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11+2+11+2+3+11+2+3+4+....+11+2+3+...+2018
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NQ
2
24 tháng 9 2018
B = 1 + 11 + 112 + 113 + 114 + ... + 112018
11B = 11(1 + 11 + 112 + 113 + 114 + ... + 112018)
11B = 11 + 112 + 113 + 114 + 114 + ... + 112019
11B - B = (11 + 112 + 113 + 114 + 114 + ... + 112019) - (1 + 11 + 112 + 113 + 114 + ... + 112018)
10B = 112019 - 1
B = \(\frac{11^{2019}-1}{10}\)
24 tháng 8 2023
1: =>(x+2018)(6x-3)=0
=>x+2018=0 hoặc 6x-3=0
=>x=1/2 hoặc x=-2018
2: x(x-11)+3(11-x)=0
=>(x-11)(x-3)=0
=>x=11 hoặc x=3
4: =>(x+5)(2x-4)=0
=>2x-4=0 hoặc x+5=0
=>x=2 hoặc x=-5
3: =>(x-3)(x+2)=0
=>x=3 hoặc x=-2
H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
Bài 1:
\(6x\left(x+2018\right)-3\left(x+2018\right)=0\)
\(\Leftrightarrow\left(x+2018\right)\left(6x-3\right)=0\)
\(\Leftrightarrow3\left(x+2018\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\x=\dfrac{1}{2}\end{matrix}\right.\)
Bài 2:
\(x\left(x-11\right)+3\left(11-x\right)=0\)
\(\Leftrightarrow x\left(x-11\right)-3\left(x-11\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=11\end{matrix}\right.\)
Câu 3:
\(x\left(x-3\right)-2\left(3-x\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Câu 4:
\(2x\left(x+5\right)-4\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\2x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
23 tháng 6 2022
b: \(\Leftrightarrow1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}+x=\dfrac{5}{3}\)
=>x+99/100=5/3
=>x=5/3-99/100=203/300
c: \(\Leftrightarrow\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
\(\Leftrightarrow\dfrac{10}{231}-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
=>5-x=7/3
hay x=8/3
DT
22 tháng 2 2022
1.
đk: \(x\ge2\)
Đặt y = \(\sqrt{x+2}\) ta biến pt về dạng pt thuần nhất bậc 3 đối vs x và y:
ta có : \(x^3-3x^2+2y^3-6x=0\)
\(\Leftrightarrow x^3-3xy^2+2y^3=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\\x=-2y\end{matrix}\right.\)
ta sẽ có nghiệm : \(x=2;x=2-2\sqrt{3}\)
22 tháng 2 2022
\(1.đk:\left(x+2\right)^3\ge0\Leftrightarrow x\ge-2\)
\(pt\Leftrightarrow x^3-3x\left(x+2\right)+2\sqrt{\left(x+2\right)^3}=0\)
\(\Leftrightarrow x^3-x\left(x+2\right)+2\sqrt{\left(x+3\right)^2}-2x\left(x+2\right)=0\)
\(\Leftrightarrow x\left[x^2-\left(x+2\right)\right]+2\left(x+2\right)\left(\sqrt{x+2}-x\right)=0\)
\(\Leftrightarrow x\left[\left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)\right]+2\left(x+2\right)\left(\sqrt{x+2}-x\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+2}-x\right)\left[-x\left(\sqrt{x+2}+x\right)+2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(\sqrt{x+2}-x\right)^2\left(2\sqrt{x+2}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}=x\left(2\right)\\2\sqrt{x+2}=-x\left(3\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=x+2\end{matrix}\right.\)\(\Leftrightarrow x=2\left(tm\right)\)
\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}-x\ge0\Leftrightarrow x\le0\\x^2=4\left(x+2\right)\end{matrix}\right.\)\(\Leftrightarrow x=2-2\sqrt{3}\left(tm\right)\)
DB
0
LM
Lê Minh Vũ
CTVHS
14 tháng 6 2023
\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)
\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)
= \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{6}{5}\)
= \(\dfrac{1}{4}-\dfrac{6}{5}\)
= \(-\dfrac{19}{20}\)
\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)
TA
14 tháng 6 2023
\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)
\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)
\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)
\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)
\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)
\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)
\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)
\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)
\(=-\dfrac{5}{11}\)
\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+2018}\)
Ý CỦA TUI Ở PHÍA DƯỚI NHA,BÊN TRÊN BỊ LỖI