tính nhanh
9764/36615+36x85x20/25x84x34+2,2+3/19/133
9764/36615+36.85.20/25.84.34+2,2+3/19/133
Tính :
a) \(\frac{9764}{36615}+\frac{36.85.20}{25.84.34}+2,2+3\frac{19}{133}\)
b) \(1\frac{40404}{70707}+\frac{244.395-151}{244+295.243}+\frac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)
bài 5: so sánh A và B, biết:
A=36x85x20/25x84x34
B=30x63x65x8/117x200x49
A\(=\frac{36x85x20}{25x84x34}=\frac{2^2x3^2x5x17x2^2x5}{5^2x2^2x3x7x2x17}=\frac{6}{7}.\)
\(B=\frac{30x63x65x8}{117x200x49}=\frac{\left(2x3x5\right)\left(3^2x7\right)\left(5x13\right)2^3}{\left(3^2x13\right)\left(2^3x5^2\right)7^2}=\frac{6}{7}\)
=>A=B
tính hợp lý \(\frac{9764}{36615}+\frac{36\cdot85\cdot20}{25\cdot84\cdot34}+2\cdot2+3\frac{19}{133}\)
thuc hien phep tinh sau 1 cach hop li
9764/36615+36.85.20/25.84.34+22/10+3 19/133(la hon so)
b, 4/6+4/30+4/70+..+4/2.99.101
Tính tổng:
1/5+2/7+3/19+4/29...+-34/146+-74/133
Câu hỏi:
\(\frac{19}{X+Y}+\frac{19}{Y+Z}+\frac{19}{Z+X}=\frac{7xX}{Z+Y}+\frac{7xY}{X+Z}+\frac{7xZ}{X+Y}=\frac{133}{10}\)\(\frac{133}{10}\)
Tính X+Y+Z=?
Mình là chủ nhân của câu hỏi này lên các bạn hãy bỏ một phân số 133phaanf 10 ra ngoài
Tính A=x+y+z biết 19/x+y + 19/y+z + 19/z+x= 7x/y+z + 7x/z+x + 7x/x+y= 133/10
Tính x+y+z nếu :
\(\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{7x}{y+z}+\frac{7y}{x+y}+\frac{7z}{x+y}=\frac{133}{10}\)\(\frac{133}{10}\)
Cần Câu TL rõ Ràng
\(\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{133}{10}\)
\(\Rightarrow19\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{133}{10}\)
\(\Rightarrow\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{7}{10}\)
\(\frac{7x}{y+z}+\frac{7y}{z+x}+\frac{7z}{x+y}=\frac{133}{10}\)
\(\Rightarrow7\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{133}{10}\)
\(\Rightarrow\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{19}{10}\)
\(\Rightarrow\left(\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1\right)=\frac{19}{10}+3\)
\(\Rightarrow\left(\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}\right)=\frac{49}{10}\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)=\frac{49}{10}\)
\(\Rightarrow\left(x+y+z\right).\frac{7}{10}=\frac{49}{10}\)
\(\Rightarrow x+y+z=7\)
Vậy x + y + z = 7