\(|\varkappa+\dfrac{1}{3}|-\dfrac{4}{3}=-\dfrac{1}{2}\)
K
Khách
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NH
28 tháng 6 2017
a
= { 1*( 1+1/2+1/3+1/4) } / { 1 * ( 1-1/2 +1/3-1/4)} : { 3*(1+1/2+1/3+1/4)} / { 2*( 1-1/2 +1/3-1/4)}
Sau đó bn tự tính ra nhé cứ tính nhu bình thường sẽ ra.
Mà mình thấy máy câu này yêu cầu tính chứ có bảo tính theo cách hợp lí đâu? Vì thế bn cứ lấy máy tính tính như bình thường là được .
T
1 tháng 10 2023
a) \(0,25-\dfrac{2}{3}+1\dfrac{1}{4}\)
\(=\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{5}{4}\)
\(=\dfrac{3}{12}-\dfrac{8}{12}+\dfrac{15}{12}\)
\(=\dfrac{10}{12}\)
\(=\dfrac{5}{6}\)
\(---\)
b) \(\dfrac{3^2}{2}:\dfrac{1}{4}+\dfrac{3}{4}\cdot2010\)
\(=\dfrac{9}{2}\cdot4+\dfrac{3015}{2}\)
\(=18+\dfrac{3015}{2}\)
\(=\dfrac{36}{2}+\dfrac{3015}{2}\)
\(=\dfrac{3051}{2}\)
\(---\)
c) \(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)
\(=\left\{\left[\left(-\dfrac{14}{25}\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-2}{6}\right)+\dfrac{3}{6}\right]\)
\(=\left\{\left[\dfrac{196}{625}\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)
\(=\left\{\dfrac{4}{5}\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)
\(=\dfrac{4}{6}-\dfrac{1}{6}\)
\(=\dfrac{3}{6}\)
\(=\dfrac{1}{2}\)
\(---\)
d) \(\left(-\dfrac{1}{2}-\dfrac{1}{3}\right)^2:\left[\left(\dfrac{-5}{36}\right)-\left(\dfrac{-5}{36}\right)^0\right]\)
\(=\left(-\dfrac{3}{6}-\dfrac{2}{6}\right)^2:\left[-\dfrac{5}{36}-1\right]\)
\(=\left(-\dfrac{5}{6}\right)^2:\left[-\dfrac{5}{36}-\dfrac{36}{36}\right]\)
\(=\dfrac{25}{36}:\left(\dfrac{-41}{36}\right)\)
\(=\dfrac{25}{36}\cdot\left(\dfrac{-36}{41}\right)\)
\(=-\dfrac{25}{41}\)
#\(Toru\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 9 2023
a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)
- \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) < \(x\) < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)
- \(\dfrac{46}{3}\) < \(x\) < - \(\dfrac{13}{7}\)
\(x\) \(\in\) {-15; -14;-13;..; -2}
23 tháng 9 2023
a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)
Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)
Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)
Suy ra \(-15\le x\le-2\), x ϵ Z
b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)
Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)
Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)
Suy ra \(-1\le x\le0\), x ϵ Z
8 tháng 3 2023
1, \(\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}+1\right)\)
\(=\dfrac{3}{4}.\dfrac{6-1+15}{15}=\dfrac{3}{4}.\dfrac{20}{15}=\dfrac{3}{4}.\dfrac{4}{3}=1\)
2, \(\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{3}.\dfrac{40}{9}=\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{9}.\dfrac{40}{3}\)
\(=\dfrac{4}{9}.\left[\left(-\dfrac{13}{3}\right)+\dfrac{40}{3}\right]=\dfrac{4}{9}.9=4\)
3, \(\dfrac{4}{9}-\dfrac{2}{3}.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)=\dfrac{2}{3}\left(\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{3}.\dfrac{20-24-15}{30}=\dfrac{2}{3}.\left(-\dfrac{19}{30}\right)=-\dfrac{19}{45}\)
TH
Thầy Hùng Olm
Manager
VIP
8 tháng 3 2023
1. \(\dfrac{3}{4}.\left(\dfrac{6}{15}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\dfrac{1}{3}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
11 tháng 8 2021
1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)
\(\Leftrightarrow5x+1-2x+4=4\)
\(\Leftrightarrow3x=-1\)
hay \(x=-\dfrac{1}{3}\)
2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
\(\Leftrightarrow9x+27+12-36x=-2x+2\)
\(\Leftrightarrow-27x+2x=2-39\)
hay \(x=\dfrac{37}{25}\)
3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x+6-10x=4-4x\)
\(\Leftrightarrow-7x+4x=4-6=-2\)
hay \(x=\dfrac{2}{3}\)
11 tháng 8 2021
4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
\(\Leftrightarrow5x-15-x-1=2x-4\)
\(\Leftrightarrow4x-2x=-4+16=12\)
hay x=6
5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
\(\Leftrightarrow12x+3-9x+5+4x-8=0\)
\(\Leftrightarrow7x=0\)
hay x=0
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 4 2020
5.
\(\lim\limits_{x\rightarrow-\infty}\frac{-3x^5+7x^3-11}{x^5+x^4-3x}=\lim\limits_{x\rightarrow-\infty}\frac{-3+\frac{7}{x^2}-\frac{11}{x^5}}{1+\frac{1}{x}-\frac{3}{x^4}}=\frac{-3}{1}=-3\)
6.
\(\lim\limits_{x\rightarrow-4}\frac{\left(x+4\right)\left(x-1\right)}{x\left(x+4\right)}=\lim\limits_{x\rightarrow-4}\frac{x-1}{x}=\frac{-5}{-4}=\frac{5}{4}\)
7.
Khi \(x< 2\Rightarrow x-2< 0\) mà \(x+2\rightarrow4\Rightarrow\lim\limits_{x\rightarrow2^-}\frac{x+2}{x-2}=\frac{4}{-0}=-\infty\)
8.
\(\lim\limits_{x\rightarrow1}\frac{9-\left(2x+7\right)}{\left(x-1\right)\left(x+1\right)\left(3+\sqrt{2x+7}\right)}=\lim\limits_{x\rightarrow1}\frac{-2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(3+\sqrt{2x+7}\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{-2}{\left(x+1\right)\left(3+\sqrt{2x+7}\right)}=\frac{-2}{2.\left(3+3\right)}=-\frac{1}{6}\)
9.
\(\lim\limits_{x\rightarrow4}\frac{\left(4-x\right)\left(16-4x+x^2\right)}{4-x}=\lim\limits_{x\rightarrow4}\left(16-4x+x^2\right)=16\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 4 2020
1.
\(\lim\limits_{x\rightarrow-\infty}\frac{x^2-7x+1-\left(x^2-3x+2\right)}{\sqrt{x^2-7x+1}+\sqrt{x^2-3x+2}}=\lim\limits_{x\rightarrow-\infty}\frac{-4x-1}{\sqrt{x^2-7x+1}+\sqrt{x^2-3x+2}}\)
\(=\lim\limits_{x\rightarrow-\infty}\frac{x\left(-4-\frac{1}{x}\right)}{-x\sqrt{1-\frac{7}{x}+\frac{1}{x^2}}-x\sqrt{1-\frac{3}{x}+\frac{2}{x^2}}}=\frac{-4}{-1-1}=2\)
2.
\(\lim\limits_{x\rightarrow0^+}\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\lim\limits_{x\rightarrow0^+}\frac{\sqrt{x}+1}{\sqrt{x}-1}=-1\)
3.
\(\lim\limits_{x\rightarrow-1}\frac{x^2-3}{x^3+2}=\frac{1-3}{-1+2}=-2\) (ko phải dạng vô định, cứ thay số tính)
4.
\(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\frac{2x^2-x-1}{x-1}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(2x+1\right)}{x-1}=\lim\limits_{x\rightarrow1}\left(2x+1\right)=3\)
Để hs có giới hạn tại \(x=1\Rightarrow m=3\)
4 tháng 3 2023
\(\dfrac{2}{5}\times\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{1}{5}\times\dfrac{3}{2}-\dfrac{1}{8}=\dfrac{3}{10}-\dfrac{1}{8}=\dfrac{24}{80}-\dfrac{10}{80}=\dfrac{14}{80}=\dfrac{7}{40}\\ \dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{3}-\dfrac{1}{5}=\dfrac{25}{15}-\dfrac{3}{15}=\dfrac{22}{15}\\ \dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\\ \dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{2}{8}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{2}{8}+\dfrac{1}{3}\times\dfrac{5}{2}=\dfrac{2}{8}+\dfrac{5}{6}=\dfrac{1}{4}+\dfrac{5}{6}=\dfrac{6}{24}+\dfrac{20}{24}=\dfrac{26}{24}=\dfrac{13}{12}\)
\(\left|x+\dfrac{1}{3}\right|-\dfrac{4}{3}=-\dfrac{1}{2}\\ =>\left|x+\dfrac{1}{3}\right|=\dfrac{4}{3}-\dfrac{1}{2}=\dfrac{5}{6}\\ =>\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{5}{6}\\x+\dfrac{1}{3}=-\dfrac{5}{6}\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{6}\end{matrix}\right.\)