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

\(=\dfrac{\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}+1}{a-b}=\dfrac{1}{a-b}\)

b: Khi a-b=1 thì A=1/1=1

a.

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

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

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

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

b.

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

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

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

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

=>\(A=\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right).\left(\dfrac{\sqrt{a}+1+\sqrt{a}-1}{a-1}\right)\left(a>0,a\ne1\right)\)

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

b, \(a=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\) thế vào A

\(=>A=\dfrac{2}{\sqrt{\left(\sqrt{2}-1\right) ^2}+1}=\dfrac{2}{\sqrt{2}}\)

tự làm nhé,dễ lắm

bài này khó đấy

Bài này dễ mà

\(=>A=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left[\dfrac{\sqrt{x}+1-2}{x-1}\right]\)

\(=>A=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}\)

b,\(=>\dfrac{1}{\sqrt{x}}=\dfrac{1}{2}=>\sqrt{x}=2=>x=\sqrt{2}\left(tm\right)\)

a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

Vậy...

b)Đk: \(x\ge-1\)

Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)

\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)

Vậy...

\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)

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

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

b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\) 

Vậy \(A_{min}=-\dfrac{1}{4}\)

a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)

\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)

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

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

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

b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)

✱ giải pt:

a.\(\sqrt{x^2-4x+4}\)\(=5\)

⇔\(\sqrt{\left(x-2\right)^2}=5\)

⇒\(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

vậy....

b.\(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

⇔ \(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

⇔ \(4\sqrt{x+1}=16\)

⇔ \(\sqrt{x+1}=16\)

⇒ \(x+1=256\)

⇔ \(x=255\)

vậy.....

a, \(A=\left(-a-b+c\right)-\left(-a-b-c\right)\)

\(=-a-b+c+a+b+c=2c\)

b, Thay c = -2 vào A ta được : \(2.\left(-2\right)=-4\)

a,A=-a-b+c+a+b+c=-a+(-b)+c+a+b+c=a+c+a+c=2x(a+c)

b,  A=2x[1+(-2)]=2x(-1)=-2

a)

A = ( -a - b + c ) - ( -a - b - c )

   = -a + b - c + a + b + c 

   = ( -a + a ) + ( -c + c ) + ( b + b ) 

   =   0 + 0 + 2b

   = 2b

b) Thay a = 1 ; b = -1 ; c = -2

A = [ -1 + ( -1 ) - ( - 2 ) ] - [ -1 - ( -1 ) - ( -2 ) ]

   = [ -1 - 1 + 2 ] - [ -1 + 1 + 2 ]

   = 0 - 2

   = -2 

a.A=-a -b -c+a+b+c =0

b/A=0 SUY RA a,b,c =0

a) A= ( -a -b -c) - ( -a -b -c)

( Bỏ dấu ngoặc, đổi dấu vế sau vì trước nó là dấu trừ )

      = -a -b -c +a +b +c 

( các số đối nhau trừ hết )

     = 0

a)A=(-a-b+c)-(-a-b-c)

   A=-a-b+c-a+b+c

   A=(-a-a)-(b+b)+(c+c)

   A=2.(-a)-2b-2c

   A=2(-a-b-c)

b)Thay a=1;b=-1;c=-2 vao bieu thuc

Ta co: A=[(-1)-(-1)+(-2)]-[(-1)-(-1)-(-2)]

          A=[-1+1+(-2)]-[-1+1+2]

          A=-2-2

          A=-4

a,A=(-a-b+c)-(-a-b-c)=-a-b+c+a+b+c=c+c=2c

b,theo câu a A=2c ma c=-2 =>A=2x(-2)=-4

a)A=(-a-b+c)-(-a-b-c)

=-a-b+c+a+b+c

=(-a+a)+(-b+b)+(c+c)

=2c

b)thay a=1;b=-1;c=-2 vào biểu thức A

ta có:2x(-2)

    =-4

vậy giá trị của biểu thức A là -4