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\(5-\dfrac{2}{3}-\dfrac{14}{15}+\dfrac{1}{35}-\dfrac{62}{63}-\dfrac{98}{99}-\dfrac{142}{143}\)
\(=5-\left(1-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{15}\right)+\dfrac{1}{35}-\left(1-\dfrac{1}{63}\right)-\left(1-\dfrac{1}{99}\right)-\left(1-\dfrac{1}{143}\right)\)
\(=5-1+\dfrac{1}{1\cdot3}-1+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}-1+\dfrac{1}{7\cdot9}-1+\dfrac{1}{9\cdot11}-1+\dfrac{1}{11\cdot13}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=1-\dfrac{1}{13}=\dfrac{12}{13}\)
a) \((-2).(-2).(-2) =4.(-2) = -8\)
b) \((-0,5).(-0,5) = 0,25\)
c)
\(\begin{array}{l}\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}\\ = \frac{{1.1.1.1}}{{2.2.2.2}}\\ = \frac{1}{{16}}\end{array}\)
a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10
=1/2x^5-3x^4-5x^3
b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x
=-15x^5+12x^4-9x^3+9x^2
c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)
d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
\(A=3\div\left(\frac{-3}{2}\right)^2+\frac{1}{9}\times\sqrt{36}\)
\(\Rightarrow A=3\div\frac{9}{4}+\frac{1}{9}\times6\)
\(\Rightarrow A=\frac{4}{3}+\frac{1}{3}\)
\(\Rightarrow A=\frac{5}{3}\)
P/s : Nếu mình tính nhầm thì bỏ qua nhé bạn :P
GIẢI
A = 3 : ( -3/2)^2 + 1/9 +\(\sqrt{ }\)36
A = 3 : ( -9/4 ) +1/9 × 6
A = 3 × 4/9 + 1/9 × 6
A = 4/3 + 1/9 × 6
A = 4/3 + 6/9
A = 4/3 + 2/3
A = 6/3
A = 2
HỌC TỐT ^_^
a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
a) Đề thiếu nhé. sửa đề:
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{195}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(=1-\frac{1}{15}\)
\(=\frac{14}{15}\)