Theo bài ra ta có:

|x+\(\frac{1}{2}\)|\(\ge\)0

|x+\(\frac{1}{6}\)|\(\ge\)0

............................

|x+\(\frac{1}{110}\)|\(\ge\)0

\(\Rightarrow\)|x+\(\frac{1}{2}\)|+|x+\(\frac{1}{6}\)|+...+|x+\(\frac{1}{110}\)|\(\ge\)0

\(\Rightarrow\)11.x\(\ge\)0

\(\Rightarrow\)x\(\ge\)0

\(\Rightarrow\)x dương.

Khi đó:|x+\(\frac{1}{2}\)|+|x+\(\frac{1}{6}\)|+...+|x+\(\frac{1}{110}\)|=11.x

\(\Rightarrow\)x+\(\frac{1}{2}\)+x+\(\frac{1}{6}\)+...+x+\(\frac{1}{110}\)=11.x

\(\Rightarrow\)27.x+\(\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)=11x

 \(\Rightarrow\)\(\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)=-16x

\(\Rightarrow\)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)=-16x

\(\Rightarrow\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)=-16x

\(\Rightarrow\)\(\frac{10}{11}\)=-16x

\(\Rightarrow\)\(\frac{10}{-176}=x\)

Vậy \(x=\frac{10}{-176}\).

\(\frac{\sqrt{x}+1}{\sqrt{x}-5}\inℤ\Leftrightarrow6⋮\sqrt{x}-5\Leftrightarrow\sqrt{x}-5\in\left\{-1;1;2;-2;-3;3;-6;6\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;6;7;3;2;8;-1;11\right\}\Leftrightarrow x\in\left\{16;36;49;9;4;64;1;121\right\}\)

a) 2-(x+3) = 1+2+3+...+99

1+2+3+...+99 → có 99 số hạng

2-(x+3) = (1+99).99 : 2 

2-(x+3) = 4950

x+3 = 2 + 4950 

x+3 = 4952

x = 4952 - 3

x = 4949

b) (x+1)+(x+2)+...+(x+100) = 5750

→ có 100 cặp

(x+x+x+...+x) + ( 1+2+3+...+100 ) = 5750

=> 100x + 5050 = 5750

100x = 5750 - 5050

100x = 700

x = 700 : 100

x = 7

 0o0 Nguyễn Đoàn Tuyết Vy 0o0 bà kêu tui học tốt có nghĩa là học giốt đúng ko

b)(x+1)+(x+2)+(x+3)+(x+4)+...............+(x+100)=5750

(x+x+x+x+...+x) + (1+2+3+4+...+100) = 5750

     100x             +           5050           = 5750

                       100x                          = 5750 - 5050

                       100x                          = 700

                         x                             = 700 : 100

                         x                             = 7

Vậy ...

pt <=> ( 2x + 3 )( x - 5 ) - 2x( 2x + 3 ) = 0

<=> ( 2x + 3 )( -x - 5 ) = 0

<=> x = -3/2 hoặc x = -5

Vậy ... 

\(\left(2x+3\right)\left(x-5\right)=4x^2+6x\Leftrightarrow\left(2x+3\right)\left(x-5\right)=2x\left(2x+3\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(-x-5\right)=0\Leftrightarrow x=-\frac{3}{2};x=-5\)

Vậy tập nghiệm của pt là S = { -5 ; -3/2 } 

bình phương rồi cauchy-schwarz

-6.258342613

\(A=\sqrt{14-x}+\sqrt{x-10}\)

Đk:\(10\le x\le14\)

\(A^2=\left(\sqrt{14-x}+\sqrt{x-10}\right)^2\)

\(=\left(14-x\right)+\left(x-10\right)+2\sqrt{\left(14-x\right)\left(x-10\right)}\)

\(=4+2\sqrt{\left(14-x\right)\left(x-10\right)}\)

\(\le4+\left(14-x\right)+\left(x-10\right)\) (BĐT AM-GM)

\(=4+4=8\Rightarrow A^2\le8\Rightarrow A\le\sqrt{8}\)

\(x^2+3x+2\) =\(x^2+2.\frac{3}{2}x+\left(\frac{3}{2}\right)^2-\frac{5}{4}\)=\(\left(x+\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

Dấu "=" xảy ra <=>\(x+\frac{3}{2}=0\)<=>\(x=-\frac{3}{2}\)

Bài 2:

a) \(x^2-4x+y^2+2y+5=0\)

=> \(\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)

=>\(\left(x-2\right)^2+\left(y+1\right)^2=0\)

Vì \(\left(x-2\right)^2+\left(y+1\right)^2\ge0\)nên:

=>\(\hept{\begin{cases}x-2=0\\y+1=0\end{cases}}\)<=>\(\hept{\begin{cases}x=2\\y=-1\end{cases}}\)

b)\(2x^2+y^2-2xy+10x+25=0\)

=>\(\left(x^2-2xy+y^2\right)+\left(x^2+10x+25\right)=0\)

=>\(\left(x-y\right)^2+\left(x+5\right)^2=0\)

Tới đây thì dễ nhá !

Mih nhầm nhá, câu a là -1/4 cơ nha bạn

phan a nghia la so nao la 1 thi thay bang 1/4 a ???