Tìm x biết:
A,|x^2+1| = |-1|
B,| x^3+1| = |1|
Giải hộ mk với!!!
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16 tháng 5 2022
Câu b,
= 50x + (1 + 3 + 5 + ..... + 99) = 2550
= 50x + 2500 = 2550
= 50x = 2550 - 2500
= 50x = 50
= x = 50 : 50
= x = 1
HP
21 tháng 9 2021
a. 9x2 - 6x - 3 = 0
<=> 3(3x2 - 2x - 1) = 0
<=> 3(3x2 - 3x + x - 1) = 0
<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)
<=> 3(3x + 1)(x - 1) = 0
<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)
b. (2x + 1)2 - 4(x + 2)2 = 9
<=> (2x + 1)2 - \(\left[2\left(x+2\right)\right]^2=9\)
<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9
<=> -3(4x + 5) = 9
<=> 4x + 5 = -3
<=> 5 + 3 = -4x
<=> -4x = 8
<=> -x = 2
<=> x = -2
21 tháng 9 2021
a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)
\(\Leftrightarrow\left(3x-1\right)^2-4=0\)
\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)
\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)
c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)
\(\Leftrightarrow9x=18\Leftrightarrow x=2\)
d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)
10 tháng 10 2021
a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)
b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)
c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Xem lại đề câu d
26 tháng 5 2022
a: \(\Rightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)
\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)
=>-4x=5
hay x=-5/4
b: \(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
=>42x=41
hay x=41/42
13 tháng 7 2021
a)Ta có: \(\dfrac{x+3}{x+1}+\dfrac{1}{3}\ge0\)
\(\Leftrightarrow\dfrac{3x+9+x+1}{3\left(x+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{4x+10}{3x+3}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1>0\\4x+10\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-1\\x\le-\dfrac{5}{2}\end{matrix}\right.\)
b) Ta có: \(\dfrac{x+2}{x+3}+\dfrac{1}{3}\le0\)
\(\Leftrightarrow\dfrac{3x+6+x+3}{3\left(x+3\right)}\le0\)
\(\Leftrightarrow\dfrac{4x+9}{3x+9}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+9>0\\4x+9\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow-3< x\le-\dfrac{9}{4}\)
LT
13 tháng 7 2021
a)\(\dfrac{x+3}{x+1}\ge-\dfrac{1}{3}\left(x\ne-1\right)\)
\(\Leftrightarrow\dfrac{x+3}{x+1}+\dfrac{1}{3}\ge0\)
\(\Leftrightarrow\dfrac{3x+9+x+1}{3x+3}\ge0\)
\(\Leftrightarrow\dfrac{4x+10}{3x+3}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+10\ge0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+10\le0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{5}{2}\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{-5}{2}\\x< -1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>-1\\x\le\dfrac{-5}{2}\end{matrix}\right.\)
b) \(\dfrac{x+2}{x+3}\le-\dfrac{1}{3}\left(x\ne-3\right)\)
\(\dfrac{x+2}{x+3}+\dfrac{1}{3}\le0\)
\(\Leftrightarrow\dfrac{3x+6+x+3}{3x+9}\le0\)
\(\Leftrightarrow\dfrac{4x+9}{3x+9}\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+9\ge0\\3x+9< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+9\le0\\3x+9>0\end{matrix}\right.\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{9}{4}\\x< -3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-\dfrac{9}{4}\\x>-3\end{matrix}\right.\end{matrix}\right.\)
TH1: loại
TH2: TM
Vậy no của BPT là :\(-\dfrac{9}{4}\ge x>-3\)
chúc bạn học tốt
30 tháng 6 2021
a) Ta có: \(8x\left(2x-3\right)-4x\left(4x+3\right)=72\)
\(\Leftrightarrow16x^2-24x-16x^2-12x=72\)
\(\Leftrightarrow-36x=72\)
hay x=-2
b) Ta có: \(\left(x+2\right)\left(x+4\right)-x\left(x+2\right)=104\)
\(\Leftrightarrow x^2+6x+8-x^2-2x=104\)
\(\Leftrightarrow4x=96\)
hay x=24
c) Ta có: \(\left(x-1\right)\left(x+4\right)-x\left(x-1\right)=308\)
\(\Leftrightarrow x^2+3x-4-x^2+x=308\)
\(\Leftrightarrow4x=312\)
hay x=78
d) Ta có: \(15x\left(2x-3\right)-\left(5x+2\right)\left(6x-5\right)=-22\)
\(\Leftrightarrow30x^2-45x-30x^2+25x-12x+10=-22\)
\(\Leftrightarrow-32x=-32\)
hay x=1