50+\(\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{67}\)
Bài 1 :Tìm a,b biết :
\(a+b=3.\left(a-b\right)=\)\(2\frac{a}{b}\)
Bài 2 :
\(A=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{5}+\frac{100}{67}+...+\frac{100}{98.99}+\frac{1}{99}\)
Bài 1 :
\(a+b=3.\left(a-b\right)=\)\(2\frac{a}{b}\)
\(\Rightarrow a+b=3.\left(a-b\right)\)
\(\Rightarrow a+b=3a-3b\)
\(\Rightarrow3a-3b-a-b=0\)
\(\Rightarrow2a-4b=0\)
\(\Rightarrow2.\left(a-2b\right)=0\)
\(\Rightarrow\hept{\begin{cases}a-2b=0\\a=2b\end{cases}}\)
Ta có : \(a+b=\frac{2a}{b}\)
Thay \(a=2b\) vào
\(\Rightarrow2b+b=\frac{2.23}{b}\)
\(\Rightarrow3b=\frac{4b}{b}\Rightarrow3b=4\)
\(\Rightarrow b=\frac{4}{3}\Rightarrow a=2.\frac{4}{3}=\frac{8}{3}\)
Vậy \(a=\frac{8}{3}\) và \(b=\frac{4}{3}\)
Chúc bạn học tốt ( -_- )
Bài 2 :
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{5}+\frac{100}{6.7}+...+\)\(\frac{100}{98.99}+\frac{1}{99}\)
\(B=\frac{100}{2}+\frac{100}{6}+\frac{100}{12}+\frac{100}{20}+\frac{100}{30}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{9900}\)
\(B=\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+\frac{100}{4.5}+\frac{100}{5.6}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{99.100}\)
\(B=100.\frac{100}{2}+\frac{100}{2}-\frac{1}{3}+\frac{100}{3}-\frac{100}{4}+\frac{100}{4}-\frac{100}{5}+\frac{100}{5}-\frac{100}{6}+\frac{100}{6}\)\(-\frac{100}{7}+...+\frac{100}{98}+\frac{100}{99}+\frac{100}{99}-1\)
\(B=100-1\)
\(B=99\)
Chúc bạn học tốt ( -_- )
Tính
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)
Tìm X:
\(\frac{50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}}{\left(2x-1\right)}\)=11
Tính tổng :
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)
\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{3}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{1}{99}\)
\(B=\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+\frac{100}{4.5}+\frac{100}{5.6}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{99.100}\)
\(B=100\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(B=100\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(B=100\left(1-\frac{1}{100}\right)\)
\(B=100.\frac{99}{100}=99\)
Tính tổng:
\(50+\frac{50}{3}+\frac{25}{3}+\frac{30}{4}+\frac{10}{3}+\frac{100}{6.7}+...+198.99\)
Tính tổng:
\(50+\frac{50}{3}+\frac{25}{3}+\frac{30}{4}+\frac{10}{3}+\frac{100}{6.7}+...+198
.
99\)
THỰC HIỆN PHÉP TÍNH SAU
50 + \(\frac{50}{3}\)+ \(\frac{25}{3}\)+ \(\frac{20}{4}\)+\(\frac{10}{3}\)+\(\frac{100}{6.7}\)+.........+ \(\frac{100}{98.99}\)+\(\frac{1}{99}\)
( GIẢI CHI TIẾT MÌNH LIKE CHO )
1 . Tìm x :
a ) \(\frac{1}{2}x+\frac{9}{4}-\frac{8}{20}=\frac{77}{20}\)
b ) \(\frac{28}{100}x+\frac{4}{25}+\frac{10}{4}=\frac{203}{50}\)
c ) \(\frac{100}{50}=\frac{1}{2}x\)
a)\(\frac{1}{2}x+\frac{9}{4}-\frac{8}{20}=\frac{77}{20}\)
<=>\(\frac{1}{2}x=\frac{77}{20}-\frac{9}{4}+\frac{8}{20}=\frac{77-45+8}{20}\)
<=>\(\frac{1}{2}x=\frac{40}{20}=2\)
<=>\(x=2:\frac{1}{2}=2.2\)
<=>x=4
Vậy x=4
TÍNH:
a)\(10.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0\)
b)\(\left(\frac{1}{3}\right)^{50}.\left(-9\right)^{25}-\frac{2}{3}:4\)
n=ghi lộn nhé !!
a)\(10.\sqrt{0,01.\sqrt{ }\frac{16}{9}}+3\sqrt{49-\frac{1}{6}}\sqrt{4}\)
a, \(10.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0\)
= 10 . 10 - \(\frac{1}{4}\) + 1
= 100 - \(\frac{1}{4}+1\)
= 99,75 + 1 = 100,75
b, \(\left(\frac{1}{3}\right)^{50}.\left(-9\right)^{25}-\frac{2}{3}:4\)
= \(\left(\frac{1}{9}\right)^{25}.\left(-9\right)^{25}-\frac{1}{6}\)
= \(\left(\frac{1}{9}.-9\right)^{25}-\frac{1}{6}\)
\(\left(-1\right)^{25}-\frac{1}{6}\)
= \(-1-\frac{1}{6}=\frac{-7}{6}\)