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KC
2
31 tháng 8 2023
`#040911`
a,
\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)
Vậy, \(x=-\dfrac{8}{21}\)
b,
\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, \(x\in\left\{-2;3\right\}\)
c,
\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)
Bạn xem lại đề có sai kh nhỉ?
31 tháng 8 2023
c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)
KV
23 tháng 8 2023
A = 2⁵.(-5)² - 8² - 7
= 32.25 - 64 - 7
= 729
= 27²
B = 2³.(-4)² + (-3)².3² - 40
= 8.16 + 9.9 - 40
= 169
= 13²
C = (1/4 - 1/2 - 1)³ . (2 - 2/5)³
= (-5/4)³ . (8/5)³
= (-5/4 . 8/5)³
= (-2)³
D = (-1/4)² : (1/2 - 1/3)
= 1/16 : 1/6
= 3/8
E = 4 . (1/4)² + 25 . [(3/4)³ : (5/4)³] : (3/2)³
= 1/4 + 25 . (3/4 . 5/4)³ : (3/2)³
= 1/4 + 25 . (15/16)³ : 27/8
= 1/4 + 25 . 3375/4096 : 27/8
= 1/4 + 84375/4096 : 27/8
= 1/4 + 3125/512
= 3253/512
F = 2³ + 3.(1/2)⁰ - 1 + [(-2)² : 1/2] - 8
= 8 + 3.1 - 1 + (4 : 1/2) - 8
= 8 + 3 - 1 + 8 - 8
= 10
22 tháng 9 2017
Nguyễn Trà My
Phần a)
\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(32-3x+13=76-x\)
\(116-3x=76-x\)
\(116-76=3x-x\)
\(46=2x\)
\(x=46\div2\)
\(x=13\)
TT
16 tháng 11 2023
a) 1/4(x-3)+2=1/5
1/4.(x-3) = 1/5-2
1/4.(x-3) = -9/5
x-3 = (-9/5):1/4
x-3 = -36/5
x = -36/5+3
x= -21/5
21 tháng 2 2021
1) Ta có: \(\dfrac{1}{7}x^2y^3\cdot\left(-\dfrac{14}{3}xy^2\right)\cdot\left(-\dfrac{1}{2}xy\right)\left(x^2y^4\right)\)
\(=\left(-\dfrac{1}{7}\cdot\dfrac{14}{3}\cdot\dfrac{-1}{2}\right)\left(x^2y^3\cdot xy^2\cdot xy\cdot x^2y^4\right)\)
\(=\dfrac{1}{3}x^6y^{10}\)
2) Ta có: \(\left(3xy\right)^2\cdot\left(-\dfrac{1}{2}x^3y^2\right)\)
\(=9xy^2\cdot\dfrac{-1}{2}x^3y^2\)
\(=-\dfrac{9}{2}x^4y^4\)
3) Ta có: \(\left(-\dfrac{1}{4}x^2y\right)^2\cdot\left(\dfrac{2}{3}xy^4\right)^3\)
\(=\dfrac{1}{16}x^4y^2\cdot\dfrac{8}{27}x^3y^{12}\)
\(=\dfrac{1}{54}x^7y^{14}\)
22 tháng 10 2021
a: \(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35-12}{20}=\dfrac{23}{20}\)
d: \(\left(-\dfrac{1}{4}\right)^2\cdot\dfrac{4}{11}+\dfrac{7}{11}\cdot\left(-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)
22 tháng 10 2021
\(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35}{20}+\dfrac{-12}{20}=\dfrac{23}{20}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9
Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9
Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)
VM
15 tháng 10 2019
3) \(\frac{x-3}{x-2}=\frac{4}{7}\)
\(\Rightarrow\left(x-3\right).7=\left(x-2\right).4\)
\(\Rightarrow7x-21=4x-8\)
\(\Rightarrow7x-4x=\left(-8\right)+21\)
\(\Rightarrow3x=13\)
\(\Rightarrow x=13:3\)
\(\Rightarrow x=\frac{13}{3}\)
Vậy \(x=\frac{13}{3}.\)
4) \(\left|x\right|+3,5=0\)
\(\Rightarrow\left|x\right|=0-3,5\)
\(\Rightarrow\left|x\right|=-3,5\)
Vì \(\left|x\right|\ge0\) \(\forall x.\)
\(\Rightarrow\left|x\right|>-3,5\)
\(\Rightarrow\left|x\right|\ne-3,5\)
Vậy \(x\in\varnothing.\)
5) \(\left|x+\frac{1}{3}\right|-4=1\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=1+4\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=5.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{3}=5\\x+\frac{1}{3}=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5-\frac{1}{3}\\x=\left(-5\right)-\frac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{14}{3}\\x=-\frac{16}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{14}{3};-\frac{16}{3}\right\}.\)
Chúc bạn học tốt!
\(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)=\frac{3}{4}\)
\(\frac{2}{3}:\left(x-1\right)=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)
\(x-1=\frac{2}{3}:\frac{1}{4}=\frac{8}{3}\)
\(\Rightarrow\)\(x=\frac{8}{3}+1=\frac{11}{3}\)