a, Với \(x\le\frac{3}{2}\)

\(2\sqrt{3-2x}=\frac{1}{2}\Leftrightarrow\sqrt{3-2x}=\frac{1}{4}\)

\(\Leftrightarrow3-2x=\frac{1}{16}\Leftrightarrow2x=\frac{47}{16}\Leftrightarrow x=\frac{47}{32}\)(tm)

b, Với \(x\ge1\)

\(4-\sqrt{x-1}=\frac{1}{2}\Leftrightarrow\sqrt{x-1}=\frac{7}{2}\Leftrightarrow x-1=\frac{49}{4}\Leftrightarrow x=\frac{53}{4}\)(tm)

c, Với \(x\ge1\)

\(\sqrt{x-1}-3=1\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\Leftrightarrow x=17\)(tm)

d, Với \(x\ge-2\)

\(\frac{1}{2}-2\sqrt{x+2}=\frac{1}{4}\Leftrightarrow2\sqrt{x+2}=\frac{1}{4}\Leftrightarrow\sqrt{x+2}=\frac{1}{8}\)

\(\Leftrightarrow x+2=\frac{1}{64}\Leftrightarrow x=-\frac{127}{64}\)(tm) 

\(\sqrt{5x^2}=2x+1\)

\(\Leftrightarrow5x=2x+1\)

\(\Leftrightarrow3x=1\)

\(\Leftrightarrow x=1\div3\Leftrightarrow x=\frac{1}{3}\)

phá căn sai rồi em nhé ;-; 

\(\sqrt{5x^2}=2x+1\)ĐK : x > = 0 

\(\Leftrightarrow5x^2=\left(2x+1\right)^2\Leftrightarrow5x^2=4x^2+4x+1\)

\(\Leftrightarrow x^2-4x-1=0\Leftrightarrow x^2-4x+4-5=0\)

\(\Leftrightarrow\left(x-2\right)^2-5=0\Leftrightarrow\left(x-2-\sqrt{5}\right)\left(x-2+\sqrt{5}\right)=0\)

\(\Leftrightarrow x=2+\sqrt{5}\left(tm\right);x=2-\sqrt{5}\left(ktm\right)\)

cũng hợp lí á, đáp án đúng

Ta có \(d\in Z\)và \(d< 5\Leftrightarrow max\left(d\right)=4\)

Ta lại có \(c< 4\left(d\right)\)mà \(max\left(d\right)=4\Leftrightarrow max\left(c\right)< 16\)mà \(c\in Z\Leftrightarrow max\left(c\right)=15\)

Tương tự \(b< 3c\Rightarrow b< 45\)mà \(b\in Z\Leftrightarrow max\left(b\right)=44\)

                 \(a< 2b\Rightarrow a< 88\)mà \(a\in Z\Leftrightarrow max\left(a\right)=87\)

Vậy giá trị lớn nhất của a là 87