Bai nay phai co dieu kien a,b >0 nha ban 

Ap dung bdt \(ab\le\frac{\left(a+b\right)^2}{4}< \frac{1}{4}\)  dau nho hon la do gia thiet nha ban

Ap dung bdt Cosi cho 2 so ko am 

 ta co A= \(ab+\frac{1}{16ab}+\frac{15}{16ab}>2\sqrt{ab.\frac{1}{16ab}}+\frac{15}{16.\frac{1}{4}}=2.\frac{1}{4}+\frac{15}{4}=\frac{17}{4}\)

Study well

Đkxđ: \(\hept{\begin{cases}x\ge-\frac{1}{4}\\y\ge2\end{cases}}\)

\(\Leftrightarrow2+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=y\Leftrightarrow2+\frac{1}{2}+\sqrt{x+\frac{1}{2}}=y\Leftrightarrow\sqrt{x+\frac{1}{2}}+\frac{5}{2}=y\)

do x,y nguyên dương nên \(\sqrt{x+\frac{1}{2}}+\frac{5}{2}\)nguyên dương\(\Leftrightarrow\sqrt{x+\frac{1}{2}}=\frac{k}{2}\)(K là số nguyên lẻ, \(k>1\))

\(\Rightarrow x=\frac{k^2-2}{4}\)

do \(k^2\)là số chính phương chia 4 dư 0,1 \(\Rightarrow x=\frac{k^2-2}{4}\notin Z\)

=> ko tồn tại cặp số nguyên dương x,y tmđkđb

Ta có: \(B=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)

do đó \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}.\frac{\sqrt{x}-6}{\sqrt{x}-1}=\frac{\sqrt{x}-6}{\sqrt{x}+1}=1-\frac{7}{\sqrt{x}+1}\)

Vì \(x\ge0\Rightarrow0< \frac{7}{\sqrt{x}+1}\le7\)

Để P nguyên thì \(\frac{7}{\sqrt{x}+1}\in Z\)

do đó \(\frac{7}{\sqrt{x}+1}\in\left\{1,2,3,4,5,6,7\right\}\)

Đến đây xét từng TH là  ra

rút gọn B ta có B=\(\frac{\sqrt{x}+6}{\sqrt{x}-1}\)\(\Rightarrow\)\(AB=\frac{\sqrt{x}+6}{\sqrt{x}+1}\in Z\)

=\(1+\frac{5}{\sqrt{x}+1}\)

Vì 1\(\in Z\) nên để P thuộc Z thì \(\frac{5}{\sqrt{x}+1}\in Z\)

\(\Rightarrow\left(\sqrt{x}+1\right)\inƯ\left(5\right)=\pm1;\pm5\)

Đến đây thì ez rồi

làm sao viết đc căn vs phân số v mấy bn

a) \(\frac{b-16}{4-\sqrt{b}}\left(b\ge0,b\ne16\right)\)

\(=\frac{\left(\sqrt{b}-4\right)\left(\sqrt{b}+4\right)}{4-\sqrt{b}}\)

\(=-\sqrt{b}-4\)

b) \(\frac{a-4\sqrt{a}+4}{a-4}\left(a\ge0;a\ne4\right)\)

\(=\frac{a-2.\sqrt{a}.2+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

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

c) \(2x+\sqrt{1+4x^2-4x}\) với \(x\le\frac{1}{2}\)

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

\(=2x+\left|1-2x\right|=2x+1-2x=1\)

d) \(\frac{4a-4b}{\sqrt{a}-\sqrt{b}}\left(a,b\ge0;a\ne b\right)\)

\(=\frac{4\left(a-b\right)}{\sqrt{a}-\sqrt{b}}=\frac{4\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)

\(=4\left(\sqrt{a}+\sqrt{b}\right)\)

ccccccccccccccccccccccccccccccccccccccccccccccccccccc

AAi giải với ạ huhuu

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

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

=\(\sqrt{3}+1+\sqrt{3}-1\)

=\(2\sqrt{3}\)

k mk nha

\(ĐK:\)\(x\ge0;x\ne1;x\ne4\)

\(P=B:A=\frac{\sqrt{x}-2}{\sqrt{x}-1}:\frac{\sqrt{x}+3}{\sqrt{x}-1}\)

   \(=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)

\(P=\frac{1}{3}\)\(\Rightarrow\)\(\frac{\sqrt{x}-2}{\sqrt{x}+3}=\frac{1}{3}\)

\(\Rightarrow\)\(3\left(\sqrt{x}-2\right)=\sqrt{x}+3\)

\(\Leftrightarrow\)\(2\sqrt{x}-9=0\)

\(\Leftrightarrow\)\(2\sqrt{x}=9\)

\(\Leftrightarrow\)\(\sqrt{x}=\frac{9}{2}\)

\(\Leftrightarrow\)\(x=\frac{81}{4}\)