Bài làm:

Ta có: \(4x^2-4x-3=0\)

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

\(\Leftrightarrow\left(2x-1\right)^2-2^2=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)

Ta có : \(4x^2-4x-3=0\)

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

\(\Leftrightarrow\left(2x-1\right)^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=2\\2x-1=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)

Vậy \(x\in\left\{\frac{3}{2};-\frac{1}{2}\right\}\)

a/ 3x^2 - 8x + 4 = 0
<=> (3x-2)(x-2)=0
<=> x=2/3 or x=2
b/ 4x^2 - 4x - 3 = 0
<=> (2x+1)(2x-3)=0
<=> x=-1/2 or x=3/2

Xin lỗi bạn nha dòng cuối mik nhầm ...

\(\Rightarrow x=-2018\)

Vậy x = -2018

\(\frac{x+4}{2014}+\frac{x+3}{2015}+\frac{x+2}{2016}+\frac{x+1}{2017}=a\)

\(\Rightarrow\frac{x+4}{2014}+1+\frac{x+3}{2015}+1+\frac{x+2}{2016}+1+\frac{x+1}{2017}+1=a+4\)

\(\frac{x+2018}{2014}+\frac{x+2018}{2015}+\frac{x+2018}{2016}+\frac{x+2018}{2017}=a+4\)

\(\left(x+2018\right).\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)=a+4\)

..

s bk

\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\frac{x+1}{2017}\)

\(\Rightarrow\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)

\(\Rightarrow\frac{x+4+2014}{2014}+\frac{x+3+2015}{2015}=\frac{x+2+2016}{2016}+\frac{x+1+2017}{2017}\)

\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)

\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}+\frac{x+2018}{2017}=0\)

\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\right)=0\)

Mà \(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\ne0\)

\(\Rightarrow x+2018=0\)

\(\Rightarrow x=2018\)

Vậy x = 2018

_Chúc bạn học tốt_

Ta có: \(x^2\ge0;\left|x+y\right|\ge0;\forall x,y\)

=> \(M=2015+3\left(x^2+1\right)^{2016}+\left|x+y\right|^{2017}\)

\(\ge2015+3\left(0+1\right)^{2016}+0^{2017}=2018\)

Dấu "=" xảy ra khi và chỉ khi: \(\hept{\begin{cases}x^2=0\\\left|x+y\right|=0\end{cases}\Leftrightarrow x=y=0}\)

Vậy gtnn của M = 2018 đạt tại x = y = 0.

D=|x-2010| + |x-2011| + |x-2012|
D=|x-2010| + |x-2011| + |2012-x|
=>D>=|x-2010+2012-x| + |x-2011|
=>D>=|2| + |x-2011|=2 + |x-2011|
Dấu = xảy ra <=> (x-2010)(2012-x)>=0<=>2010<=x<=2012(1)
                           x-2011=0 => x =2011(2)
Từ 1,2 => x=2011
Vậy Bmin=2 khi x=2011
 

  1. A = 9/(2/x-1) + 2/x = 9/(y-1) + y (với y = 2/x > 1).
Sử dụng BĐT Cauchy (Cô-si): A = 1+ 9/(y-1) + (y-1) >= 1+ 2*căn9 = 7 (vì y - 1 > 0 do y > 1). Dấu = xảy ra khi 9/(y-1) = (y-1) tương đương y-1 = 3 hay y = 4 tức x = 1/2. 

2. B = 3(1-x + x)/(1-x) + 4/x = 3 + 3x/(1-x) + 4/x = 3 + 12/(4/x - 4) + 4/x = 7 + 12/(4/x - 4) + (4/x - 4) >= 7 + 4căn3. Dấu = khi 12/(4/x - 4) = (4/x - 4) hay 4/x - 4 = 2căn3 (bạn tự tìm x nhé). 

3. Sử dụng BĐT Bunhi: Q*2 = [x²/(y+z) + y²/(z+x) + z²/(x+y)]*[(y+z) + (z+x) + (x+y)] >= [(x/căn(y+z))*căn(y+z) + y/căn(x+z))*căn(x+z) + z/căn(y+x))*căn(y+x)]^2 = (x+y+z)^2 = 4 hay Q>=1/2. 
Dấu = xảy ra khi x = y = z = 2/3. 

4. Sử dụng BĐT Bunhi: (x²)² + (y²)² >= [(x²) + (y²)]²/2 >= [(x+y)²/2]²/2 = 1/8. 
 

cảm ơn bạn đã giúp nha =)))) 

mà mấy bài đó ko phải bài mà mình muốn hỏi bạn ạ 

Ta có:  \(\frac{x-18}{2018}=\frac{x-17}{2017}\)

\(\Rightarrow\left(x-18\right).2017=\left(x-17\right).2018\)( tính chất của 2 tỉ số bằng nhau )

\(2017x-2017.18=2018x-2018.17\)

\(2018.17-2017.18=2018x-2017x\)

\(\left(2017+1\right).17-2017.\left(17+1\right)=x\)

\(2017.17+17-2017.17-2017=x\)

\(x=-2000\)

Vậy \(x=-2000\)

\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x-1}{101}+\frac{x-2}{102}\)

\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)=\left(\frac{x-1}{101}+1\right)+\left(\frac{x-2}{102}+1\right)\) ( cộng cả 2 vế thêm 2 )

\(\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{101}+\frac{x+100}{102}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{101}-\frac{x+100}{102}=0\)

\(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{100}\right)=0\)

Ta có: \(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{100}\ne0\)

\(\Rightarrow x+100=0\)

​​​\(x=-100\)​

Vậy \(x=-100\)

a, \(\frac{x-18}{2018}=\frac{x-17}{2017}\)

=>\(\frac{x-18}{2018}+1=\frac{x-17}{2017}+1\)

=>\(\frac{x-18+2018}{2018}=\frac{x-17+2017}{2017}\)

=>\(\frac{x+2000}{2018}=\frac{x+2000}{2017}\)

=>\(\frac{x+2000}{2018}-\frac{x+2000}{2017}=0\)

=>\(\left(x+2000\right)\left(\frac{1}{2018}-\frac{1}{2017}\right)=0\)

Mà \(\frac{1}{2018}-\frac{1}{2017}\ne0\)

=>x+2000=0 => x=-2000

b, 

=>\(\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x-1}{101}+1+\frac{x-2}{102}+1\)

=>\(\frac{x+1+99}{99}+\frac{x+2+98}{98}=\frac{x-1+101}{101}+\frac{x-2+102}{102}\)

=>\(\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{101}+\frac{x+100}{102}\)

=>\(\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{101}-\frac{x+100}{102}=0\)

=>\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{102}\right)=0\)

Mà \(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{102}\ne0\)

=>x+100=0 => x=-100

\(\frac{x-1}{2018}=\frac{x-17}{2017}\)

\(\Rightarrow\frac{x-18}{2018}=\frac{x-17}{2017}=\frac{\left(x-18\right)-\left(x-17\right)}{2018-2017}=-\frac{1}{1}=-1\)

\(\Rightarrow x-18=-2018\)

\(\Rightarrow x=-2000\)

Uyên ơi m đừng mất dậy đừng chửi con gái h đú vc