Cho ac+by=5c, by+cz=5a, cz+ac=5b.
Tính M=1/x+5 + 1/y+5 + 1/z+5
Ax+by=5c, by+cz=5a, cz+ax-5b. Tính giá trị biểu thức M= 1/x+5 + 1/y+5 + 1/z+5?
ax + by = 5c (1); by + cz = 5a (2); cz + ax = 5b (3);
Lấy (1) - (2) + (3) về theo vế có : 2ax = - 5a + 5b + 5c => 2a(x + 5) = 5(a + b + c)
=> 1/(x + 5) = 2a/5(a + b + c) (4)
Tương tự :
1/(y + 5) = 2b/5(a + b + c) (5)
1/(z + 5) = 2c/5(a + b + c) (6)
Từ (4) + (5) + (6) :
M = 1/(x + 5) + 1/(y + 5) + 1/(z + 5) = 2/5
1 ,
Cho x2 - 9x + 1 = 0 . Tính V = \(\dfrac{x4+x2+1}{5x2}\)
2.
Cho : ax + by = 5c ( 1 )
by + cz = 5a ( 2 )
cz + ax = 5b ( 3 )
Tính M = \(\dfrac{1}{x+5}\)+ \(\dfrac{1}{y+5}\)+\(\dfrac{1}{z+5}\)
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:....
\(x^2-9x+1=0\Rightarrow x=9x-1\)
Ta có:
\(V=\dfrac{x^4+x^2+1}{5x^2}\)
\(=\dfrac{\left(x^2\right)^2+x^2+1}{5x^2}\)
\(=\dfrac{\left(9x-1\right)^2+9x-1+1}{5\left(9x-1\right)}=\dfrac{81x^2-18x+1+9x-1+1}{5\left(9x-1\right)}=\dfrac{81\left(9x-1\right)-9x+1}{5\left(9x-1\right)}=\dfrac{729x-81-9x+1}{5\left(9x-1\right)}\)\(=\dfrac{720x-80}{5\left(9x-1\right)}=\dfrac{80\left(9x-1\right)}{5\left(9x-1\right)}=16\)
Cho \(ax+by=5c\)
\(by+cz=5a\)
\(cz+ax=5b\)
Và \(x\ne-5\), \(y\ne-5\), \(z\ne-5\)
\(a+b+c\ne0\)
Tính
\(A=\frac{1}{x+5}+\frac{1}{y+5}+\frac{1}{z+5}\)
Biet ax+by+cz=0 va a+b+c=1/2003
Tinh ax^2+by^2+cz^2 / bc(y-z)^2+ac(x-z)^2+ab(x-y)^2
cho x = by + cz , y= ax + cz , z = ax + by , x + y + z khác 0
tính Q = 1/(a+1) + 1/(1+b) + 1/(1+c)
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 \(Q=2.\)
Tim x toa man: |x-22|+|x-3|+|x-2017|=2014
Cho các số a,b,c,x,y,z t/m:x=by+cz ;y=cz+ax ;z=ax+by và a,b,c khác 0.Tính M=1/1+a+1/1+b+1/1+c
Cho \(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\\ax^5+by^5=cz^5\end{cases}}\)
Chứng minh \(\sqrt[5]{ax^4+by^4+cz^4}=\sqrt[5]{a}+\sqrt[5]{b}+\sqrt[5]{c}\)
cho ax+by+cz=0 và a+b+c =2019.Tính
A=bc(x-y)^2+ac(x-z)^2+ab(x-y)^2/ax^2+by^2+cz^2
Cho x= by+cz , y= ax+cz z= ax +by và x+ +y + z =0
Tính Q = 1/a+1 + 1/b+1 + 1/c+1
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
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
\(x+y+z=0\) sao tính được, Sửa lại thành: \(x+y+z\)khác \(0\)
Ta có: \(x+y=by+cz+ax+cz=2cz+z\Leftrightarrow2cz=x+y-z\Leftrightarrow c=\frac{x+y-z}{2z}\Leftrightarrow c+1=\)\(\frac{x+y+z}{2z}\Leftrightarrow\frac{1}{c+1}=\frac{2z}{x+y+z}\)(1)
Tương tự, ta có: \(\frac{1}{a+1}=\frac{2x}{x+y+z}\)(2)\(;\frac{1}{b+1}=\frac{2y}{x+y+z}\)(3)
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 \(Q=2\)