\(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)

Vì GTTĐ luôn lớn hơn hoặc bằng 0 với mọi x 

\(\Rightarrow\left|x+1,1\right|+...+\left|x+1,4\right|\ge0\forall x\)

\(\Rightarrow5x\ge0\forall x\)

\(\Rightarrow x\ge0\)

\(\Leftrightarrow x+1,1+x+1,2+x+1,3+x+1,4=5x\)

\(\Leftrightarrow4x+5=5x\)

\(\Leftrightarrow5x-4x=5\)

\(\Rightarrow x=5\)

Vậy x = 5

vì các giá trị tuyệt đối lớn hơn hoặc bằng 0=>5x  lớn hơn hoặc bằng 0

=>x lớn hơn hoặc bằng 0

=>ta có thể phá dấu giá trj tuyệt đối

ta có x+1.1+x+1.2+x+1.3+x+1.4=5x

5=5x-4x=x

=>x=5

(x+1)+(x+2)+(x+3)+(x+4)=5x

x.(1+2+3+4)=5x

x.5=5x

=>x chỉ có thể là 0 hoặc 1

Nha!!!!!!!!!!!

\(\Leftrightarrow x^2+10x-2000=0\)

\(\Leftrightarrow x^2+10x+25-2025=0\)

\(\Leftrightarrow\left(x+5\right)^2=2025\)

=>x+5=45 hoặc x+5=-45

=>x=40 hoặc x=-50

1.1 / 3x(x-2005)-x+2005=0

  <=>3x(x-2005)-(x-2005)=0

  <=>(x-2005)(3x-1)=0

  <=>x-2005=0 hoặc 3x-1=0

  <=>x=2005 hoặc x=1/3

1.2/ x+1 =(x+1)²

  <=>(x+1) - (x+1)²=0

  <=>(x+1) (1-x+1)=0

  <=> (x+1) (2-x) =0

  <=>x+1=0 hoặc 2-x =0

<=> x=-1 hoặc x=2

1.3/x³=5x 

<=>x³-5x=0

<=>x(x²-5)=0

<=>x=0 hoặc x²-5=0

<=>x=0 hoặc x²=5

<=>x=0 hoặc x=\(\sqrt{5}\)và \(-\sqrt{5}\)

1.4/x²(x² -2)-4(2-x²)=0

<=>x²(x²-2) +4(x²-2)=0

<=> (x² -2)(x²+4)=0

<=>x²-2=0 hoặc x²+4=0

<=>x²=2 hoặc x²=-4(vô lí)

<=>x=\(\sqrt{2}\)hoặc \(-\sqrt{2}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{\left(5x+1\right)\left(5x+3\right)}\right)=\frac{11}{23}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{5x+1}-\frac{1}{5x+3}\right)=\frac{11}{23}\)

\(\Leftrightarrow1-\frac{1}{5x+3}=\frac{22}{23}\)

\(\Leftrightarrow\frac{1}{5x+3}=\frac{1}{23}\)

\(\Leftrightarrow5x+3=23\Leftrightarrow x=4\) ( TM )

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(5x+1\right).\left(5x+3\right)}=\frac{11}{23}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(5x+1\right)\left(5x+3\right)}\right)=\frac{11}{23}\)

\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{\left(5x+1\right)}-\frac{1}{\left(5x+3\right)}\right)=\frac{11}{23}\)

\(\Rightarrow1-\frac{1}{\left(5x+3\right)}=\frac{11}{23}:\frac{1}{2}\)

\(\Rightarrow\frac{1}{5x+3}=\frac{1}{23}\)

\(\Rightarrow5x+3=23\)

\(\Rightarrow5x=23-3\)

\(\Rightarrow x=20:5\)

\(\Rightarrow x=4\)

Đề: \(\frac{\left(x-1\right)}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{3}{5x}-\frac{7}{15}\)

\(\Leftrightarrow\frac{\left(x-1\right)}{2}\left(1-\frac{1}{15}\right)=\frac{3}{5x}-\frac{7}{15}\)

\(\Leftrightarrow\frac{\left(x-1\right)}{2}.\frac{14}{15}=\frac{3}{5x}-\frac{7}{15}\Leftrightarrow\frac{7\left(x-1\right)+7}{3}=\frac{3}{x}\)

\(\Leftrightarrow\frac{7x}{3}=\frac{3}{x}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{7}}{3}\\x=-\frac{\sqrt{7}}{3}\end{cases}}\)

`(5x-1)^2-(5x-4)(5x+4)=7`

`\Leftrightarrow 25x^2-10x+1-25x^2+16-7=0`

`\Leftrightarrow  -10x+10=0`

`\Leftrightarrow  x=1`

\(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)

\(25x^2+10x+1-\left(25x^2-9\right)=30\)

\(25x^2+10x+1-25x^2+9=30\)

\(10x+10=30\)

\(10x=20\)

\(x=2\)

T_T tôi chưa hc lp 8

(5x+1)²-(5x+3)(5x-3)=30

=>25x²+10x+1-25x²+9=30

=>(25x²-25x²)+10x+1+9=30

=>10x+10=30

=>10x=20

=>x=2