a.giá trị nhỏ nhất hả bạn?
ta có: B = x⁴-x²+2x+7

=x⁴-2x²+1+x²+2x+1+5

=(x²-1)²+(x+1)²+5\(\ge5\)

vậy min B=5
dấu "=" xảy ra \(\Leftrightarrow x=-1\)

b.\(\frac{x+6}{1005}+2+\frac{x+132}{471}+4\frac{x+1008}{168}+6=0\)

\(\Leftrightarrow\frac{x+2016}{1005}+\frac{x+2016}{471}+\frac{x+2016}{168}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{1005}+\frac{1}{471}+\frac{1}{168}\right)=0\)

dễ thấy x+2016=0 =>x=-2016

vậy...

a, \(\frac{x+1006}{1000}+\frac{x+1007}{999}+\frac{x+1008}{998}+\frac{x+1009}{997}+\frac{x+2022}{4}=0\)

\(\Leftrightarrow\frac{x+1006}{1000}+1+\frac{x+1007}{999}+1+\frac{x+1008}{998}+1+\frac{x+1009}{997}+1+\frac{x+2022}{4}-4=0\)

\(\Leftrightarrow\frac{x+2006}{1000}+\frac{x+2006}{999}+\frac{x+2006}{998}+\frac{x+2006}{997}+\frac{x+2006}{4}=0\)

\(\Leftrightarrow\left(x+2006\right)\left(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+\frac{1}{997}+\frac{1}{4}\right)=0\)

Mà \(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+\frac{1}{997}+\frac{1}{4}\ne0\)

\(\Rightarrow x+2006=0\Leftrightarrow x=-2006\)

x-1009/1001+x-4/1003+x+2010/1005=7

((x-1009/1001)-1))+((x-4/1003)-2)+((x+2010/1005)-4))=0

(x-2010/1001)+(x-2010/1003)+(x-2010/1005)=0

(x-2010)*(1/1001+1/1003+1/1005)=0

okk!!!!!!!!!!!!!!!

Thanks bingodeo nhé :))

\(\Rightarrow\frac{861.\left(x-214\right)}{75768}+\frac{902.\left(x-132\right)}{75768}+\frac{924.\left(x-54\right)}{75768}=6\)

\(\Rightarrow\frac{861x-184254}{75768}+\frac{902x-119064}{75768}+\frac{924x-49896}{75768}=6\)

\(\Rightarrow861x-184254+902x-119064+924x-49896=6\)

tự làm tiếp nhé!!!!!!!!!!!!!!

\(\frac{x-214}{88}+\frac{x-132}{84}+\frac{x-54}{82}=6\)

\(\frac{6888x-1474032+7216x-952512+7392x-399168}{606144}=\frac{3636864}{606144}\)

6888x+7216x+7392x=1474032+952512+399168+3636864

21496x=6462576

x=300,6408634

xl mk chi bt lm theo kieu thu cong thoi

nhưng mà bạn Đặng Nguyễn Khánh Uyên ơi,cái bước thứ 3 đó hình như phải quy đồng cả số 6 nữa thì phải

\(PT\Leftrightarrow\frac{14x}{84}+\frac{7x}{84}+\frac{12x}{84}+\frac{420}{84}+\frac{42x}{84}+\frac{336}{84}=\frac{84x}{84}\)

=> 14x + 7x + 12x + 420 + 42x + 336 = 84x

<=> 14x + 7x + 12x + 42x - 84x = -336 - 420

<=> -9x = -756

<=> x = 84

Vậy S = {84}.

Lời giải:

Tập xác định của phương trình

Sử dụng tính chất tỉ lệ thức, có thể biến đổi phương trình như sau

Lời giải thu được

\(\frac{x+2012}{2}+\frac{x+2010}{3}+\frac{x+2011}{5}=\frac{x}{1008}+\frac{x-2}{1009}+\frac{x+1}{2015}\)

\(\Leftrightarrow\frac{x+2012}{2}+\frac{x+2010}{3}+\frac{x+2011}{5}-\frac{x}{1008}-\frac{x-2}{1009}-\frac{x+1}{2015}=0\)

\(\Leftrightarrow\frac{x+2012}{2}+2+\frac{x+2010}{3}+2+\frac{x+2011}{5}+1-\frac{x}{1008}-2-\frac{x-2}{1009}-2-\frac{x+1}{2015}-1=0\)

\(\Leftrightarrow\frac{x+2016}{2}+\frac{x+2016}{3}+\frac{x+2016}{5}-\frac{x+2016}{1008}-\frac{x+2016}{1009}-\frac{x+2016}{2015}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}-\frac{1}{1008}-\frac{1}{1009}-\frac{1}{2015}\right)=0\)

Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}-\frac{1}{1008}-\frac{1}{1009}-\frac{1}{2015}\ne0\)

\(\Leftrightarrow x+2016=0\)

\(\Leftrightarrow x=-2016\)

Vậy tập nghiệm của phương trình là \(S=\left\{-2016\right\}\)

\(\frac{1}{2-x}+1=\frac{1}{x+2}-\frac{6-x}{3x^2-12}\)ĐKXĐ : \(x\ne\pm2\)

\(\Leftrightarrow\frac{-3\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}+\frac{3\left(x-2\right)\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}=\frac{3\left(x-2\right)}{3\left(x-2\right)\left(x+2\right)}+\frac{x-6}{3\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{-3x-6+3\left(x^2-4\right)}{3\left(x-2\right)\left(x+2\right)}-\frac{3x-6+x-6}{3\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{-3x-6+3x^2-12-3x+6-x+6}{3\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{-7x-6+3x^2}{3\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow3x^2-7x-6=0\)

\(\Leftrightarrow3x^2-9x+2x-6=0\)

\(\Leftrightarrow3x\left(x-3\right)+2\left(x-3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-2}{3}\end{cases}}\)( thỏa mãn )

Vậy....