\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)

+) chứng minh 1/ab+b+1 + 1/bc+c+1 + 1/ac+a+1=1

<=> abc/ab+b+abc + abc/bc+c+abc + 1/ac+a+1

<=> ac/ac+a+1 + ab/b+1+ab + 1/ac+a+1

<=> ac+a+1/ac+a+1

<=> 1

+) xét: a^2+2b^2+3=(a^2+b^2)+(b^2+1)+2 >= 2ab+2b+2<=1/2(ab+b+1) (1)

chứng minh tương tự:1/ b^2+2c^2+3 <= 1/2(bc+c+1) (2)

                                    1/ c^2+2a^2+3 <= 1/2(ac+a+1) (3)

cộng các vế của (1),(2),(3) ta duoc: 1/(a^2+2b^2+3) + 1/(b^2+2c^2+3) + 1/(c62+2a^2+3) <= 1/2.(1/ab+b+1 + 1/bc+c+1 + 1/ac+a+1)=1/2 (đpcm)

mình làm rồi, bạn vào đây tham khảo nha: http://olm.vn/hoi-dap/question/559729.html

\(a+b+c=\frac{1}{abc}\)

\(\Leftrightarrow abc\left(a+b+c\right)=1\)(*)

\(\Leftrightarrow a^2bc+ab^2c+abc^2=1\)

Ta có :

\(1+b^2c^2=a^2bc+ab^2c+abc^2+b^2c^2\)

\(=abc\left(a+b\right)+bc^2\left(a+b\right)\)

\(=bc\left(a+b\right)\left(a+c\right)\)

Tương tự ta cũng có \(1+a^2c^2=ac\left(a+b\right)\left(b+c\right)\)

Khi đó : \(\left(1+b^2c^2\right)\left(1+a^2c^2\right)=abc^2\left(a+b\right)^2\left(b+c\right)\left(a+c\right)\)(1)

Xét \(c^2+a^2b^2c^2\)

\(=a^2b^2c^2+\frac{abc^3}{abc}\)

\(=a^2b^2c^2+abc^3\left(a+b+c\right)\)( theo giả thiết )

\(=a^2b^2c^2+a^2bc^3+ab^2c^3+abc^4\)

\(=abc^2\left(ab+bc+ca+c^2\right)\)

\(=abc^2\left(b+c\right)\left(a+c\right)\)(2)

Từ (1) và (2) ta suy ra :

\(\sqrt{\frac{\left(1+b^2c^2\right)\left(1+a^2c^2\right)}{c^2+a^2b^2c^2}}=\sqrt{\frac{abc^2\left(a+b\right)^2\left(b+c\right)\left(a+c\right)}{abc^2\left(b+c\right)\left(a+c\right)}}\)

\(=\sqrt{\left(a+b\right)^2}=\left|a+b\right|=a+b\)( vì \(a,b\in Z^+\) )

Ta có đpcm.

a, \(15:\left(x+2\right)=3\Leftrightarrow x+2=3\Leftrightarrow x=1\)

b, \(20:\left(x+1\right)=2\Leftrightarrow x+1=10\Leftrightarrow x=9\)

c, \(240:\left(x-5\right)=2^2.5^2-20=80\Leftrightarrow x-5=3\Leftrightarrow x=8\)

d, \(96-3\left(x+1\right)=42\Leftrightarrow3\left(x+1\right)=54\Leftrightarrow x+1=18\Leftrightarrow x=17\)

a) 15 : (x + 2) = 3

x + 2 = 15 : 3

x + 2 = 5

x = 5 – 2 = 3

b) 20 : (1 + x) = 2

1 + x = 20 : 2

1 + x = 10

x = 10 – 1 = 9

c) 240 : (x – 5) = 2².5² – 20

240 : (x – 5) = 4.25 – 20

240 : (x - 5) = 100 – 20

240 : (x - 5) = 80

x – 5 = 240 : 80

x – 5 = 3

x = 3 + 5 = 8

d) 96 - 3(x + 1) = 42

3(x + 1) = 96 – 42

3(x + 1) = 54

x + 1 = 54 : 3

x + 1 = 18

x = 18 – 1

x = 17

\(15\div\left(x+2\right)=3\)                                           \(20\div\left(1+x\right)=2\)

\(x+2=15\div3\)                                               \(1+x=20\div2\)

\(x+2=5\)                                                       1+x=10

x = 5 - 2                                                           x=9

x = 3

\(240\div\left(x-5\right)=2^2\cdot5^2-20\)                                \(96-3\left(x+1\right)=42\)

\(240\div\left(x-5\right)=80\)                                              \(3\left(x+1\right)=96-42\)

\(x-5=240\div80\)                                                 \(3\left(x+1\right)=54\)

x-5=3                                                                    \(x+1=54\div3\)

x=8                                                                        x+1=18

                                                                              x=17

het thoirui pan oi

Áp dụng bđt cosi ta có:

\(M=\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}=\frac{1}{a^2+b^2+b^2+1+2}+\frac{1}{b^2+c^2+c^2+1+2}+\frac{1}{c^2+a^2+a^2+1+2}\le\frac{1}{2\sqrt{ab}+2\sqrt{b}+2}+\frac{1}{2\sqrt{bc}+2\sqrt{c}+2}+\frac{1}{2\sqrt{ac}+2\sqrt{a}+2}=\frac{1}{2}\left(\frac{1}{\sqrt{ab}+\sqrt{b}+1}+\frac{1}{\sqrt{bc}+\sqrt{c}+1}+\frac{1}{\sqrt{ac}+\sqrt{a}+1}\right)=\frac{1}{2}\left(\frac{1}{\sqrt{ab}+\sqrt{b}+1}+\frac{\sqrt{abc}}{\sqrt{bc}+\sqrt{c}+\sqrt{abc}}+\frac{\sqrt{b}}{\sqrt{abc}+\sqrt{ab}+\sqrt{b}}\right)=\frac{1}{2}\left(\frac{1}{\sqrt{ab}+\sqrt{b}+1}+\frac{\sqrt{ab}}{\sqrt{b}+1+\sqrt{ab}}+\frac{\sqrt{b}}{1+\sqrt{ab}+\sqrt{b}}\right)=\frac{1}{2}\left(\frac{1+\sqrt{ab}+\sqrt{b}}{\sqrt{ab}+\sqrt{b}+1}\right)=\frac{1}{2}\Rightarrow M\le\frac{1}{2}\)

Vậy GTLN của M là \(\frac{1}{2}\)

\(M=\sum\frac{1}{a^2+b^2+b^2+1+2}\le\frac{1}{2}\sum\frac{1}{ab+b+1}\)

Maặt khác, ta có bài toán quen thuộc, cho \(abc=1\Rightarrow\sum\frac{1}{ab+b+1}=1\)

\(\Rightarrow M\le\frac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=1\)