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Ta có :
\(B=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\)
\(B=\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+\left(1+\frac{1}{27}\right)+...+\left(1+\frac{1}{3^{98}}\right)\)
\(B=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^{98}}\right)\)
\(B=97+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^{98}}\right)\)
gọi A là biểu thức trong ngoặc
Lại có :
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^{98}}\)
\(\Leftrightarrow A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
\(2A=1-\frac{1}{3^{98}}< 1\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{98}}}{2}< \frac{1}{2}< 1\)
\(\Rightarrow A< 1\)
\(\Rightarrow B< 97+1=98< 100\)
vậy \(B< 100\)
Bài đây tính nhanh nhé ミ★ʟuғғʏ☆мũ☆ʀơм★彡 chứ không phải quy đồng lên đâu :)
a) \(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(A=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=35-\frac{167}{32}=\frac{953}{32}\)
b) \(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}:\frac{-7}{3}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}\cdot\frac{-3}{7}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}\)
\(B=\frac{-3}{7}+\frac{17}{7}=\frac{14}{7}=2\)
c) \(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right)\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{2}{8}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=6\frac{3}{8}\cdot\frac{4}{5}=\frac{51}{8}\cdot\frac{4}{5}=\frac{51}{2}\cdot\frac{1}{5}=\frac{51}{10}\)
d) \(D=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)\cdot107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
\(M=1+\frac{1}{3}+1+\frac{1}{9}+1+\frac{1}{27}+...+1+\frac{1}{3^{98}}\)
\(=1.98+\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
Đặt A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{98}}\)
=>\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)
=>3A-A=2A=\(1-\frac{1}{3^{98}}\Rightarrow A=\frac{1-\frac{1}{3^{98}}}{2}< 1\)
=>M=98+A<98+1<99<100
=>đpcm
Bài 2:
a: =>x/7=1/21
=>x=1/3
c: =>x(3x-2)=0
=>x=0 hoặc x=2/3
Bài1:
a: \(=\left(-\dfrac{7}{3}\right)^{3-2}=\dfrac{-7}{3}\)
b: \(=\left(-\dfrac{4}{9}\right)^{1-3}=\left(-\dfrac{4}{9}\right)^{-2}=\dfrac{81}{16}\)
c: \(=\left(\dfrac{1}{5}\right)^{10-7}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}\)
\(5,\left(x\cdot0,5-\frac{3}{7}\right):\frac{1}{2}=1\frac{1}{7}\)
\(\Leftrightarrow x\cdot0,5:\frac{1}{2}-\frac{3}{7}:\frac{1}{2}=1\frac{1}{7}\)
\(\Leftrightarrow x-\frac{6}{7}=\frac{8}{7}\)
\(\Leftrightarrow x=2\)
\(6,x\cdot1,75=1\frac{3}{10}+45\%\)
\(\Leftrightarrow x\cdot\frac{7}{4}=\frac{13}{10}+\frac{9}{20}\)
\(\Leftrightarrow x\cdot\frac{7}{4}=\frac{7}{4}\)
\(\Leftrightarrow x=1\)
\(7,\frac{5-x}{15}+\frac{5}{12}-\frac{1}{8}=\frac{3}{8}\)
\(\Leftrightarrow\frac{5-x}{15}=\frac{3}{8}-\frac{5}{12}+\frac{1}{8}\)
\(\Leftrightarrow\frac{5-x}{15}=\frac{1}{12}\)
\(\Leftrightarrow60-12x=15\)
\(\Leftrightarrow12x=45\)
\(\Leftrightarrow x=\frac{15}{4}\)
\(8,\left|x-\frac{25}{33}\right|-\frac{3}{11}=\frac{2}{3}\)
\(\Leftrightarrow\left|\frac{x-25}{33}\right|=\frac{31}{33}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{25}{33}=\frac{31}{33}\\x-\frac{25}{33}=-\frac{31}{33}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{56}{33}\\x=-\frac{2}{11}\end{cases}}\)
\(9,-\frac{9}{8}+\frac{-3}{8}\cdot x=-\frac{1}{8}\)
\(\Leftrightarrow\frac{-9}{8}+\frac{-3}{8}\cdot x+\frac{1}{8}=0\)
\(\Leftrightarrow-1-\frac{3}{8}x=0\)
\(\Leftrightarrow\frac{3}{8}x=-1\)
\(\Rightarrow x=-\frac{8}{3}\)
a) \(\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-9}{4}\right):\frac{3}{7}\)
= \(\left(-\frac{3}{4}+\frac{2}{5}\right)\cdot\frac{7}{3}+\left(\frac{3}{5}+\frac{-9}{4}\right)\cdot\frac{7}{3}\)
= \(\left(-\frac{15}{20}+\frac{8}{20}\right)\cdot\frac{7}{3}+\left(\frac{12}{20}-\frac{45}{20}\right)\cdot\frac{7}{3}\)
= \(-\frac{7}{20}\cdot\frac{7}{3}-\frac{33}{20}\cdot\frac{7}{3}\)
=\(\frac{7}{3}\cdot\left(-\frac{7}{20}-\frac{33}{20}\right)\)
=\(\frac{7}{3}\cdot\left(-2\right)\)
=\(-\frac{14}{3}\)
phân số cuối sai thì phải