tính
a, A= 2+2^2+2^3+....+2^10
b, B= 2+2^2+2^3+....+2^n
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C
23 tháng 5 2022
\(\left(\dfrac{1}{6}+4\right)\times\dfrac{3}{10}=\left(\dfrac{1}{6}+\dfrac{24}{6}\right)=\dfrac{25}{6}\times\dfrac{3}{10}=\dfrac{5}{4}\)
\(\left(\dfrac{1}{8}-\dfrac{1}{9}\right)\times\dfrac{3}{2}=\left(\dfrac{9}{72}-\dfrac{8}{72}\right)\times\dfrac{3}{2}=\dfrac{1}{72}\times\dfrac{3}{2}=\dfrac{1}{48}\)
22 tháng 9 2021
Bài 3:
a: Ta có: \(3x^2=75\)
\(\Leftrightarrow x^2=25\)
hay \(x\in\left\{5;-5\right\}\)
b: Ta có: \(2x^3=54\)
\(\Leftrightarrow x^3=27\)
hay x=3
22 tháng 9 2021
Bài 2:
b: Ta có: \(30-3\cdot2^n=24\)
\(\Leftrightarrow3\cdot2^n=6\)
\(\Leftrightarrow2^n=2\)
hay n=1
c: Ta có: \(40-5\cdot2^n=20\)
\(\Leftrightarrow5\cdot2^n=20\)
\(\Leftrightarrow2^n=4\)
hay n=2
d: Ta có: \(3\cdot2^n+2^n=16\)
\(\Leftrightarrow2^n\cdot4=16\)
\(\Leftrightarrow2^n=4\)
hay n=2
24 tháng 11 2023
a: \(\lim\limits_{n\rightarrow+\infty}\dfrac{n^5+n^2-n+2}{\left(2n^3-1\right)\left(n^2+n+1\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{1+\dfrac{1}{n^3}-\dfrac{1}{n^4}+\dfrac{2}{n^5}}{\left(\dfrac{2n^3}{n^3}-\dfrac{1}{n^3}\right)\left(\dfrac{n^2+n+1}{n^2}\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{1+\dfrac{1}{n^3}-\dfrac{1}{n^4}+\dfrac{2}{n^5}}{\left(2-\dfrac{1}{n^3}\right)\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}\)
\(=\dfrac{1}{2\cdot1}=\dfrac{1}{2}\)
b: \(\lim\limits_{n\rightarrow+\infty}\dfrac{\sqrt{n^2-n+2}}{n+2}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n\sqrt{1-\dfrac{1}{n}+\dfrac{2}{n^2}}}{n\left(1+\dfrac{2}{n}\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\sqrt{1-\dfrac{1}{n}+\dfrac{2}{n^2}}}{1+\dfrac{2}{n}}=\dfrac{\sqrt{1-0+0}}{1+0}=\dfrac{1}{1}=1\)
c: \(\lim\limits_{n\rightarrow+\infty}\dfrac{n-\sqrt[3]{n^2-n^3}}{n^2+n+1}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\dfrac{n}{n^2}-\dfrac{\sqrt[3]{n^2-n^3}}{n^2}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\dfrac{1}{n}-\sqrt[3]{\dfrac{1}{n^4}-\dfrac{1}{n^3}}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}=\dfrac{0}{1}=0\)
d: \(\lim\limits_{n\rightarrow+\infty}\left(n-\sqrt{n^2+n+1}\right)\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2-n^2-n-1}{n+\sqrt{n^2+n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-n-1}{n+\sqrt{n^2+n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-1-\dfrac{1}{n}}{1+\sqrt{1+\dfrac{1}{n}+\dfrac{1}{n^2}}}=-\dfrac{1}{1+1}=-\dfrac{1}{2}\)
24 tháng 11 2023
\(\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+n^2+n+1}-n\right)\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^3+n^2+n+1-n^3}{\sqrt[3]{\left(n^3+n^2+n+1\right)^2}+n\cdot\sqrt[3]{n^3+n^2+n+1}+n}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2+n+1}{\sqrt[3]{\left[n^3\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)\right]^2}+n^2\cdot\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+n^2}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2+n+1}{n^2\cdot\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+n^2\cdot\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+n^2}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{1+\dfrac{1}{n}+\dfrac{1}{n^2}}{\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2+\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+1}}\)
\(=\dfrac{1}{1+1+1}=\dfrac{1}{3}\)
b: \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt{n^2+n}-\sqrt{n^2-n+1}\right)\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2+n-n^2+n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{2n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{2-\dfrac{1}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}}=\dfrac{2}{\sqrt{1}+\sqrt{1}}=1\)
KB
6 tháng 9 2018
\(5a^2+10b^2-6ab-4a+2b+3\)
\(=\left(a^2-6ab+9b^2\right)+\left(4a^2-4a+1\right)+\left(b^2+2b+1\right)+1\)
\(=\left(a-3b\right)^2+\left(2a-1\right)^2+\left(b+1\right)^2+1>0\left(đpcm\right)\)
25 tháng 11 2021
\(A=\dfrac{x^3}{9y^2}-\dfrac{1}{8}x^2y+\dfrac{2}{15}xy^2\\ B=\dfrac{2a-b}{a+1}-\dfrac{\left(a-1\right)^2}{b-2}\cdot\dfrac{\left(b-2\right)\left(b+2\right)}{\left(a-1\right)\left(a+1\right)}\\ B=\dfrac{2a-b}{a+1}-\dfrac{\left(a-1\right)\left(b+2\right)}{a+1}\\ B=\dfrac{2a-b-\left(a-1\right)\left(b+2\right)}{a+1}\\ B=\dfrac{2a-b-ab-2a+b+2}{a+1}=\dfrac{2-ab}{a+1}\)
BT
2
KV
16 tháng 9 2023
4/5 - (-2/7) - 7/10
= 4/5 + 2/7 - 7/10
= 8/10 - 7/10 + 2/7
= 1/10 + 2/7
= 7/70 + 20/70
= 27/70
16 tháng 9 2023
\(\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)-\dfrac{7}{10}\)
\(=\dfrac{4}{5}+\dfrac{2}{7}-\dfrac{7}{10}\)
\(=\dfrac{8-7}{10}+\dfrac{2}{7}\)
\(=\dfrac{1}{10}+\dfrac{2}{7}\)
\(=\dfrac{7+20}{70}\)
\(=\dfrac{27}{70}\)
TN
5 tháng 7 2017
\(a,36-4a^2+20ab-25b^2\)
\(=6^2-\left(2a-5b\right)^2=\left(6-2a+5b\right)\left(6+2a-5b\right)\)\(b,x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
\(d,5a^2-10a^2b+5ab^2-10a+10b\)
\(=5a^2-5a^2b-5a^2b+5ab^2-10a+10b\)
\(=5a\left(a-b\right)-5ab\left(a-b\right)-10\left(a-b\right)\)
\(=\left(a-b\right)\left(5a-5ab-10\right)\)
a) A = 2 + 22 + 23 + ... + 210
2A = 22 + 23 + 24 + ... + 211
2A - A = ( 22 + 23 + 24 + ... + 211 ) - ( 2 + 22 + 23 + ... + 210 )
A = 211 - 2