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Bài 1:
a, 2x + 35 = x - 27
2x - x = -27 - 35
x = -62
b, 2x - 41 = 3x + (-23)
2x - 3x = -23 + 41
-x = 18
x = -18
c, 4 . (x - 3) - 3 . (x - 5) = 45 . (-2) - 34
4x - 12 - 3x + 15 = -90 - 34
4x - 3x = -90 - 34 + 12 - 15
x = -127
Bài 2:
- ( -1234 + 345 - 29 ) - ( 1234 + 135 - 59 )
= 1234 - 345 + 29 - 1234 - 135 + 59
= (1234 - 1234) - (345 + 135) + (29 + 59)
= 0 - 480 + 88
= -392
a) ta có: 3x + 5 chia hết cho x + 1
=> 3x + 3 + 2 chia hết cho x + 1
3.(x+1) + 2 chia hết cho x + 1
mà 3.(x+1) chia hết cho x + 1
=> 2 chia hết cho x + 1
...
bn tự làm tiếp nha! phần b làm tương tự
Ta có: \(\dfrac{x+1}{2018}+\dfrac{x+1}{2019}+\dfrac{x+1}{2020}+\dfrac{x+1}{2021}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
\(\left(5x-29\right)-\left(2x-29\right)=-21\)
\(5x-29-2x+29=-21\)
\(5x-2x=-21+29-29\)
\(3x=-21\)
\(x=-7\)
\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
Câu a) xem lại đề giùm nhé em
b) \(\left(x-1\right)^3=9^3\)
\(x-1=9\)
\(x=10\)
Vậy \(x=10\)
c) \(\left(x-1\right)^2=25\)
\(x-1=5\) hoặc \(x-1=-5\)
* \(x-1=5\)
\(x=6\)
* \(x-1=-5\)
\(x=-4\)
Vậy \(x=-4\); \(x=6\)
d) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy \(x=2\)
e) Sửa đề: \(\left(2x+4\right)^3=64\)
\(\left(2x+4\right)^3=4^3\)
\(2x+4=4\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)
\(a,2^x+2^{x+3}=144\\ 2^x.\left(1+2^3\right)=144\\ 2^x.9=144\\ 2^x=144:9\\ 2^x=16=2^4\\ vậy:x=4\)
\(b,\left(x-5\right)^{2022}=\left(x-5\right)^{2021}\\ Vì:\left[{}\begin{matrix}0^{2022}=0^{2021}\\1^{2022}=1^{2021}\end{matrix}\right.\\ Vậy:\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
2x-x=\(\frac{-1}{6}\)-\(\frac{1}{4}\)+\(\frac{1}{3}\)
x=\(\frac{-1}{12}\)
\(\frac{2x-1}{3}=\frac{x-1}{6}-\frac{1}{4}\)
\(\frac{8x-4}{12}=\frac{2x-2}{12}-\frac{3}{12}\)
\(8x-4=2x-2-3\)
\(8x-4-2x+2+3=0\)
\(6x+1=0\Leftrightarrow x=-\frac{1}{6}\)
\(x\left(2x-4\right)-2x\left(x+3\right)-3\left(x-1\right)-29=0\)
\(\Leftrightarrow2x^2-4x-2x^2-6x-3x+3-29=0\)
\(\Leftrightarrow-13x-26=0\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=26:-13\)
\(\Leftrightarrow x=-2\)
Vậy ...
\(x\left(2x-4\right)-2x\left(x+3\right)-3\left(x-1\right)-29=0\)
\(2x^2-4x-2x^2-6x-3x+3-29=0\)
\(2x^2-4x-2x^2-6x-3x=0+29-3\)
\(\left(2x^2-2x^2\right)+\left(-4x-6x-3x\right)=26\)
\(0+\left(-4-6-3\right)x=26\)
\(\Rightarrow-13x=26\rightarrow x=-2\)