Bạn Kiên giải đúng nhưng chưa rõ nên mình giải lại.

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{202}{201}\)

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{202}{201}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{202}{201}\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=\frac{202}{201}\)

\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{202}{201}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{202}{201}:2=\frac{202}{402}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{202}{402}=-\frac{1}{402}=\frac{-1}{402}=\frac{1}{-402}\)

\(\Rightarrow\frac{1}{x+1}=\hept{\begin{cases}\frac{-1}{402}\\\frac{1}{-402}\end{cases}}\Rightarrow x+1=\hept{\begin{cases}402\\-402\end{cases}}\Rightarrow\hept{\begin{cases}x=402-1\\x=\left(-402\right)-1\end{cases}}\Rightarrow x=\hept{\begin{cases}401\\-403\end{cases}}\)

\(\Rightarrow A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{202}{201}\)\(\Rightarrow A=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{202}{201}\)

\(\Rightarrow A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{202}{201}\)

\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{202}{201}\)

\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{202}{201}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{202}{402}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{202}{402}=\frac{-1}{402}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{-402}\)

\(\Rightarrow x+1=-402\)

\(\Rightarrow x=-403\)

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

\(\Rightarrow\left(x+1\right).3=\left(x-2\right).2\)

\(3x+3=2x-4\)

\(\Rightarrow3x-2x=-4-3\)

\(x=-7\)

KL: x= -7

Học tốt nhé bn !!

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

=> (x+1).3 = 2(x-2)

=> 3x + 3 = 2x - 2

=> 3 + 2 = 2x - 3x

=> 5 = -x

=> x = -5

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

\(\Rightarrow\left(x+1\right)3=2\left(x-2\right)\)

\(\Rightarrow3x+3=2x-4\)

\(\Rightarrow3x-2x=-4-3\)

\(\Rightarrow x=-7\)

vậy \(x=-7\)

a) Đặt  \(A=-x^2+9x-12\)

\(-A=x^2-9x+12\)

\(-A=\left(x^2-9x+\frac{81}{4}\right)-\frac{33}{4}\)

\(-A=\left(x-\frac{9}{2}\right)^2-\frac{33}{4}\)

Mà  \(\left(x-\frac{9}{2}\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge-\frac{33}{4}\Leftrightarrow A\le\frac{33}{4}\)

Dấu "=" xảy ra khi :  \(x-\frac{9}{2}=0\Leftrightarrow x=\frac{9}{2}\)

Vậy  \(A_{Max}=\frac{33}{4}\Leftrightarrow x=\frac{9}{2}\)

b) Đặt \(B=2x^2+10x-1\)

\(B=2\left(x^2+5x+\frac{25}{4}\right)-\frac{29}{4}\)

\(B=2\left(x+\frac{5}{2}\right)^2-\frac{29}{4}\)

Mà  \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\Rightarrow2\left(x+\frac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow B\ge-\frac{29}{4}\)

Dấu "=" xảy ra khi :  \(x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy  \(B_{Min}=-\frac{29}{4}\Leftrightarrow x=-\frac{5}{2}\)

c) Đặt  \(C=\left(2x+6\right)\left(x-1\right)\)

\(C=2x^2-2x+6x-6\)

\(C=2x^2+4x-6\)

\(C=2\left(x^2+2x+1\right)-8\)

\(C=2\left(x+1\right)^2-8\)

Mà  \(\left(x+1\right)^2\ge0\forall x\Rightarrow2\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow C\ge-8\)

Dấu "=" xảy ra khi :  \(x+1=0\Leftrightarrow x=-1\)

Vậy  \(C_{Min}=-8\Leftrightarrow x=-1\)

d) Đặt  \(D=3x-2x^2\)

\(-2D=4x^2-6x\)

\(-2D=\left(4x^2-6x+\frac{9}{4}\right)-\frac{9}{4}\)

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

Mà  \(\left(2x-\frac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow-2D\ge-\frac{9}{4}\)

\(\Leftrightarrow D\le\frac{9}{8}\)

Dấu "=" xảy ra khi :  \(2x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{4}\)

Vậy  \(D_{Max}=\frac{9}{8}\Leftrightarrow x=\frac{3}{4}\)

giúp mk vs các bn ui, mai mk nộp bài rùi, mk cần gấp lắm lắm,...giúp mk nha....

B = | x + 1 | + | x - 2 | lớn hơn hoặc bằng | x + 1 + 2 - x | = 3

Dấu "=" xảy ra <=>\(\hept{\begin{cases}x+1\ge0\\2-x\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge-1\\x\le2\end{cases}\Rightarrow}-1\le x\le2}\)

Vậy,..........

À mà  \(B=|x-1|+|x-2|\) chứ ko phải \(B=|x+1|+|x-2|\) nha bn

B = | x - 1 | + | x + 2 | \(\ge\)| x + 1 + 2 - x  | = 3

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1\ge0\\2-x\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge-1\\x\le2\end{cases}\Rightarrow}-1\le x\le2}\)

Vậy ....................................

1.2 

<=> x³ - 3x²+ 3x -1 <=> (x-1)³

=> x= 1

bài 1.2 làm như sau:

x³ - 3x²+3x-1=0

x³-3x².1+3x.1²-1³=0

áp dụng HĐT số 5 trong sách ta có

(x-1)³=0

=> x-1=0

x=1

Bài 1 , 2

\(x^3-3x^2+3x-1=\left(x-1^3\right)\)

\(Suy\)\(ra\)\(:x=1\)

a) \(\left|2x-1\right|+\frac{1}{3}=0\)

\(\Leftrightarrow\left|2x-1\right|=-\frac{1}{3}\)

=> vô lý

=> PT vô nghiệm

b) \(\left|x+2\right|+\left|x-3\right|=0\)

\(\Leftrightarrow\left|x+2\right|=-\left|x-3\right|\)

Vì \(\hept{\begin{cases}\left|x+2\right|\ge0\\-\left|x-3\right|\le0\end{cases}\left(\forall x\right)}\) nên dấu "=" xảy ra khi: 

\(\left|x+2\right|=-\left|x-3\right|=0\Rightarrow\hept{\begin{cases}x=-2\\x=3\end{cases}}\) (vô lý)

=> PT vô nghiệm

Bài 1: 

a: \(A=-\left|x-\dfrac{4}{9}\right|+\dfrac{7}{33}\le\dfrac{7}{33}\forall x\)

Dấu '=' xảy ra khi x=4/9

b: \(B=-\left|x+\dfrac{11}{9}\right|+\dfrac{101}{90}\le\dfrac{101}{90}\forall x\)

Dấu '=' xảy ra khi x=-11/9

Bài 2:

=>2x-8/33=0 và 3y+7/45=0

=>2x=8/33 và 3y=-7/45

=>x=8/66=4/33 và y=-7/135

a. (9x + 2).3 = 60

<=> 9x + 2 = 20

<=> 9x = 18

<=> x = 2

b. 71 + (26 - 3x):5 = 75

<=> (26 - 3x) : 5 = 4

<=> 26 - 3x = 4/5

<=> 3x = 26 - 4/5 

<=> x = 42/5

c. 2^x = 32

<=> 2^x = 2⁵

<=> x = 5

d. (x - 6)² = 9

<=> x - 6 = 3

<=> x = 9

a) \(\left(9x+2\right)\times3=60\)

\(\Rightarrow9x+2=60:3\)

\(\Rightarrow9x+2=20\)

\(\Rightarrow9x=20-2\)

\(\Rightarrow9x=18\)

\(\Rightarrow x=18:9\)

\(\Rightarrow x=2\)

Vậy x = 2

b) \(71+\left(26-3x\right):5=75\)

\(\Rightarrow\left(26-3x\right):5=75-71\)

\(\Rightarrow\left(26-3x\right):5=4\)

\(\Rightarrow26-3x=4\times5\)

\(\Rightarrow26-3x=20\)

\(\Rightarrow3x=26-20\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=6:3\)

\(\Rightarrow x=2\)

Vậy x = 2

c) \(2^x=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

Vậy x = 5

d) \(\left(x-6\right)^2=9\)

\(\Rightarrow\orbr{\begin{cases}x-6=3\\x-6=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=9\\x=3\end{cases}}\)

Vậy x = 9 hoặc x = 3

_Chúc bạn học tốt_

a) (9x+2).3=60

       9x+2  = 20

           9x  = 18

             x  = 2

b) 71 + (26-3x) : 5 = 75

   (26 - 3x) : 5         = 4

    26 - 3x                = 20

          3x                  = 26 - 20

            x                 = 6 : 3

            x                  = 2

c) 2^x = 32

   2^x  = 2⁵

     x = 5

d) (x-6)² = 9

   (x-6)²  = 3²

      x - 6 = 3

         x =  9

bao đúng bn nhé