tìm x
a) ( 3x - 4 )3 = 8
b) 2 . ( 2x + 3 )3 = 250
c) A . 3x - 2 = 36
d) 2x + 2x + 1 + 2x + 2 = 56
K
Khách
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5 tháng 9 2021
a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Rightarrow\left[2x+1-2\left(x+2\right)\right]\left[2x+1+2\left(x+2\right)\right]=9\\ \Rightarrow\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\\ \Rightarrow-3\left(4x+5\right)=9\\ \Rightarrow4x+5=-3\\ \Rightarrow4x=-8\\ \Rightarrow x=-2\)
K
30 tháng 6 2021
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
28 tháng 10 2023
a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)
=>\(\left|2x+1\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
c: (2x-3)2=36
=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
d: \(7^{x+2}+2\cdot7^x=357\)
=>\(7^x\cdot49+7^x\cdot2=357\)
=>\(7^x=7\)
=>x=1
T
28 tháng 10 2023
a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
\(---\)
b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)
\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)
\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
\(---\)
c) \(\left(2x-3\right)^2=36\)
\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(---\)
d) \(7^{x+2}+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)
\(\Rightarrow7^x\cdot\left(49+2\right)=357\)
\(\Rightarrow7^x\cdot51=357\)
\(\Rightarrow7^x=357:51\)
\(\Rightarrow7^x=7\)
\(\Rightarrow x=1\)
NT
1
8 tháng 9 2021
a: Ta có: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)
\(\Leftrightarrow-2x+1-x-2=8\)
\(\Leftrightarrow-3x=9\)
hay x=-3
b: Ta có: \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)
\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1
3 tháng 10 2023
a) Ta có 2x + 2x + 1 + 2x + 2 = 56
⇒ 2x ( 1 + 21 + 22 ) = 56
⇒ 2x . 7 = 56
⇒ 2x = 56 : 7 = 8 = 23
Vậy x = 3
b) Ta có 3x + 3x + 2 + 3x + 3 = 111
⇒ 3x ( 1 + 32 + 33 ) = 111
⇒ 3x . 37 = 111
⇒ 3x = 111 : 37 = 3 = 31
Vậy x = 1
31 tháng 8 2021
a: ta có: \(2\left(4-3x\right)+2x=5\left(2x-3\right)\)
\(\Leftrightarrow8-6x+2x-10x+15=0\)
\(\Leftrightarrow-14x=-23\)
hay \(x=\dfrac{23}{14}\)
b: Ta có: \(\dfrac{1}{2}-\left(2x-\dfrac{1}{3}\right)^2=\dfrac{7}{18}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{3}\\2x-\dfrac{1}{3}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{2}{3}\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=0\end{matrix}\right.\)
TA
1
12 tháng 10 2021
\(a,\Rightarrow3x^2-3x+6-2x-3x^2=0\\ \Rightarrow-5x=-6\Rightarrow x=\dfrac{6}{5}\\ b,\Rightarrow\left(x-1\right)\left(x-1+x+2\right)=0\\ \Rightarrow\left(x-2\right)\left(2x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\\ c,\Rightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\\ \Rightarrow\left(x^2+1\right)\left(2x+3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\2x+3=0\end{matrix}\right.\\ \Rightarrow x=-\dfrac{3}{2}\\ d,\Rightarrow2x^2+x-6=0\\ \Rightarrow2x^2+4x-3x-6=0\\ \Rightarrow2x\left(x+2\right)-3\left(x+2\right)=0\\ \Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
9 tháng 11 2021
Bài 1:
\(a,=6x^2+19x-7-6x^3-4x^2+7x=-6x^3+2x^2+26x-7\\ b,B=26\cdot\left(63^2+63\cdot37+37^2\right):26+63\cdot37\\ =63^2+63\cdot37+37^2+63\cdot37\\ =\left(63+37\right)^2=100^2=10000\)
Bài 2:
\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=\left(x-y\right)\left(x+2\right)\\ c,=\left(x-3\right)\left(x^2-4\right)=\left(x-2\right)\left(x-3\right)\left(x+2\right)\)
H9
HT.Phong (9A5)
CTVHS
3 tháng 8 2023
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)
\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)
\(\Rightarrow x\left(6x-2-15-6x\right)\)
\(\Rightarrow-16x=0\)
\(\Rightarrow x=0\)
d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)
\(\Rightarrow9x^2-4-4x+4=0\)
\(\Rightarrow9x^2-4x=0\)
\(\Rightarrow x\left(9x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
câu c có A là sao
a) ( 3x - 4 )3 = 8
(3x-4)3=23
=>3x-4=2
3x=2+4
3x=6
x=6:2
x=3
b) 2 . ( 2x + 3 )3 = 250
(2x+3)3=250:2
(2x+3)3=125
(2x+3)3=53
=>2x+3=5
2x=5-3
2x=2
x=2:2
x=1
d) 2x + 2x + 1 + 2x + 2 = 56
2x.1+2x.21+2x.22=56
2x.(1+21+22)=56
2x.(1+2+4)=56
2x.7=56
2x=56:7
2x=8
2x=23
=>x=3
x=3