dtydudjgbjbjbjvjkkdxkdiuryyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyykkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhrrrrrrrrrrrrrrrrrrrrrrrrrrnmdchytfegttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttdyyyyyyyyyyyyyyyyyyyyyyyyrrrrrrrrrrrrrrrrrrrrrrrrr

ASDFGHJKL;''\\\\\\\\\\\\\\09876212EFGNM,///////////////,HHVSZZCCCCCCCCCCCCCCBBBBBBBBBBBBBBMMMMMMMMMMMMJJXGGJBDU.LH7UJKI,M MYN YBRROP

IJUL[

-PIIGDAAQWRTYUIOLP;LNBF1954DGW22568997TVV32V456

Câu a hình như sai đề mk sửa nha

a)\(A=\left(2x+\frac{1}{3}\right)^4-1\)

         Vì \(\left(2x+\frac{1}{3}\right)^4\ge0\)

      Suy ra:\(\left(2x+\frac{1}{3}\right)^4-1\ge-1\)

                   Dấu = xảy ra khi \(2x+\frac{1}{3}=0\)

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

                                                \(x=-\frac{1}{6}\)

Vậy Min A=-1 khi \(x=-\frac{1}{6}\)

b)\(B=-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\)

    \(B=3-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\)

           Vì \(-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\le0\)

                     Suy ra:\(3-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\le3\)

Dấu = xảy ra khi \(\frac{4}{9}x-\frac{2}{15}=0\)

                            \(\frac{4}{9}x=\frac{2}{15}\)

                            \(x=\frac{3}{10}\)

     Vậy Max B=3 khi \(x=\frac{3}{10}\)

Bài làm:

a) | 5x - 4 | = | x + 2 |

=> 5x - 4 = x + 2

=> 5x - x = 2 + 4

=> x . (5 - 1) = 6

=> x . 4 = 6

=> x = 6 : 4 = 1,5

b) | x + 2/5 | = 2x

=> x + 2/5 = 2x hoặc x + 2/5 = -2x

* x + 2/5 = 2x

=> x - 2x = -2/5

=> x . (1 - 2) = -2/5

=> x .(-1) = -2/5

=> x = -2/5 : (-1)

=> x = 2/5

* x + 2/5 = -2x

=> x + 2x = 2/5

=> x . (1 + 2) = 2/5

=> x . 3 = 2/5

=> x = 2/5 : 3

=> x = 2/15

mk chỉ làm 2 bài này thôi, còn 2 bài kia mk ko có pít làm. Sorry!

a/ ĐKXĐ : \(x\ge0;x\ne1\)

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

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

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

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

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

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

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

Vậy...

b/ Ta có :

\(P>0\)

\(\Leftrightarrow-\sqrt{x}\left(x-1\right)>0\)

\(\Leftrightarrow\sqrt{x}\left(x-1\right)< 0\)

Mà \(\sqrt{x}\ge0\)

\(\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ

Vậy \(0< x< 1\) thì P > 0

c/ Ta có :

\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) thỏa mãn \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{x}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

Thay vào P rồi bạn tự tính ra nhé :>

Bài 1 :

\(P=\frac{x+16}{\sqrt{x}+3}=\frac{x-9+25}{\sqrt{x}+3}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+25}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)

Áp dụng BĐT Cô-si ta có :

\(\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\ge2\sqrt{\frac{25\left(\sqrt{x}+3\right)}{\sqrt{x}+3}}=10\)

\(\Rightarrow P\ge10-6=4\)

Vậy \(MIN_P=4\) . Dấu \("="\) xảy ra khi \(x=4\)

Bài 2 : Đặt \(\sqrt{x}=a\)

\(M=\frac{2}{a^2+a+1}\Leftrightarrow Ma^2+Ma+M-2=0\)

\(\Delta=M^2-4M\left(M-2\right)=-3M^2+8M\)

Để phương trình có hai nghiệm phân biệt :

\(\Rightarrow-3M\left(M-\frac{8}{3}\right)\ge0\Rightarrow0\le M\le\frac{8}{3}\)

Vậy GTLN của M là \(-\frac{8}{3}\)

Ta có /x+1/ >/ 0 với mọi x

=> A>/ 5 với mọi x

=>A_max=5

Dấu "=" xảy ra<=>x+1=0<=>x=-1

B=(x^2+3)+12/(x^2+3)=1+(12/x^2+3)

 ta có x^2+3 >/ 3 với mọi x

=>12/x^2+3 </ 12/3=4 với mọi x

=>B </ 1+4=5 với mọi x

Dấu "=" xảy ra<=>x=0

Vậy...