x+16/9=y-25/16=z+9/25 và 9-x/7+11-x/9=2.tìm x+y+z
cho x+16 / 9 = y -25 / 16 = z + 9 / 25 và 9-x / 7 + 11-x / 9 = 2 . Tìm x+ y+z
cho x+16/9=y-25/16=z+9/25 va (9-x)/7+(11-x)/9=2.Tinh x+y+z
Cho x+16/9 = y-25/16 = z+9/25 và 9-x/7 + 11-x/9=2. Khi đó x+y+z= .....
Cho \(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}và\dfrac{9-x}{7}+\dfrac{11-x}{9}=2\).Tìm x+y+z
theo bài ra ta có:
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{x+16+y-25+z+9}{9+16+25}=\dfrac{x+y+z}{50}\\ \Rightarrow\dfrac{x+16}{9}=\dfrac{x+y+z}{50}\left(1\right)\)ta lại có:
\(\dfrac{9-x}{7}+\dfrac{11-x}{9}=2\\ \Rightarrow\dfrac{7+2-x}{7}+\dfrac{9+2-x}{9}=2\\ \Rightarrow\left(1+\dfrac{2-x}{7}\right)+\left(1+\dfrac{2-x}{9}\right)=2\\ \Rightarrow\left(1+1\right)+\left(\dfrac{2-x}{7}+\dfrac{2-x}{9}\right)=2\\ \Rightarrow2+\left(2-x\right)\left(\dfrac{1}{7}+\dfrac{1}{9}\right)=2\\ \Rightarrow\left(2-x\right)\left(\dfrac{1}{7}+\dfrac{1}{9}\right)=0\\ \Rightarrow2-x=0\\ \Rightarrow x=2\)
thay x = 2 vào 1 ta có:
\(\Rightarrow\dfrac{2+16}{9}=\dfrac{x+y+z}{50}\\ \Rightarrow\dfrac{18}{9}=\dfrac{x+y+z}{50}\\ \Rightarrow2=\dfrac{x+y+z}{50}\\ \Rightarrow x+y+z=2.50\\ \Rightarrow x+y+z=100\)
vậy x + y + z = 100
Cho \(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\) và \(\frac{9-x}{7}+\frac{11-x}{9}=2\).Tính x+y+z?
Từ \(\frac{9-x}{7}+\frac{11-x}{9}=2\)
\(=>\frac{9-x}{7}+\frac{11-x}{9}-2=0\)
\(=>\frac{9-x}{7}+\frac{11-x}{9}-1-1=0\)
\(=>\left(\frac{9-x}{7}-1\right)+\left(\frac{11-x}{9}-1\right)=0\)
\(=>\frac{2-x}{7}+\frac{2-x}{9}=0=>\left(2-x\right).\left(\frac{1}{7}+\frac{1}{9}\right)=0\)
Vì \(\frac{1}{7}+\frac{1}{9}\) khác 0=>2-x=0=>x=2
Theo T/c dãy tỉ số=nhau:
\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{x+16+y-25+z+9}{9+16+25}\)\(=\frac{\left(x+y+z\right)+\left(16-25+9\right)}{9+16+25}=\frac{x+y+z}{50}\)
Thay x=2 vào \(\frac{x+16}{9}=>\frac{2+16}{9}=\frac{x+y+z}{50}=>\frac{x+y+z}{50}=2=>x+y+z=100\)
Vậy x+y+z=100
Cho \(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\)và\(\frac{9-x}{7}+\frac{11-x}{9}=2\)
Tìm x,y,z
Cho \(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\)và \(\frac{9-x}{7}+\frac{11-x}{9}=2\),Tính x + y + z
Ta có : \(\frac{9-x}{7}=\frac{11-x}{9}=1+\frac{2-x}{7}+1+\frac{2-x}{9}=2=>\left(2-x\right)\left(\frac{1}{7}+\frac{1}{9}\right)=0=>2-x=0=>x=2\)
Thế vào tìm đc y và z rồi ra x+y+z nha bạn
Cho \(A=\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\) và \(\frac{9-x}{7}+\frac{11-x}{9}=2\)
tìm x+y+z
Cho \(\frac{x+16}{9}\)=\(\frac{y-25}{16}\)=\(\frac{z+9}{25}\)và \(\frac{9-x}{7}\)+\(\frac{11-x}{9}\)= 2.Tìm x+y+z.