Tim x,biết:
(3x + 1 ) mũ 2 : (-1/4) = -49/4
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F
5 tháng 10 2020
a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
F
5 tháng 10 2020
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)
NN
7 tháng 1 2018
a) \(2^3:\left|x-2\right|=2\)
\(\Leftrightarrow8:\left|x-2\right|=2\)
\(\Leftrightarrow\left|x-2\right|=8:2\)
\(\Leftrightarrow\left|x-2\right|=4\)
Xét trường hợp 1: \(x-2=4\)
\(\Rightarrow x=4+2\)
\(\Rightarrow x=6\)
Xét trường hợp 2: \(x-2=-4\)
\(\Rightarrow x=-4+2\)
\(\Rightarrow x=-\left(4-2\right)\)
\(\Rightarrow x=-2\)
Vậy \(x=6\) hoặc \(x=-2\)
b)
VD
3
12 tháng 8 2021
( 3x + 4 )2 - ( 3x - 1 )( 3x + 1 ) = 49
<=> 9x2 + 24x + 16 - ( 9x2 - 1 ) - 49 = 0
<=> 9x2 + 24x - 33 - 9x2 + 1 = 0
<=> 24x - 32 = 0 <=> x = 4/3
AD
12 tháng 8 2021
\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)
\(\Leftrightarrow9x^2+24x+16-9x^2+1-49=0\)
\(\Leftrightarrow24x=32\)
\(\Leftrightarrow x=\frac{3}{2}\)
#H
YN
4 tháng 10 2021
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)
NL
0
SN
2
12 tháng 7 2017
\(c.\:\left(3x+4\right)^2-\left(3x+1\right)\left(3x-1\right)\\ =9x^2+24x+16-9x^2+1\\ 40x=-1\\ x=-\dfrac{1}{40}\)
\(d.\:\left(3x-1\right)^2-\left(3x-2\right)^2=0\\ \left(3x-1+3x-2\right)\left(3x-1-3x+2\right)=0\\ \left(6x-3\right)=0\\ x=\dfrac{1}{2}\)
\(g.\:\left(2x+1\right)^2-\left(x-1\right)^2=0\\ \left(2x+1+x-1\right)\left(2x+1-x+1\right)=0\\ 3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
DH
12 tháng 7 2017
c,\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)
\(\Rightarrow9x^2+24x+16-\left(9x^2-1\right)=49\)
\(\Rightarrow9x^2+24x+16-9x^2+1=49\)
\(\Rightarrow24x=49-1-16\)
\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)
d, \(\left(3x-1\right)^2-\left(3x-2\right)^2=0\)
\(\Rightarrow\left(3x-1-3x+2\right).\left(3x-1+3x-2\right)=0\)
\(\Rightarrow6x-3=0\Rightarrow6x=3\Rightarrow x=\dfrac{1}{2}\)
e, \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)
\(\Rightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)
\(\Rightarrow\left(x+2\right).3x=0\Rightarrow x.\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Chúc bạn học tốt!!!
KN
21 tháng 8 2020
a. ( 2x + 1 )2 = 49
<=> ( 2x + 1 )2 = 72
<=> 2x + 1 = 7
<=> x = 3
b. ( 2x - 1 )4 = 81
<=> ( 2x - 1 )4 = 34
<=> 2x - 1 = 3
<=> x = 2
c. ( x + 1 )3 = 2x3
<=> x + 1 = 2x
<=> x = 1
d. ( 2x + 1 )3 = 3x3
<=> 2x + 1 = 3x
<=> x = 1
21 tháng 8 2020
( 2x + 1 )2 = 49
<=> ( 2x + 1 )2 = ( ±7 )2
<=> \(\orbr{\begin{cases}2x+1=7\\2x+1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
( 2x - 1 )4 = 81
<=> ( 2x - 1 )4 = ( ±3 )4
<=> \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
( x + 1 )3 = ( 2x )3
<=> x + 1 = 2x
<=> x - 2x = -1
<=> -x = -1
<=> x = 1
( 2x + 1 )3 = ( 3x )3
<=> 2x + 1 = 3x
<=> 2x - 3x = -1
<=> -x = -1
<=> x = 1
\(\left(3x+1\right)^2:\left(-\frac{1}{4}\right)=-\frac{49}{4}\)
\(\Leftrightarrow\left(3x+1\right)^2=-\frac{49}{4}\cdot\left(-\frac{1}{4}\right)\)
\(\Leftrightarrow\left(3x+1\right)^2=\frac{49}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(3x+1\right)^2=\left(\frac{7}{4}\right)^2\\\left(3x+1\right)^2=\left(-\frac{7}{4}\right)^2\end{cases}\Leftrightarrow\orbr{\begin{cases}3x+1=\frac{7}{4}\\3x+1=-\frac{7}{4}\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=\frac{3}{4}\\3x=-\frac{11}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{11}{12}\end{cases}}}\)
(3x+1)2 =49/16
(=)3x+1=7/4
(=)3x=3/4
(=)x=1/4