a: Khi x=25 thì \(A=\dfrac{5-2}{5-3}=\dfrac{3}{2}\)

b: P=A*B

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\left(\dfrac{6x+6\sqrt{x}-12}{x+5\sqrt{x}+4}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\left(\dfrac{6x+6\sqrt{x}-12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\dfrac{6x+6\sqrt{x}-12-5x-5\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

c: \(\sqrt{P}< =\dfrac{1}{2}\)

=>0<=P<=1/4

=>\(\left\{{}\begin{matrix}P>=0\\P-\dfrac{1}{4}< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{x}-2}{\sqrt{x}-1}>=0\\\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{1}{4}< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{4\left(\sqrt{x}-2\right)-\sqrt{x}+1}{4\left(\sqrt{x}-1\right)}< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{3\sqrt{x}-7}{\sqrt{x}-1}< =0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< \sqrt{x}< =\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< x< \dfrac{49}{9}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\x=\dfrac{49}{9}\end{matrix}\right.\)

=>\(4< =x< =\dfrac{49}{9}\)

mà x nguyên

nên \(x\in\left\{4;5\right\}\)

a) giải pt ra ta được  : x=-1

b) giải pt ra ta được  : x=2

c)giải pt ra ta được  : x vô ngiệm

d)giải pt ra ta được  : x=vô ngiệm

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

\(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)

\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}\)

\(=\left|3x-1\right|+\left|5-3x\right|\)

\(\ge\left|3x-1+5-3x\right|=4\)

\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.5\sqrt{7}+7}+\sqrt{25-2.5\sqrt{7}+7}=5+\sqrt{7}+5-\sqrt{7}=10\)

\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{25+2.5.3\sqrt{2}+18}=5+3\sqrt{2}\) \(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{3-\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}=\dfrac{1}{3+\sqrt{x}}\)

\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)

\(e.\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-1}=\sqrt{x}-2\)

\(f.\dfrac{x\sqrt{x}+64}{\sqrt{x}+4}=\dfrac{\left(\sqrt{x}+4\right)\left(x-4\sqrt{x}+16\right)}{\sqrt{x}+4}=x-4\sqrt{x}+16\)

\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)

Còn 2 con cuối làm tương tự nhé ( đăng dài quá ).

\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.\sqrt{25}.\sqrt{7}+7}+\sqrt{25-2.\sqrt{25}.\sqrt{7}+7}=\sqrt{\left(5+\sqrt{7}\right)^2}+\sqrt{\left(5-\sqrt{7}\right)^2}=5+\sqrt{7}+5-\sqrt{7}=10\)\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.\sqrt{8}.1}+1}}=\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}=\sqrt{13+30\sqrt{3+2\sqrt{2}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}}=\sqrt{13+30\sqrt{2}+30}=\sqrt{\sqrt{25}+2.\sqrt{25}.\sqrt{18}+18}=\sqrt{\left(5+\sqrt{18}\right)^2}=5+\sqrt{18}\)

\(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{9-x}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{1}{3+\sqrt{x}}=\dfrac{3-\sqrt{x}}{9-x}\)\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{x-2\sqrt{x}-3\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)}=\sqrt{x}-2\)\(e.\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-1}=\sqrt{x}-2\)

\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(x\sqrt{x}-y\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{x^2+x\sqrt{xy}-y\sqrt{xy}-y^2}{x-y}=\dfrac{\sqrt{xy}\left(x-y\right)+\left(x-y\right)\left(x+y\right)}{x-y}=\dfrac{\left(x-y\right)\left(\sqrt{xy}+x+y\right)}{x-y}=x+y+\sqrt{xy}\)\(h.6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(x-3\right)^2}=6-2x-\left|x-3\right|=6-2x-3+x=3-x\)

\(i.\sqrt{x+2+2\sqrt{x+1}}=\sqrt{x+1+2\sqrt{x+1}+1}=\sqrt{\left(\sqrt{x+1}+1\right)^2}=\sqrt{x+1}+1\)

a) \(\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}\)

\(=\sqrt{\left(5+\sqrt{7}\right)^2}+\sqrt{\left(5-\sqrt{7}\right)^2}\)

\(=5+\sqrt{7}+5-\sqrt{7}\)

=10

b)\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{\left(1+2\sqrt{2}\right)^2}}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{13+30\sqrt{2}+30}\)

\(=\sqrt{43+30\sqrt{2}}\)

\(=\sqrt{\left(5+3\sqrt{2}\right)^2}\)

\(=5+3\sqrt{2}\)

Ta có :

\(\sqrt{9x^2-6x+2}=\sqrt{\left(9x^2-6x+1\right)+1}=\sqrt{\left(3x-1\right)^2+1}\ge\sqrt{1}=1\)

\(\sqrt{45x^2-30x+9}=\sqrt{5\left(9x^2-6x+1\right)+4}=\sqrt{5\left(3x-1\right)^2+4}\ge\sqrt{4}=2\)

\(\sqrt{6x-9x^2+8}=\sqrt{-\left(9x^2-6x+1\right)+9}=\sqrt{-\left(3x-1\right)^2+9}\le3\)

\(\Rightarrow VT\ge3\ge VP\)

mÀ đề lại cho \(VT=VP\) \(\Rightarrow\hept{\begin{cases}\sqrt{\left(3x-1\right)^2+1}=1\\\sqrt{\left(3x-1\right)^2+4}=2\\\sqrt{-\left(3x-1\right)^2+9}=3\end{cases}\Rightarrow x=\frac{1}{3}}\)

Vậy \(x=\frac{1}{3}\)

x=1/3 nha

X=1/3 đấy !!!!

Đăng 1 lúc mà nhiều thế. Lần sau đăng 1 câu thôi b.

b/ \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)

Ta có: \(VT\ge1+2+\sqrt{5}=3+\sqrt{5}\)

Dấu = xảy ra khi \(x=2\)

c/ \(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x-8}=\sqrt{3-\left(x-1\right)^2}+\sqrt{1-\left(x+3\right)^2}\)

\(\le1+\sqrt{3}\)

Dấu = không xảy ra nên pt vô nghiệm

Câu d làm tương tự

\(a,\sqrt{x^2-4}-x^2+4=0\) 

\(\Leftrightarrow\sqrt{x^2-4}=x^2-4\) 

\(\Leftrightarrow x^2-4=\left(x-4\right)^2\) 

\(\Leftrightarrow x^2-4-x^4+8x^2-16=0\)  

\(\Leftrightarrow-x^4-7x^2-20=0\) 

\(\Leftrightarrow-\left(x^4+7x^2+\frac{49}{4}\right)-\frac{31}{4}=0\) 

\(\Leftrightarrow-\left(x^2+\frac{7}{2}\right)^2=\frac{31}{4}\) 

\(\Leftrightarrow\left(x^2+\frac{7}{2}\right)=-\frac{31}{4}\) 

\(\Rightarrow\)pt vô nghiệm

a/ĐK: \(x\ge2\)

 \(PT\Leftrightarrow x^2-4=\sqrt{x^2-4}\)

Đặt \(x^2-4=t\Rightarrow x^2=t+4\)

Thay vào,phương trình đã cho tương đương với: 

\(t=\sqrt{t}\Leftrightarrow t^2=t\Rightarrow\orbr{\begin{cases}t=1\\t=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-4=1\\x^2-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=5\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}\\x=2\end{cases}}\)  (t/m)

6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)

Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)

Phương trình sẽ trở thành là: a^2+a-42=0

=>(a+7)(a-6)=0

=>a=-7(loại) hoặc a=6(nhận)

=>2x^2+3x+9=36

=>2x^2+3x-27=0

=>2x^2+9x-6x-27=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)