Ta thấy : \(\left|x-1\right|\ge0\)

\(\left|y+2007\right|\ge0\)

\(\Rightarrow B=\left|x-1\right|=2\left|y+2007\right|-2010\ge-2010\)

\(MaxB=-2010\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2007=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2007\end{cases}}}\)

a)có ng` lm r`

b)Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có:

\(C-10\ge\left|x-2+2009-x\right|=2007\)

\(\Rightarrow C\ge2017\)

Dấu = khi x=2 hoặc x=2009

Vậy MinC=2017 khi x=2 hoặc x=2009

c)Xét từng trường hợp và ta có:

MinD=-1 khi \(x\ge1\)

d)\(\left|x-1\right|+\left|x-5\right|+\left|x-7\right|\)

\(\ge\left|x-1+0+7-x\right|=6\)

\(\Rightarrow E\ge6\)

Dấu = khi \(\hept{\begin{cases}x-1\ge0\\x-5=0\\x-7\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x=5\\x\le7\end{cases}}\Leftrightarrow x=5\)

Vậy MinE=6 khi x=5

a/ \(x^2-2x=-1\)

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

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

\(\Leftrightarrow x-1=0\Rightarrow x=1\)

Vậy..............

b/ \(x^2+2x+1=0\)

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

\(\Leftrightarrow x+1=0\Rightarrow x=-1\)

Vậy.......

c/ \(4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)

\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)=3x^2\)

\(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)

\(\Leftrightarrow-4x=0\Rightarrow x=0\)

Vậy...................

d/ \(x\left(x-2017\right)-x^2\left(2017-x\right)=0\)

\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)

\(\Leftrightarrow x^3-2016x^2-2017x=0\)

\(\Leftrightarrow x^3+x^2-2017x^2-2017x=0\)

\(\Leftrightarrow x\left(x^2+x\right)-2017\left(x^2+x\right)=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x-2017\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\\x-2017=0\Rightarrow x=2017\end{matrix}\right.\)

Vậy pt có 3 nghiệm là.....(tự ghi ra)

\(a,x^2-2x=-1\)

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

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

\(\Rightarrow x-2=0\Rightarrow x=2\)

\(b,x^2+2x+1=0\)

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

\(\Rightarrow x+2=0\Rightarrow x=-2\)

\(c,4\left(x-1\right)^2-\left(x-2\right)^2=3x^2\)

\(\Leftrightarrow4\left(x^2-2x+1\right)-\left(x^2-4x+4\right)-3x^2=0\) \(\Leftrightarrow4x^2-8x+4-x^2+4x-4-3x^2=0\)

\(\Leftrightarrow-4x=0\Rightarrow x=0\)

\(d,x\left(x-2017\right)-x^2\left(2017-x\right)=0\)

\(\Leftrightarrow x^2-2017x-2017x^2+x^3=0\)

\(\Leftrightarrow x^3+x^2-2017x-2017=0\)

\(\Leftrightarrow x^2\left(x+1\right)-2017\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-2017\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-2107=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x^2=2017\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\\left[{}\begin{matrix}x=\sqrt{2017}\\x=-\sqrt{2017}\end{matrix}\right.\end{matrix}\right.\)

làm bừa thui,ai tích mình mình tích lại

Số số hạng là : 

Có số cặp là :

50 : 2 = 25 ( cặp )

Mỗi cặp có giá trị là :

99 - 97 = 2 

Tổng dãy trên là :

25 x 2 = 50

Đáp số : 50

\(\frac{x-2017}{5}-\frac{x-2017}{6}=\frac{x-2017}{7}-\frac{x-2017}{8}\)

\(\Leftrightarrow\frac{x-2017}{5}-\frac{x-2017}{6}-\frac{x-2017}{7}+\frac{x-2017}{8}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\right)=0\)

\(\Rightarrow x-2017=0\Rightarrow x=2017\)

Vậy x=2017

giúp mk làm cầu b đi

x² + 2y + 1 = y² + 2z + 1 = z² + 2x + 1 = 0

=> x² + 2y + 1 + y² + 2z + 1 + z² + 2x + 1 = 0

=> (x + 1)² + (y + 1)² + (z + 1)² = 0

=> x = y = z = - 1

=> A = (-1)²⁰¹⁷ + (-1)²⁰¹⁷ + (-1)²⁰¹⁷ = -3