cái phần A là thiếu dấu cộng đấy ạ

Em thử nha!Sai thì thôi:((

\(A=\left|m+1\right|+\left|m-1\right|=\left|m+1\right|+\left|1-m\right|\ge\left|m+1+1-m\right|=2\)

Dấu"=" xảy ra khi \(\left(m+1\right)\left(1-m\right)\ge0\Leftrightarrow-m^2+1\Leftrightarrow-1\le m\le1\)

\(B=\sqrt{\left(2a\right)^2-2.2a.1+1}+\sqrt{4a^2-2.2a.3+9}\)

\(=\left|2a-1\right|+\left|2a-3\right|=\left|2a-1\right|+\left|3-2a\right|\ge2\)

Dấu "=" xảy ra khi...

a) x + \(\sqrt{\left(x-2^{ }\right)^2}\)= x +\(|x-2|\)= x +2-x (vì x<2)

b) \(\sqrt{\left(x-3\right)^2}\)-x = \(|x-3|-x=x-3-x\) (vì x>3)

c) m- \(\sqrt{m^2-2m+1}=m-\sqrt{\left(m-1\right)^2}\)

Những con còn lại bạn làm như trên và rút gọn đi là được

d: \(=x+y-\left|x-y\right|\)

=x+y-x+y=2y

e: \(=\left|5a-1\right|-4a=\left|5\cdot\dfrac{1}{2}-1\right|-2\)

\(=\dfrac{5}{2}-1-2=\dfrac{5}{2}-3=-\dfrac{1}{2}\)

f: \(=\left|2a-3\right|-4a-1\)

\(=\left|-10-3\right|-4\cdot\left(-5\right)-1=13+20-1=32\)

\(\sqrt{4a^2+12a+9}+\sqrt{4a^2-12a+9}\) với \(-\dfrac{3}{2}\le a\le\dfrac{3}{2}\)

\(\sqrt{\left(2a+3\right)^3}+\sqrt{\left(2a-3\right)^3}\)

\(\left|2a+3\right|+\left|2a-3\right|\)

\(2a+3-2a+3\)

\(6\)

\(a)\sqrt{-9a}-\sqrt{9+12a+4a^2}\)

\(==\sqrt{3^2.\left(-a\right)}-\sqrt{3^2-2.3.2a+\left(2a\right)^2}\)

\(=3\sqrt{-a}-\sqrt{\left(3+2a\right)^2}\)

\(=3\sqrt{a}-\left|3+2a\right|\)

\(b)1+\frac{3m}{m-2}\sqrt{m^2-4m+4}\)

\(=1+\frac{3m}{m-2}\sqrt{\left(m\right)^2-2.2m+2^2}\)

\(=1+\frac{3m}{m-2}\sqrt{\left(m-2\right)^2}\)

\(=1+\frac{3m}{m-2}|m-2|\)

\(c)4x-\sqrt{9x^2+6x+1}\)

\(=4x-\sqrt{\left(3x\right)^2+2.3x+1}\)

\(=4x-\sqrt{\left(3x+1\right)^2}\)

\(=4x-|3x+1|\)

biến b để làm gì thế bạn???

a/ \(\sqrt{4a^4-12a^2+9}-\sqrt{a^4-8a^2+16}\)

= \(\sqrt{\left(2a^2-3\right)^2}-\sqrt{\left(a^2-4\right)^2}\)

= \(|2a^2-3|-|a^2-4|\)

= \(2a^2-3+a^2-4\)
= \(3a^2-7\)

Thay a=\(\sqrt{3}\).Ta có:

\(3.\left(\sqrt{3}\right)^2-7\)

= 3.3-7=2

b/ \(\sqrt{10a^2-12a\sqrt{10}+36}\)

= \(\sqrt{\left(a\sqrt{10}\right)^2-2.a\sqrt{10}.6+6^2}\)

= \(\sqrt{\left(a\sqrt{10}-6\right)^2}\)

= \(|a\sqrt{10}-6|\)

=  \(-a\sqrt{10}+6\)

Thay  a= \(\sqrt{\frac{5}{2}}-\sqrt{\frac{2}{5}}\)=\(\frac{3}{\sqrt{10}}\),Ta có:

\(-\frac{3}{\sqrt{10}}.\sqrt{10}+6\)

= -3+6 =3

a: \(A=\dfrac{1}{2a-1}\cdot\sqrt{5a^2}\cdot\left|2a-1\right|\)

\(=\dfrac{2a-1}{2a-1}\cdot a\sqrt{5}=a\sqrt{5}\)(do a>1/2)

b: \(A=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x-1}+1}\)

\(=\dfrac{\left|\sqrt{x-1}-1\right|}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1}+1}{\sqrt{x-1}+1}\)

\(=\dfrac{\sqrt{x-1}-1}{\sqrt{x-1}-1}+1=1+1=2\)

c:

\(=\dfrac{a+b}{b^2}\cdot\dfrac{ab^2}{a+b}=a\)

d: Sửa đề: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\left(\dfrac{1}{1+\sqrt{a}}\right)^2\)

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

e:

\(A=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^4}}\)

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

f:

\(A=\sqrt{\dfrac{m}{\left(1-x\right)^2}\cdot\dfrac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\sqrt{\dfrac{m}{\left(x-1\right)^2}\cdot\dfrac{4m\left(x-1\right)^2}{81}}\)

\(=\sqrt{\dfrac{4m^2}{81}}=\dfrac{2m}{9}\)