Sửa lại đề bài là giải PT và biện luận nhé các bạn

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.....

\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)

=>\(\dfrac{sin^2a+1}{cos^2a}+\dfrac{cos^2a+1}{sin^2a}=7\)

=>\(\dfrac{sin^4a+sin^2a+cos^4a+cos^2a}{sin^2a\cdot cos^2a}=7\)

=>\(sin^4a+cos^4a+1=7\cdot sin^2a\cdot cos^2a\)

=>\(\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a+1=7\cdot sin^2a\cdot cos^2a\)

=>\(2=9\cdot sin^2a\cdot cos^2a\)

=>\(8=9\cdot sin^22a\)

=>16=9(1-cos4a)

=>1-cos4a=16/9

=>cos4a=-7/9

Đặt 2a/5b=5b/6c=6c/7d=7d/2a=k

=> k^4=2a/5b.5b/6c.6c/7d.7d/2a=1

=>k=1 hoặc k=-1

Với k=1 thì B=4

Với k=-1 thì B=-4

Vậy B=4 hoặc B=-4

Hả chỉ cho bài này r mak

\(A=a+\dfrac{\left(2a+x\right)\left(2+x\right)-\left(2a-x\right)\left(2-x\right)-4a}{\left(2-x\right)\left(2+x\right)}\)

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

\(=a+\dfrac{4ax+4x-4a}{\left(2-x\right)\left(2+x\right)}\)

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

\(=\dfrac{4a-ax^2+4ax+4x-4a}{4-x^2}\)

\(=\dfrac{-ax^2+4ax+4x}{4-x^2}\)

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

\(=\left[\dfrac{a}{a+1}\cdot\left(-a\cdot\dfrac{a}{a+1}+4a+4\right)\right]:\left[4-\dfrac{a^2}{\left(a+1\right)^2}\right]\)

\(=\left(\dfrac{a}{a+1}\cdot\left(\dfrac{-a^2+4a^2+8a+4}{a+1}\right)\right):\left[\dfrac{4\left(a+1\right)^2-a^2}{\left(a+1\right)^2}\right]\)

\(=\dfrac{a\left(3a^2+8a+4\right)}{4\left(a+1\right)^2-a^2}=\dfrac{a\left(3a^2+6a+2a+4\right)}{\left(2a+2\right)^2-a^2}\)

\(=\dfrac{a\left(a+2\right)\left(3a+2\right)}{\left(a+2\right)\left(3a+2\right)}=a\)

a là nghiệm nên \(\sqrt{2}a^2+a-1=0\Rightarrow\sqrt{2}a^2=1-a\)

\(\Rightarrow2a^4=\left(1-a\right)^2=a^2-2a+1\)

\(\Rightarrow2a^4-2a+3=a^2-4a+4=\left(a-2\right)^2\)

Mặt khác \(1-a=\sqrt{2}a^2>0\Rightarrow a< 1\)

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

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

\(\Rightarrow C=\dfrac{2a-3}{\sqrt{2}\left(3-2a\right)}=-\dfrac{\sqrt{2}}{2}\)

Tham khảo: 

\(x=\dfrac{1}{a}.\sqrt{\dfrac{2a}{b}-1}\Rightarrow ax=\sqrt{\dfrac{2a}{b}-1}\)

\(\Rightarrow\left\{{}\begin{matrix}1+ax=\dfrac{\sqrt{2a-b}+\sqrt{b}}{\sqrt{b}}\\1-ax=\dfrac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1-ax}{1+ax}=\dfrac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}+\sqrt{2a-b}}=\dfrac{\left(\sqrt{b}-\sqrt{2a-b}\right)^2}{2\left(b-a\right)}\)

Lại có:

\(\dfrac{1+bx}{1-bx}=\dfrac{a+\sqrt{2ab-b^2}}{a-\sqrt{2ab-b^2}}=\dfrac{a^2-\left(2ab-b^2\right)}{\left(a-\sqrt{2ab-b^2}\right)^2}=\dfrac{\left(a-b\right)^2}{\left(a-\sqrt{2ab-b^2}\right)^2}\)

\(\Rightarrow\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{b-a}{a-\sqrt{2ab-b^2}}\)

\(\Rightarrow A=\dfrac{1-ax}{1+ax}.\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{\left(\sqrt{b}-\sqrt{2a-b}\right)^2}{2a-2\sqrt{2ab-b^2}}=\dfrac{2a-2\sqrt{2ab-b^2}}{2a-2\sqrt{2ab-b^2}}=1\)