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2 tháng 1 2019

Ghi đề nhầm rồi bạn ơi

2 tháng 1 2019

Nhầm chỗ nào

5 tháng 7 2017

Từ (1); (2) và (3) ta được:

\(ax+by+by+cz+cz+ax=5a+5b+5c\)

\(\Leftrightarrow2\left(ax+by+cz\right)=5\left(a+b+c\right)\)

\(\Rightarrow a+b+c=\dfrac{2\left(ax+by+cz\right)}{5}\)

Ta có:

\(ax+by=5a\)

\(\Leftrightarrow ax+by+cz=5c+cz\)

\(\Leftrightarrow ax+by+cz=c\left(z+5\right)\)

\(\Rightarrow\dfrac{1}{z+5}=\dfrac{c}{ax+by+cz}\) (3)

Tượng tự ta có:

\(\dfrac{1}{x+5}=\dfrac{a}{ax+by+cz}\) (4)

\(\dfrac{1}{y+5}=\dfrac{b}{ax+by+cz}\)(5)

Từ (3);(4)và (5) \(\Rightarrow\dfrac{1}{x+5}+\dfrac{1}{y+5}+\dfrac{1}{z+5}=\dfrac{a+b+c}{ax+by+cz}\)

\(=\dfrac{\dfrac{2\left(ax+by+cz\right)}{5}}{ax+by+cz}=\dfrac{2}{5}\)

Vậy:....

5 tháng 7 2017

\(x^2-9x+1=0\Rightarrow x=9x-1\)

Ta có:

\(V=\dfrac{x^4+x^2+1}{5x^2}\)

\(=\dfrac{\left(x^2\right)^2+x^2+1}{5x^2}\)

\(=\dfrac{\left(9x-1\right)^2+9x-1+1}{5\left(9x-1\right)}=\dfrac{81x^2-18x+1+9x-1+1}{5\left(9x-1\right)}=\dfrac{81\left(9x-1\right)-9x+1}{5\left(9x-1\right)}=\dfrac{729x-81-9x+1}{5\left(9x-1\right)}\)\(=\dfrac{720x-80}{5\left(9x-1\right)}=\dfrac{80\left(9x-1\right)}{5\left(9x-1\right)}=16\)

22 tháng 10 2016

Vì \(x=by+cz\)

\(\Rightarrow by=x-cz\)

Mà \(z=ax+by\)

\(\Rightarrow by=z-ax\)

\(\Rightarrow x-cz=z-ax\left(=by\right)\)

\(\Rightarrow x+ax=z+cz\)

\(\Rightarrow x\left(a+1\right)=z\left(c+1\right)\)

Cũng có :

\(z=ax+by\)

\(\Rightarrow ax=z-by\)

\(y=ax+cz\)

\(\Rightarrow ax=y-cz\)

\(\Rightarrow z-by=y-cz\left(=ax\right)\)

\(\Rightarrow z+cz=y+by\)

\(\Rightarrow z\left(c+1\right)=y\left(b+1\right)\)

\(\Rightarrow x\left(a+1\right)=y\left(b+1\right)=z\left(c+1\right)\)

Đặt \(x\left(a+1\right)=y\left(b+1\right)=z\left(c+1\right)=k\)

\(\Rightarrow3k=x\left(a+1\right)+y\left(b+1\right)+z\left(c+1\right)\)

Có :

\(Q=\frac{1}{a+1}+\frac{1}{1+b}+\frac{1}{c+1}\)

\(=\frac{x}{x\left(a+1\right)}+\frac{y}{y\left(b+1\right)}+\frac{z}{z\left(c+1\right)}\)

\(=\frac{x}{k}+\frac{y}{k}+\frac{z}{k}\)

\(=\frac{x+y+z}{k}\)

\(=\frac{3\left(x+y+z\right)}{3k}\)

Mà \(3k=x\left(a+1\right)+y\left(b+1\right)+z\left(c+1\right)\)

\(\Rightarrow Q=\frac{3\left(x+y+z\right)}{x\left(a+1\right)+y\left(b+1\right)+z\left(c+1\right)}\)

\(=\frac{3\left(x+y+z\right)}{xa+x+by+y+zc+z}\)

\(=\frac{3\left(x+y+z\right)}{\left(x+y+z\right)+\left(xa+by+zc\right)}\)

\(=\frac{3\left(x+y+z\right)}{\left(x+y+z\right)+\frac{1}{2}\left[\left(xa+by\right)+\left(xa+zc\right)+\left(by+zc\right)\right]}\)

Có \(x+y+z=\left(ax+by\right)+\left(by+cz\right)+\left(ax+cz\right)\)

\(\Rightarrow Q=\frac{3\left(x+y+z\right)}{\left(x+y+z\right)+\frac{1}{2}\left(x+y+z\right)}\)

\(=\frac{3\left(x+y+z\right)}{\frac{3}{2}\left(x+y+z\right)}\)

\(=\frac{3}{\frac{3}{2}}\)

\(=2\)

Vậy \(Q=2.\)

29 tháng 3 2017

Tim x toa man: |x-22|+|x-3|+|x-2017|=2014

27 tháng 7 2018

1 la sai ; 2 cung sai ; xin loi cho ming ting xiu ; aaaaa! 3 la ......................................sai; chan chan 4 la ..............................................................................................d...........................sai ; 1000000000000000000000000000000000000000000000000000000000000000000000000000 la ..................................................................................................sai

27 tháng 7 2018

x+y+z=0 sao tính được. sửa đề: x+y+z khác 0

Ta có: \(x+y=by+cz+ax+cz=2cz+z\Leftrightarrow2cz=x+y-z\Leftrightarrow c=\frac{x+y-z}{2z}\Leftrightarrow c+1=\frac{x+y+z}{2z}\Leftrightarrow\frac{1}{c+1}=\frac{2z}{x+y+z}\left(1\right)\)

Tương tự, ta có: \(\frac{1}{a+1}=\frac{2x}{x+y+z}\left(2\right);\frac{1}{b+1}=\frac{2y}{x+y+z}\left(3\right)\)

Cộng (1),(2),(3) vế với vế ta được:

\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2\left(x+y+z\right)}{x+y+z}=2\) hay Q = 2

Vậy Q=2

2 tháng 12 2016

cộng 3 cái lại nhe bạn =))

2 tháng 12 2016

Có nhiều cách làm bài này.

Có \(2a+2b+2c=by+cz+a.x+cz+a.x+by\)

\(2\left(a+b+c\right)=2\left(a.x+by+cz\right)\)

\(\Rightarrow a+b+c=a.x+by+cz\)

  • \(a+b+c=a.x+\left(by+cz\right)=a.x+2.a=a\left(x+2\right)\)

\(\Rightarrow\frac{1}{x+2}=\frac{a}{a+b+c}\)

  • \(a+b+c=\left(a.x+by\right)+cz=2c+cz=c\left(z+2\right)\)

\(\Rightarrow\frac{1}{z+2}=\frac{c}{a+b+c}\)

  • \(a+b+c=by+\left(a.x+cz\right)=by+2b=b\left(y+2\right)\)

\(\Rightarrow\frac{1}{y+2}=\frac{b}{a+b+c}\)

\(\Rightarrow M=\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{a+b+c}{a+b+c}=1\)

Vậy ...