Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dễ thế mà không làm được thì bạn nên xem lại nhé,một hai câu thì còn được chứ cả 10 câu thế kia rõ là ỷ lại rồi bạn ạ.Thân!
Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
\(-\dfrac{3}{5}\times x+\dfrac{1}{2}=\dfrac{4}{5}\)
\(-\dfrac{3}{5}\times x=\dfrac{4}{5}-\dfrac{1}{2}\)
\(-\dfrac{3}{5}\times x=\dfrac{3}{10}\)
\(x=\dfrac{3}{10}:-\dfrac{3}{5}\)
\(x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
\(\dfrac{1}{2}\times x+\dfrac{3}{8}=\dfrac{7}{16}\)
\(\dfrac{1}{2}\times x=\dfrac{7}{16}-\dfrac{3}{8}\)
\(\dfrac{1}{2}\times x=\dfrac{1}{16}\)
\(x=\dfrac{1}{8}\)
Vậy \(x=\dfrac{1}{8}\)
\(\dfrac{2}{3}x-\dfrac{5}{6}x=-\dfrac{1}{2}\)
\(x\times\left(\dfrac{2}{3}-\dfrac{5}{6}\right)=-\dfrac{1}{2}\)
\(x\times\dfrac{-1}{6}=-\dfrac{1}{2}\)
\(x=3\)
Vậy \(x=3\)
\(\dfrac{2}{5}x+\dfrac{3}{10}x=-\dfrac{1}{5}\)
\(x\times\left(\dfrac{2}{5}+\dfrac{3}{10}\right)=-\dfrac{1}{5}\)
\(x\times\dfrac{7}{10}=-\dfrac{1}{5}\)
\(x=-\dfrac{2}{7}\)
Vậy \(x=-\dfrac{2}{7}\).
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
1: Ta có: 7x+6(3-x)=27-20+73
\(\Leftrightarrow7x+18-6x=80\)
\(\Leftrightarrow x=80-18=62\)
Vậy: x=62
2: Ta có: \(6x-5\left(x-7\right)=\left(27-514\right)-486-73\)
\(\Leftrightarrow6x-5x+35=27-514-486-73\)
\(\Leftrightarrow x+35=-1046\)
\(\Leftrightarrow x=-1081\)
Vậy: x=-1081
(9x - 2³) : 5=2
3x . [ 8² -2 .( 2⁵-1)]=2022
[ 3⁴ - ( 8² +14) :13] . x=5²+10²
2²: ( 6x - 3²) - 3 = 33
(9\(x\) - 23) : 5 = 2
9\(x\) - 8 = 2 x 5
9\(x\) - 8 = 10
9\(x\) = 18
\(x\) = 2
3\(x\) .(82 - 2.(25 - 1)] = 2022
3\(x\) .(64 - 2.31) = 2022
3\(x\).(64 - 62) = 2022
3\(x\).2 = 2022
\(x\) = 2022 : 2
\(x\) = 1011
8)4(x-2)-2(x-1)=12
4x-8-(2x-2)=12
4x-8-2x+2=12
2x-8+2=12
2x=12-2+8
2x=18
x=18/2=9
9)6x-13-2-(x×5)=7
6x-5x=7+13+2
x=22
10)7(x-5)-3(x+5)=30
7x-35-(3x+15)=30
7x-35-3x-15=30
4x=30+35+15
4x=80
x=20
8)4(x-2)-2(x-1)=12
\(\Rightarrow4x-8-2x+2=12\)
\(\Rightarrow2x=18\)
\(=>x=9\)
9)6x-13-2-(x×5)=7
\(\Rightarrow6x-15-\left(x+5\right)=7\)
\(\Rightarrow6x-15-x-5=7\)
\(\Rightarrow5x=27\)
\(\Rightarrow x=\frac{27}{5}\)
10)7(x-5)-3(x+5)=30
\(\Rightarrow7x-35-3x-15=30\)
\(\Rightarrow4x=80\)
\(\Rightarrow x=20\)
* \(6x+3-2\left(x-5\right)=30\)
\(\Rightarrow6x+3-2x+10=30\)
\(\Rightarrow4x=17\Leftrightarrow x=\frac{17}{4}\)
* \(8\left(x-\frac{1}{8}\right)-5\left(x-\frac{1}{5}\right)=10\)
\(\Rightarrow3\left(x-\frac{1}{5}\right)=10\)
\(\Rightarrow x-\frac{1}{5}=\frac{10}{3}\)
\(\Rightarrow x=\frac{10}{3}+\frac{1}{5}=\frac{53}{15}\)
* \(13x-5-2\left(x+3\right)=1\)
\(\Rightarrow13x-5-2x-6=1\)
\(\Rightarrow11x=12\Leftrightarrow x=\frac{12}{11}\)