a) ĐKXD: \(\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)

b) Với \(a>0;a\ne1;a\ne4\), ta có:

\(B=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\\ =\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

c)\(B\le\dfrac{1}{3}\rightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}\le\dfrac{1}{3}\rightarrow\dfrac{-2}{\sqrt{a}}\le0\) (đúng với mọi a thoả ĐKXĐ).

a, ĐKXĐ: 

\(\left\{{}\begin{matrix}\left|a\right|>1^2\\\left|a\right|>0\\\left|a\right|>2^2\end{matrix}\right.\Leftrightarrow a>4\)

b,

 \(B=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\\ B=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left[\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)\right]}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ B=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(a-1\right)-\left(a-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ B=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\\ B=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

\(c,B\le\dfrac{1}{3}\\ \Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}\le\dfrac{1}{3}\\ \Leftrightarrow3\left(\sqrt{a}-2\right)\le3\sqrt{a}\\ \Leftrightarrow\sqrt{a}-2\le\sqrt{a}\\ \Leftrightarrow\sqrt{a}-\sqrt{a}\le2\\ \Leftrightarrow0\le2\left(luôn.đúng\right)\)

Vậy: Với a>4 thì \(B\le\dfrac{1}{3}\)

Bài 2: 

a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)

\(=\dfrac{4+6-3}{n-1}\)

\(=\dfrac{7}{n-1}\)

Để A là số tự nhiên thì \(7⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(7\right)\)

\(\Leftrightarrow n-1\in\left\{1;7\right\}\)

hay \(n\in\left\{2;8\right\}\)

Vậy: \(n\in\left\{2;8\right\}\)

ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2                                                   Để B là STN thì 4n+10⋮n+2                          4n+8+2⋮n+2                                  4n+8⋮n+2                                                      ⇒2⋮n+2                                     n+2∈Ư(2)                                                        Ư(2)={1;2}                                  Vậy n=0                                                                                  

Ta có: =4+6−3n−1=4+6−3�−1

1) Áp dụng bđt Cauchy cho 3 số dương ta có

 \(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)

\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)

\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)

Cộng (1);(2);(3) theo vế ta được

\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)

\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)

\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)

2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)

Dấu"=" khi a = 4b

nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)

Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)

Đặt \(\sqrt{a+b}=t>0\) ta được

\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)

\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)

Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)

nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)

khi đó a + b = 1

mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

a) \(...\dfrac{11}{4}-a+\dfrac{1}{4}=\dfrac{3}{2}\)

\(\dfrac{11}{4}+\dfrac{1}{4}-a=\dfrac{3}{2}\)

\(3-a=\dfrac{3}{2}\)

\(a=3-\dfrac{3}{2}\)

\(a=\dfrac{6}{2}-\dfrac{3}{2}\)

\(a=\dfrac{3}{2}\)

b) \(...\dfrac{13}{4}-a-\dfrac{13}{4}=\dfrac{7}{8}\)

\(\dfrac{13}{4}-\dfrac{13}{4}-a=\dfrac{7}{8}\)

\(0-a=\dfrac{7}{8}\)

\(a=-\dfrac{7}{8}\) (ra số âm lớp 5 chưa học nên bạn xem lại đề)

c) \(...\dfrac{17}{6}-\dfrac{3}{2}-a=\dfrac{1}{6}\)

\(\dfrac{17}{6}-\dfrac{9}{6}-a=\dfrac{1}{6}\)

\(\dfrac{8}{6}-a=\dfrac{1}{6}\)

\(a=\dfrac{8}{6}-\dfrac{1}{6}\)

\(a=\dfrac{7}{6}\)

a, 2\(\dfrac{3}{4}\) - a + \(\dfrac{1}{4}\) = 1\(\dfrac{1}{2}\)

     a = 2 + \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) - 1 - \(\dfrac{1}{2}\)

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

     a  = 2 - \(\dfrac{1}{2}\)

     a = \(\dfrac{3}{2}\)

b, 3\(\dfrac{1}{4}\) - a - 3\(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)

    (3\(\dfrac{1}{4}\) - 3\(\dfrac{1}{4}\)) - a = \(\dfrac{7}{8}\)

                     a = - \(\dfrac{7}{8}\)

c,    2\(\dfrac{5}{6}\) - 1\(\dfrac{1}{2}\) - a  = \(\dfrac{1}{6}\)

    a =  2 + \(\dfrac{5}{6}\) - 1 - \(\dfrac{1}{2}\)  - \(\dfrac{1}{6}\) 

     a =  (2-1) + (\(\dfrac{5}{6}\) - \(\dfrac{1}{6}\)) - \(\dfrac{1}{2}\)

     a = 1 +  \(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)

     a = \(\dfrac{7}{6}\)

`#040911`

`a)`

\(2\dfrac{3}{4}-a+\dfrac{1}{4}=1\dfrac{1}{2}\\ \left(2\dfrac{3}{4}+\dfrac{1}{4}\right)-a=1\dfrac{1}{2}\\ 3-a=1\dfrac{1}{2}\\ a=3-1\dfrac{1}{2}\\ a=\dfrac{3}{2}\\ \text{Vậy, a = }\dfrac{3}{2}\)

`b)`

\(3\dfrac{1}{4}-a-3\dfrac{1}{4}=\dfrac{7}{8}\\ \left(3\dfrac{1}{4}-3\dfrac{1}{4}\right)-a=\dfrac{7}{8}\\0-a=\dfrac{7}{8}\\ a=0-\dfrac{7}{8} \\ a=\dfrac{-7}{8}\)

Bạn xem lại đề, lớp 5 chưa học dấu âm.

`c)`

\(2\dfrac{5}{6}-1\dfrac{1}{2}-a=\dfrac{1}{6}\\ \dfrac{4}{3}-a=\dfrac{1}{6}\\ a=\dfrac{4}{3}-\dfrac{1}{6}\\ a=\dfrac{7}{6}\\ \text{Vậy, a = }\dfrac{7}{6}.\)

a) Trước hết ta chứng minh \(a^2-1=\left(a-1\right)\left(a+1\right)\text{tự chứng minh }\)

Áp dụng bổ đề trên ta có:

\(-A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right) =\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot...\cdot\dfrac{100^2-1}{100^2}=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}=\dfrac{1\cdot2\cdot3^2\cdot...\cdot99^2\cdot100\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\dfrac{1\cdot101}{2\cdot100}>\dfrac{1}{2}\\ \Rightarrow A< -\dfrac{1}{2}\)

b)

TH1: x chẵn  mà x là số nguyên tố => x=2

=> y^2 = 117+4=121 => y=11 (thỏa mãn)

TH2:  x lẻ => x^2 lẻ  . Mà 117 lẻ

=> x^2+117 chẵn => y^2 chẵn => y chẵn mà y là số nguyên tố

=> y=2 

=>x^2+117= 4=> x^2 = -113 (vô lý)

Vậy x=2;y=11

có thể coi a=b=c=d từ đó thì ra 2 nghiệm đều thỏa mãn biểu thức là:

x = {-2;2}