −   2 . a + 2 .     a   − 4 a   =    a −   4 a   +   4 −   a −   4 a   −   4   a   − 4 .     a   − 4 a   =     − 8 a a   =   − 8

giúp mình giải bài này vs

a) A=(4-5x)²-(3+5x)²=(4-5x-3-5x)(4-5x+3+5x)=(-25x+1)1=-25x+1

B=(3x-1)(1+3x)-(3x+1)²=9x²-1-(3x+1)²=9x²-1-(9x²+6x+1)=9x²-1-9x²-6x-1=-6x-2=-2(3x+1)

ădadawdw

Ta có

B   =   2 a − 3 a + 1 − a − 4 2 − a a + 7   =   2 a 2   +   2 a   –   3 a   –   3   –   ( a 2   –   8 a   +   16 )   –   ( a 2   +   7 a )     =   2 a 2   +   2 a   –   3 a   –   3   –   a 2   +   8 a   –   16   –   a 2   –   7 a     =   -   19

Đáp án cần chọn là: D

a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)

\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)

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

=-a-1

b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)

\(=\left|3a-5\right|-2a+4\)

\(=5-3a-2a+4\)

=9-5a

c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)

\(=4a-3-\left|2a-1\right|\)

\(=4a-3-2a+1\)

\(=2a-2\)

d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)

\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)

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

\(=-a^2\)