\(E=\frac{1}{1\times101}+\frac{1}{2\times102}+\frac{1}{3\times103}+.........+\frac{1}{10\times110}\) và \(F=\frac{1}{1\times11}+\frac{1}{2\times12}+\frac{1}{3\times13}+.......+\frac{1}{100\times110}\)
tìm tỉ số \(\frac{E}{F}\)
\(A=\frac{1}{1\times300}+\frac{1}{2\times301}+...+\frac{1}{101\times400}\)
\(B=\frac{1}{1\times102}+\frac{1}{2\times103}+....+\frac{1}{298\times399}+\frac{1}{299\times400}\)
\(Tính:\frac{A}{B}\)
Giups mk voi mk can gap
A=\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
299.A= 299.(\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\))
299.A=\(\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101.400}=\frac{1}{1}-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\)
A= \(=\frac{1}{299}\left(1+\frac{1}{2}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)
Tương tự
B=\(\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{400}\right)\)
B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)
Hai dấu ngoặc ở biểu thức A và biểu thức B như nhau
Vậy \(A:B=\frac{1}{299}:\frac{1}{101}=\frac{101}{299}\)
Bài 1: Thực hiện phép tính:
a) \(\left(\frac{3}{5}\right)^2-[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2]-\left(10\times12-2014\right)^0\)
b)\(|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
c) \(\frac{\left(-5\right)^{32}\times20^{43}}{\left(-8\right)^{29}\times125^{25}}\)
Bài 2: Tìm x biết
a)\(\frac{2}{3}-\left(\frac{3}{4}-x\right)=\sqrt{\frac{1}{9}}\)
b) \(\left(\frac{1}{2}-x\right)^2=\left(-2\right)^2\)
c) \(|3x+\frac{1}{2}|-\frac{2}{3}=1\)
d) \(3^{2x+1}=81\)
Bài 2:
a)
\(\frac{2}{3}-(\frac{3}{4}-x)=\sqrt{\frac{1}{9}}=\frac{1}{3}\)
\(\frac{3}{4}-x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\)
b)
\((\frac{1}{2}-x)^2=(-2)^2=2^2\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}-x=-2\\ \frac{1}{2}-x=2\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{5}{2}\\ x=\frac{-3}{2}\end{matrix}\right.\)
c)
\(|3x+\frac{1}{2}|-\frac{2}{3}=1\)
\(|3x+\frac{1}{2}|=\frac{2}{3}+1=\frac{5}{3}\)
\(\Rightarrow \left[\begin{matrix} 3x+\frac{1}{2}=\frac{5}{3}\\ 3x+\frac{1}{2}=-\frac{5}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{7}{18}\\ x=\frac{-13}{18}\end{matrix}\right.\)
d)
\(3^{2x+1}=81=3^4\)
\(\Rightarrow 2x+1=4\Rightarrow x=\frac{3}{2}\)
Bài 1:
a)
\((\frac{3}{5})^2-[\frac{1}{3}:3-\sqrt{16}.(\frac{1}{2})^2]-(10.12-2014)^0\)
\(=\frac{9}{25}-(\frac{1}{9}-1)-1\)
\(=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}\)
b)
\(|-\frac{100}{123}|:(\frac{3}{4}+\frac{7}{12})+\frac{23}{123}:(\frac{9}{5}-\frac{7}{15})\)
\(=\frac{100}{123}:\frac{4}{3}+\frac{23}{123}:\frac{4}{3}=(\frac{100}{123}+\frac{23}{123}):\frac{4}{3}=1:\frac{4}{3}=\frac{3}{4}\)
c)
\(\frac{(-5)^{32}.20^{43}}{(-8)^{29}.125^{25}}=\frac{5^{32}.(2^2.5)^{43}}{(-2)^{3.29}.(5^3)^{25}}=\frac{5^{32}.2^{86}.5^{43}}{-2^{87}.5^{75}}\)
\(=\frac{5^{32+43}.2^{86}}{-2^{87}.5^{75}}=\frac{5^{75}.2^{86}}{-2^{87}.5^{75}}=-\frac{1}{2}\)
Bài 2:
a)
\(\frac{2}{3}-(\frac{3}{4}-x)=\sqrt{\frac{1}{9}}=\frac{1}{3}\)
\(\frac{3}{4}-x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\)
b)
\((\frac{1}{2}-x)^2=(-2)^2=2^2\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}-x=-2\\ \frac{1}{2}-x=2\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{5}{2}\\ x=\frac{-3}{2}\end{matrix}\right.\)
c)
\(|3x+\frac{1}{2}|-\frac{2}{3}=1\)
\(|3x+\frac{1}{2}|=\frac{2}{3}+1=\frac{5}{3}\)
\(\Rightarrow \left[\begin{matrix} 3x+\frac{1}{2}=\frac{5}{3}\\ 3x+\frac{1}{2}=-\frac{5}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{7}{18}\\ x=\frac{-13}{18}\end{matrix}\right.\)
d)
\(3^{2x+1}=81=3^4\)
\(\Rightarrow 2x+1=4\Rightarrow x=\frac{3}{2}\)
Bài 1: Thực hiện phép tính:
a) \(\left(\frac{3}{5}\right)^2-[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2]-\left(10\times12-2014\right)^0\)
b) \(|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
c) \(\frac{\left(-5\right)^{32}\times20^{43}}{\left(-8\right)^{29}\times125^{25}}\)
Bài 2: Tìm x biết
a) \(\frac{2}{3}-\left(\frac{3}{4}-x\right)=\sqrt{\frac{1}{9}}\)
b) \(\left(\frac{1}{2}-x\right)^2=\left(-2\right)^2\)
c) \(|3x+\frac{1}{2}|-\frac{2}{3}=1\)
d) \(3^{2x+1}=81\)
Bài 1
\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)
\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)
\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)
\(=\frac{9}{25}+\frac{8}{9}-1\)
\(=\frac{56}{225}\)
\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)
\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)
\(=1:\frac{4}{3}=\frac{3}{4}\)
Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v
\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)
\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)
\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)
\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)
\(=-\frac{1}{2}\)
Bài 2
\(a,\frac{2}{3}-\left(\frac{3}{4}-x\right)=\sqrt{\frac{1}{9}}\)
\(\frac{2}{3}-\left(\frac{3}{4}-x\right)=\frac{1}{3}\)
\(\frac{3}{4}-x=\frac{2}{3}-\frac{1}{3}\)
\(\frac{3}{4}-x=\frac{1}{3}\)
\(x=\frac{3}{4}-\frac{1}{3}\)
\(x=\frac{5}{12}\)
\(b,\left(\frac{1}{2}-x\right)^2=\left(-2\right)^2\)
\(\Rightarrow\frac{1}{2}-x=-2\) hoặc \(\frac{1}{2}-x=2\)
\(\Rightarrow x=\frac{1}{2}-\left(-2\right)\) \(x=\frac{1}{2}-2\)
\(\Rightarrow x=\frac{5}{2}\) \(x=-\frac{3}{2}\)
\(c,|3x+\frac{1}{2}|-\frac{2}{3}=1\)
\(|3x+\frac{1}{2}|=1+\frac{2}{3}\)
\(|3x+\frac{1}{2}|=\frac{5}{3}\)
\(\Rightarrow\orbr{\begin{cases}3x+\frac{1}{2}=\frac{5}{3}\\3x+\frac{1}{2}=-\frac{5}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}3x=\frac{5}{3}-\frac{1}{2}=\frac{7}{6}\\3x=-\frac{5}{3}-\frac{1}{2}=-\frac{13}{6}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{18}\\x=-\frac{13}{18}\end{cases}}}\)
\(d,3^{2x+1}=81\)
\(\Rightarrow3^{2x+1}=3^4\)
\(\Rightarrow2x+1=4\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\frac{3}{2}\)
cho A =\(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+\frac{1}{3\cdot103}+....+\frac{1}{25\cdot125}\)
B = \(\frac{1}{1\cdot26}+\frac{1}{2\cdot27}+\frac{1}{3\cdot28}+...+\frac{1}{100\cdot125}\)
Trong đó A có 25 số hạng , B có 100 số hạng . tính A:B
thực hiện phép tính
A=\(\frac{3}{1\times5}+\frac{3}{5\times10}+....+\frac{3}{100\times105}\)
B=
\(\dfrac{5}{1\times3\times5}+\dfrac{5}{3\times5\times7}+...+\dfrac{5}{99\times101\times103}\)
Ta có:
\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)
\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)
\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)
\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)
1\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+............+\frac{1}{99\times100}+\frac{1}{100\times101}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}.\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
Cho A=\(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+\frac{1}{3\cdot103}+....+\frac{1}{25\cdot125}\)
B=\(\frac{1}{1\cdot26}+\frac{1}{2\cdot27}+\frac{1}{3\cdot28}+...+\frac{1}{100\cdot125}\)
Trong đó A có 25 số hạng, B có 100 số hạng. Tìm thương A:B
Tính:
\(A=\frac{1^2}{1\times2}\times\frac{2^2}{2\times3}\times\frac{3^2}{3\times4}\times...\times\frac{99^2}{99\times100}\times\frac{100^2}{100\times101}\)
Ta có:
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1}{2}.\frac{4}{6}.\frac{9}{12}....\frac{9801}{9900}.\frac{10000}{10100}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}=\frac{1.2.3...99.100}{2.3.4...100.101}=\frac{1}{101}\)(Tối giản)
