A = 11.....1 ( 2013 chữ số 1) × 100....05 ( 2012 chữ số 0) - 66....6 ( 2013 chữ số 6)

A = 11.....1 ( 2013 chữ số 1) × 100....05 ( 2012 chữ số 0) - 6 × 11....1 ( 2013 chữ số 6)

A = 11.....1 ( 2013 chữ số 1) × ( 100....05 ( 2012 chữ số 0) - 6)

A = 11.....1 ( 2013 chữ số 1) × 99....9 ( 2013 chữ số 9)

A = 11....1 ( 2013 chữ số 1) × 3 × 33....3 ( 2013 chữ số 3)

A = 33....3 ( 2013 chữ số 3) × 33....3 ( 2013 chữ số 3)

A = 33....3² ( 2013 chữ số 3)

\(\frac{2013x}{xy+2013x+2013}+\frac{y}{yz+y+2013}+\frac{z}{xz+z+1}\)

\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)

\(=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\)

\(=\frac{xz+z+1}{xz+z+1}=1\)

=>đpcm

2013x/xy+2013x+2013 + y/yz+y+2013 + z/xz+z+1

= xyz.x/xy+xyz.x+xyz + y/yz+y+xyz + z/xz+z+1

= xz/1+xz+z + 1/z+1+xz + z/xz+z+1

= xz+1+x/1+xz+x = 1 (đpcm)

Thay xyz=2013 vào ta có:

\(\frac{xyz\cdot x}{xy+xyz\cdot x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)

\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+z+1}\)

\(=\frac{xy\cdot xz}{xy\left(xz+z+1\right)}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)

\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)

\(=\frac{xz+1+z}{xz+z+1}=1\) (Đpcm)

Ta có \(x+y+z=0\Leftrightarrow\left(x+y+z\right)^2=0\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)mà xy+yz+zx=0

\(\Rightarrow x^2+y^2+z^2=0\left(1\right)\)

Lại có: \(x^2,y^2,z^2\ge0\Rightarrow x^2+y^2+z^2\ge0\)Kết hợp (1)

\(\Leftrightarrow x^2=y^2=z^2=0\Leftrightarrow x=y=z=0\)

Vậy \(T=\left(0-1\right)^{2013}+0^{2013}+\left(0+1\right)^{2013}=-1+0+1=0\)

Ta có : \(x+y+z=0\)

\(\Rightarrow\left(x+y+z\right)^2=0\)

\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)

\(\Rightarrow x^2+y^2+z^2=0\) ( Do \(xy+yz+zx=0\) )

\(\Rightarrow x^2+y^2+z^2=xy+yz+zx\)

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)=0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Rightarrow x=y=z\)

Khi đó : \(x+y+z=3x=0\)

\(\Rightarrow x=0\Rightarrow x=y=z=0\)

Nên \(T=\left(0-1\right)^{2013}+0^{2013}+\left(0+1\right)^{2013}=0\)

Vậy : \(T=0\).

Ta có: \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)

mà xy+yz+zx=0 => \(x^2+y^2+z^2=0\)vì \(x^2>0;y^2>0;z^2>0\)

Suy ra: x=y=z=0. Thế số ta được T=0

tk mk , mk tk 

+) Ta có: P(x) = 0 có 3 nghiệm phân biệt 

=> Gọi 3 nghiệm đó là m; n ; p. 

=> P(x) = ( x - m ) ( x - p ) (x - n) 

=> P(Q(x)) = ( x^2 + x + 2013 -m )( x^2 + x + 2013 -n )( x^2 + x + 2013 - p )

Vì P(Q(x)) =0 vô nghiệm nên: x^2 + x + 2013 - m = 0 ;x^2 + x + 2013 - m = 0; x^2 + x + 2013 - m = 0 đều vô nghiệm 

=> \(\Delta_m=1^2-4\left(2013-m\right)< 0;\Delta_n=1^2-4\left(2013-n\right)< 0;\Delta_p=1^2-4\left(2013-p\right)< 0\)​​

=> \(2013-m>\frac{1}{4};2013-n>\frac{1}{4};2013-p>\frac{1}{4}\)

=> P(2013) = ( 2013 - m) (2013 -n ) (2013 - p) >\(\frac{1}{4}.\frac{1}{4}.\frac{1}{4}=\frac{1}{64}\)

a) ( 2x - 1 )( 2x + 1 ) - 4( x² + x ) = 16

⇔ 4x² - 1 - 4x² - 4x = 16

⇔ -4x - 1 = 16

⇔ -4x = 17

⇔ x = -17/4

b) 5x( x - 2013 ) - x + 2013 = 0

⇔ 5x( x - 2013 )  - ( x - 2013 ) = 0

⇔ ( x - 2013 )( 5x - 1 ) = 0

⇔ \(\orbr{\begin{cases}x-2013=0\\5x-1=0\end{cases}}\)

⇔ \(\orbr{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}\)

a) \(\left(2x-1\right)\left(2x+1\right)-4.\left(x^2+x\right)=16\)

\(4x^2-1-4x^2-4x=16\)

\(-1-4x=16\)

\(-4x=17\)

\(x=-\frac{17}{4}\)

b) \(5x\left(x-2013\right)-x+2013=0\)

\(\left(x-2013\right)\left(5x-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-2013=0\\5x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}}\)

a) \(\left(2x-1\right)\left(2x+1\right)-4\left(x^2+x\right)=16\)

\(\Leftrightarrow4x^2-1-4x^2-4x=16\)

\(\Leftrightarrow-1-4x=16\)\(\Leftrightarrow4x=-17\)\(\Leftrightarrow x=-\frac{17}{4}\)

Vậy \(x=-\frac{17}{4}\)

b) \(5x\left(x-2013\right)-x+2013=0\)

\(\Leftrightarrow5x\left(x-2013\right)-\left(x-2013\right)=0\)

\(\Leftrightarrow\left(x-2013\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2013=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2013\\5x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}\)

Vậy \(x=2013\)hoặc \(x=\frac{1}{5}\)

(x - 2013)^x+1 - (x - 2013)^x+10 = 0

=> (x - 2013)^x+1 = (x - 2013)^x+10 (1)

Mà 2 vế có x - 2013 chung (2)

=> x + 1 = x + 10

=> x - x = 10 - 1

=> 0 = 9 (vô lí)

=> x không tồn tại

Bạn kia giải sai rồi

Giải:

Ta có:

\(\left(x-2013\right)^{x+1}-\left(x-2013\right)^{x+10}=0\)

\(\Leftrightarrow\left(x-2013\right)^{x+1}\left[1-\left(x-2013\right)^9\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2013\right)^{x+1}=0\\1-\left(x-2013\right)^9=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-2013=0\\\left(x-2013\right)^9=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2013\\x=2014\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=2013\\x=2014\end{matrix}\right.\)