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a) Câu này đề chưa rõ rành lắm nên mk k làm nhé.
b) Đặt \(A=1+3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A=3+3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-1\)
\(\Rightarrow A=\frac{3^{101}-1}{2}\)
a) \(\frac{2015x\left(1-\frac{1}{2016}+\frac{1}{2017}\right)}{5x\left(1-\frac{1}{2016}+\frac{1}{2017}\right)}\)
\(=\frac{2015x}{5x}\)
\(=\frac{2015}{5}=403\)
\(1000-\left\{\left(-5\right)^3.\left(-2\right)^3-11.[7^2-5.2^3+8.\left(11^2-121\right)]\right\}\)
=\(1000-\left\{[\left(-5\right).\left(-2\right)]^3-11.[7^2-5.2^3+2^3.\left(11^2-11^2\right)]\right\}\)
= \(1000-\left\{1000-11.[7^2-2^3.\left(5+0\right)]\right\}\)
= \(1000-[1000-11.\left(7^2-2^3.5\right)\)
= \(1000-[1000-11.\left(49-40\right)]\)
= \(1000-\left(1000-11.9\right)\)
= \(1000-\left(1000-99\right)=1000-1000+99\)
= 0 + 99 = 99
\(A=1500-\left\{5^3.2^3-11.\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
\(A=1500-\left\{125.8-11.\left[49-5.8+8\left(121-121\right)\right]\right\}\)
\(A=1500-\left\{1000-11\left[49-40+8.0\right]\right\}\)
\(A=1500-\left\{1000-11.9\right\}\)
\(A=1500-\left\{1000-99\right\}\)
\(A=1500-901=599\)
Ta tính các số mũ thành số hết
\(A=-1500-\left\{125.8-11.\left[49-40+8\left(121-121\right)\right]\right\}.2\)
\(A=-1500-\left\{1000-11.\left(9+0\right)\right\}.2\)
\(A=-1500-\left(1000-99\right).2\)
\(A=-1500-901.2\)
\(A=-1500-1802=-3302\)
d) ( 4x+5):3-121:11=4
(4x+5):3-11=4
(4x+5):3=4+11
(4x+5):3=15
(4x+5)=15x3
(4x+5)=45
4x=45-5
4x=40
x=40:4
x=10
1: \(=-\dfrac{7}{80}\cdot\dfrac{4}{7}-\dfrac{2}{9}:\dfrac{16}{3}+\dfrac{5}{24}\cdot\left(\dfrac{-50+38}{15}\right)^2\)
\(=\dfrac{-1}{20}-\dfrac{2}{9}\cdot\dfrac{3}{16}+\dfrac{5}{24}\cdot\dfrac{16}{25}\)
\(=\dfrac{-1}{20}-\dfrac{1}{24}+\dfrac{2}{15}\)
\(=\dfrac{-6-5+16}{120}=\dfrac{5}{120}=\dfrac{1}{24}\)
2: \(=1500-\left\{10^3-11\cdot\left[49-5\cdot8\right]\right\}\)
\(=1500-\left\{1000-11\cdot9\right\}\)
\(=1500-1000+99=599\)
\(2017-\left\{5^2.2^2-11\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
Đặt : \(A=2017-\left\{5^2.2^2-11\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-5.8+8\left(121-121\right)\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-5.8+0\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-40\right]\right\}\)
\(A=2017-\left\{25.4-11.9\right\}\)
\(A=2017-\left\{25.4-99\right\}\)
\(A=2017-\left\{100-99\right\}\)
\(A=2017-1=2016\)
Vậy A = 2016