Bài toán phụ : cho 0<x<y; z>0

CMR: \(\frac{x+z}{y+z}>\frac{x}{y}\)

giải: \(0< x< y;z>0\Rightarrow zy>zx\Rightarrow zx+zy>xz+xy\)

                                          \(\Rightarrow y\left(x+z\right)>x\left(y+z\right)\Rightarrow\frac{x+z}{y+z}\Rightarrow\frac{x+z}{y+z}>\frac{x}{y}\)

Áp dụng bài toán phụ ta có:

                                    \(\frac{A+a+B+b}{A+a+B+b+c+d}>\frac{A+a}{A+a+c+d}\)

Tương tự : \(\frac{B+b+C+c}{B+b+C+c+a+d}>\frac{C+c}{C+c+A+a+b+d}\)

Mà: 

\(\frac{A+a}{A+a+c+d}+\frac{C+c}{C+c+a+d}>\frac{A+a}{C+c+A+a+b+d}+\frac{C+c}{C+c+A+a+b+d}\)

Do đó:

\(\frac{A+a+B+b}{A+a+B+b+c+d}+\frac{B+b+C+c}{B+b+C+c+a+d}>\frac{C+c+A+a}{C+c+A+a+b+d}\)

ko dùng máy tính

trả lời

75+82-99=58

hok tốt( mik ko dùng máy tính)

Ta có : 

\(|x^2+|x-1||=x^2+2\)

\(\Rightarrow\orbr{\begin{cases}x^2+|x-1|=x^2+2\left(1\right)\\x^2+|x-1|=-x^2-2\left(2\right)\end{cases}}\)

  Xét (1) ta có 

   \(x^2+|x-1|=x^2+2\)

   \(|x-1|=2+\left(x^2-x^2\right)\)

   \(|x-1|=2\)

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

   Xét (2) ta có 

   \(x^2+|x-1|=-x^2-2\)

   \(|x-1|=(-x^2-x^2)-2\)

   \(|x-1|=-2x^2-2\)

       Vì \(|x|\ge0\forall x\inℤ\)nên \(|x-1|\ge1\forall x\inℤ\)

            mà \(-2x^2-2< 0\)

         Suy ra \(x\in\varnothing\)

Vậy x = -1 ; x = 1

Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+...+\frac{2019}{3^{2019}}\)

=>\(3A=1+\frac{2}{3}+...+\frac{2019}{3^{2018}}\)

=>\(2A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)

Đặt \(B=1+\frac{1}{3}+...+\frac{1}{3^{2018}}\)

=>\(2B=3-\frac{1}{3^{2018}}\)=>\(B=\frac{3-\frac{1}{3^{2018}}}{2}\)

=>\(2A=\frac{3-\frac{1}{3^{2018}}}{2}-\frac{2019}{3^{2019}}=\frac{\frac{3^{2019}-1}{3^{2018}}}{2}-\frac{2019}{3^{2019}}\)

\(=\frac{3^{2019}-1}{3^{2018}.2}-\frac{2019}{3^{2019}}=\frac{3\left(3^{2019}-1\right)-2019.2}{3^{2019}.2}\)

Nhầm tí

dòng thứ 2 từ dưới lên cm bé hơn 0,75 luôn nhá

a) \(2x^2-3x+1\)

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

\(=2x\left(x-1\right)-\left(x-1\right)\)

\(=\left(x-1\right)\left(2x-1\right)\)

b) \(x^3-3x^2-4\)

\(=x^2-4x^3+x^3-4\)

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

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

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

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