Tìm x
4(x+3)-2(7-3x)=-3
cần rất gấp ạ
K
Khách
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C
3
7 tháng 8 2021
a) Ta có: \(36x^3-4x=0\)
\(\Leftrightarrow4x\left(9x^2-1\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)
b) Ta có: \(3x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
14 tháng 8 2023
A=-4x^2+4x-1+4
=-(4x^2-4x+1)+4
=-(2x-1)^2+4<=4
Dấu = xảy ra khi x=1/2
5 tháng 10 2021
a: \(\left|3x-2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b: Ta có: \(\left|5x-3\right|=\left|x-7\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)
Q
1
AH
Akai Haruma
Giáo viên
13 tháng 12 2022
Lời giải:
$\frac{xy+3x-2y-6}{y+3}=3$
$\Rightarrow xy+3x-2y-6=3y+9$
$\Rightarrow xy+3x-5y-15=0$
$\Rightarrow x(y+3)-5(y+3)=0$
$\Rightarrow (y+3)(x-5)=0$
$\Rightarrow y+3=0$ hoặc $x-5=0$
Mà $y$ tự nhiên nên $y+3>0$. Do đó $x-5=0$
$\Rightarrow x=5$
Vậy $x=5$ và $y$ là số tự nhiên tùy ý.
28 tháng 12 2021
\(1,3x-7=19\\ \Rightarrow3x=26\\ \Rightarrow x=\dfrac{26}{3}\\ 2,\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\\ 3,3x+\dfrac{2}{4}+1=5x-\dfrac{1}{3}\\ \Rightarrow5x-\dfrac{1}{3}-3x-\dfrac{2}{4}-1=0\\ \Rightarrow2x-\dfrac{11}{6}=0\\ \Rightarrow2x=\dfrac{11}{6}\\ \Rightarrow x=\dfrac{11}{12}\)
\(4,\dfrac{x}{15}+\dfrac{1}{2}-\dfrac{x}{50}=\dfrac{5}{6}\\ \Rightarrow\dfrac{x}{15}-\dfrac{x}{50}=\dfrac{5}{6}-\dfrac{1}{2}\\ \Rightarrow x\left(\dfrac{1}{15}-\dfrac{1}{50}\right)=\dfrac{1}{3}\\ \Rightarrow\dfrac{7}{150}x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{50}{7}\)
6 tháng 8 2023
Tính:
\(\dfrac{10}{21}-1\cdot251+\dfrac{11}{21}+2\cdot251\\ 251\cdot\left(\dfrac{10}{21}+\dfrac{11}{21}\right)+\left(2-1\right)\\ 251\cdot1+1=252\)
Tìm x
c,\(\left(3x-1\right)^2:7=7\\ \left(3x-1\right)^2=7\cdot7=14\\ \left(3x-1\right)^2=\left(\pm7\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x-1=7\\3x-1=-7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=7+1=8\\3x=-7+1=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{6}{3}=-2\end{matrix}\right.\)
Vậy x={8/3;-2}
d; \(\left(7-2x\right)^3=\left(\dfrac{1}{2}\right)^2\cdot\left(-\dfrac{1}{2}\right)\\ \left(7-2x\right)^3=\left(-\dfrac{1}{2}\right)^3\\ 7-2x=-\dfrac{1}{2}\\ 2x=7--\dfrac{1}{2}=\dfrac{15}{2}\\ x=\dfrac{15}{2}:2=\dfrac{15}{4}\)
C
2
14 tháng 8 2021
a:Ta có: \(x\left(x-1\right)+x=4\)
\(\Leftrightarrow x^2-x+x=4\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
b: Ta có: \(3x\left(x-5\right)-2x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c: Ta có: \(5x^2-3x-2=0\)
\(\Leftrightarrow5x^2-5x+2x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d: Ta có: \(x^4-11x^2+18=0\)
\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
14 tháng 8 2021
a) x(x-1)+x=4
⇔x2=4⇔\(x=\pm2\)
b)3x(x-5)-2x+10=0
⇔3x(x-5)-2(x-5)=0
⇔(x-5)(3x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
c)5x2-3x-2=0
⇔ 5x(x-1)+2(x-1)=0
⇔ (x-1)(5x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d)x4-11x2+18=0
⇔ x2(x2-2)-9(x2-2)=0
⇔ (x2-2)(x2-9)=0
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)
KV
22 tháng 9 2023
a) (3x - 1)³ + 17 = 710 : 5
(3x - 1)³ + 17 = 142
(3x - 1)³ = 142 - 17
(3x - 1)³ = 125
(3x - 1)³ = 5³
3x - 1 = 5
3x = 5 + 1
3x = 6
x = 6 : 3
x = 2
4(x+3)-2(7-3x)=-3
<=> 4x + 12 - 14 + 6x = -3
<=> 4x + 6x = -3 - 12 + 14
<=> 10x = -1
<=> x = \(-\frac{1}{10}\)
\(4\left(x+3\right)-2\left(7-3x\right)=-3\)
\(4x+12-14+6x=-3\)
\(10x-2=-3\)
\(10x=-3+2\)
\(10x=1\)
\(x=1:10\)
\(x=\frac{1}{10}\)