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Đây là toán lớp 6.
=>1/5B= 4/7.5.31 +6/7.5.41+9/5.10.41+7/5.10.57+13/57.5.14
=>1/5B=4/31.35+6/35.41+....+13/57.70
=>1/5B=1/31-1/35+1/35-1/41+...+1/57-1/70
=>1/5B=1/31-1/70
=>1/5B=39/2170
=>B=39/2170:1/5
=>B=39/424
Ta có:
\(\frac{B}{5}=\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}+\frac{13}{57.70}\)
\(=\frac{35-31}{35.31}+\frac{41-35}{35.41}+\frac{50-41}{50.41}+\frac{57-50}{50.57}+\frac{70-57}{57.70}\)
\(=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}+\frac{1}{57}-\frac{1}{70}\)
\(=\frac{1}{31}-\frac{1}{70}\)
\(\rightarrow B=5\cdot\left(\frac{1}{31}-\frac{1}{70}\right)\)
\(=5\cdot\frac{39}{2170}\)
\(=\frac{39}{434}\)
Vậy B=\(\frac{39}{434}\)
B1:
Ta có: a - b = ab => a = ab + b = b(a + 1)
Thay a = b(a + 1) vào a - b = a : b ta có: \(a-b=\frac{b\left(a+1\right)}{b}=a+1\)
=> a - b = a + 1 => a - a - b = 1 => -b = 1 => b = -1
Lại có: ab = a - b
<=> a x (-1) = a - (-1) <=> -a = a + 1 <=> -a - a = 1 <=> -2a = 1 <=> a = -1/2
Vậy...
B2:
a, \(3y\left(y-\frac{2}{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3y=0\\y-\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0\\y=\frac{2}{5}\end{cases}}}\)
b, \(7\left(y-1\right)+2y\left(y-1\right)=0\)
\(\Rightarrow\left(y-1\right)\left(7+2y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\7+2y=0\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\2y=7\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\y=\frac{7}{2}\end{cases}}\)
B3: \(K=\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}\)
\(K=\left(-\frac{2}{3}+\frac{1}{6}\right)+\left(\frac{3}{4}-\frac{2}{5}\right)\)
\(K=\left(\frac{-4}{6}+\frac{1}{6}\right)+\left(\frac{15}{20}-\frac{8}{20}\right)\)
\(K=\frac{-1}{2}+\frac{7}{20}=\frac{-10}{20}+\frac{7}{20}=\frac{-3}{20}\)
thương A chia B là \(\frac{A}{B}\)
ta có :
\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
Ta có:
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)
=> \(\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)
=> \(\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
+) Nếu a + b + c = 0 => a + b = -c; b + c = -a; c + a = -b
=> \(\frac{a+b}{c}=-1\);\(\frac{b+c}{a}=-1\); \(\frac{c+a}{b}=-1\)
=> M = (-1)3 = -1
+) Nếu a + b + c khác 0 => a = b = c => a + b = 2c; b + c = 2a; c + a = 2b
=> M \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=2.2.2=8\)
Vậy M = -1 hoặc M = 8
\(B=\frac{3^{12}.13+3^{12}.3}{3^{11}.2^{24}}\)
\(B=\frac{3^{12}.\left(13+3\right)}{3^{11}.2^{24}}\)
\(B=\frac{3^{12}.16}{3^{11}.2^{24}}\)
\(B=\frac{3^{12}.2^4}{3^{11}.2^{24}}\)
\(B=\frac{3}{2^{20}}\)
=>\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
=>\(\left(x+2\right).\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)
vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\ne0\)
=>x+2=0 =>x=-2
Vậy x=-2
ta co: 6x-2y=x+y(nhan cheo)
\(\Rightarrow\)5x=3y
\(\Rightarrow\)x/y=3/5
\(B=\frac{20}{35.31}+\frac{30}{35.41}+\frac{45}{50.41}+\frac{35}{50.57}+\frac{65}{57.70}\)
\(B=5.\left(\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}+\frac{13}{57.70}\right)\)
\(B=5.\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}+\frac{1}{57}-\frac{1}{70}\right)\)
\(B=5.\left(\frac{1}{31}-\frac{1}{70}\right)\)
\(B=5.\frac{39}{2170}\)
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