(3x 2 + 10x – 8) 2 = (5x 2 – 2x + 10) 2
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TA
28 tháng 5 2021
`(3x^2+10x-8)=(5x^2-2x+10)^2`
`<=>(3x^2+10x-8+5x^2-2x+10)(3x^2+10x-8-5x^2+2x-10)=0`
`<=> (8x^2+8x+2)(-2x^2+12x-18)=0`
\(\Leftrightarrow\left[{}\begin{matrix}8x^2+8x+2=0\\-2x^2+12x-18=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy `S={-1/2 ; 3}`.
DH
5 tháng 1 2020
\(\left(5x^2-2x+19\right)^2=\left(3x^2+10x-8\right)^2\)
\(\Leftrightarrow\left(5x^2-2x+19\right)^2-\left(3x^2+10x-8\right)^2=0\)
\(\Leftrightarrow\left(5x^2-2x+19-3x^2-10x+8\right)\left(5x^2-2x+19+3x^2+10x-8\right)=0\)
\(\Leftrightarrow\left(2x^2-12x+27\right)\left(8x^2+8x+11\right)=0\)
....
VM
5 tháng 1 2020
\(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)
\(\Leftrightarrow\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\)
\(\Leftrightarrow\left(5x^2-2x+10-3x^2-10x+8\right).\left(5x^2-2x+10+3x^2+10x-8\right)=0\)
\(\Leftrightarrow\left(2x^2-12x+18\right).\left(8x^2+8x+2\right)=0\)
\(\Leftrightarrow2.\left(x^2-6x+9\right).2.\left(4x^2+4x+1\right)=0\)
\(\Leftrightarrow2.\left(x-3\right)^2.2.\left(2x+1\right)^2=0\)
\(\Leftrightarrow4.\left(x-3\right)^2.\left(2x+1\right)^2=0\)
Vì \(4\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(2x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{3;-\frac{1}{2}\right\}.\)
Chúc bạn học tốt!
6 tháng 2 2022
\(\Leftrightarrow\left(5x^2-2x+10-3x^2-10x+8\right)\left(5x^2-2x+10+3x^2+10x-8\right)=0\)
\(\Leftrightarrow\left(2x^2-12x+18\right)\left(8x^2+8x+2\right)=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
hay x=-1/2
6 tháng 2 2022
\(PT\Leftrightarrow\left(3x^2+10x-8\right)^2-\left(5x^2-2x+10\right)^2=0\)
\(\Leftrightarrow\left(3x^2+10x-8-5x^2+2x-10\right)\left(3x^2+10x-8+5x^2-2x+10\right)=0\)
\(\Leftrightarrow\left(-2x^2+12x-18\right)\left(8x^2+8x+2\right)=0\)
\(\Leftrightarrow-4\left(x-3\right)^2\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(2x+1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{3;-\dfrac{1}{2}\right\}\)
2 tháng 1 2023
2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25
OP
29 tháng 1 2022
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
DT
1
14 tháng 2 2016
TH1: 5x2 - 2x + 10 = 3x2 + 10x - 8
=> 2x2 - 12x + 18 = 0
=> x2 - 6x + 9 = 0
=> (x - 3)2 = 0
=> x = 3
TH2: 5x2 - 2x + 10 = - 3x2 - 10x + 8
=> 8x2 + 8x + 2 = 0
=> 4x2 + 4x + 1 = 0
=> (2x + 1)2 = 0
=> x = -1/2
Vậy x = 3 , x = -1/2
10 tháng 8 2017
\(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)
\(\Leftrightarrow\left(5x^2-2x+10\right)^2-\left(3x^2+10x-8\right)^2=0\)
\(\Leftrightarrow\left(5x^2-2x+10+3x^2+10x-8\right)\left(5x^2-2x+10-3x^2-10x+8\right)=0\)
\(\Leftrightarrow\left(8x^2+8x+2\right)\left(2x^2-12x+18\right)=0\)
\(\Leftrightarrow2\left(4x^2+4x+1\right).2\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow4\left(2x+1\right)^2\left(x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy ..........
19 tháng 2 2020
=> 5x^2 - 2x + 10=3x^2 + 10x - 8
=> 2x^2 -12x +18 = 0
=> 2(x^2 - 6x +9) = 0
=> 2(x - 3)^2 = 0
=> x - 3 = 0
=> x = 3
7 tháng 10 2021
a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
c: Ta có: \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
8 tháng 10 2021
b.
PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$
$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$
$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$
$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$
$\Leftrightarrow (x-3)^2(2x+1)^2=0$
$\Leftrightarrow (x-3)(2x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$
d.
$x^2-2x=24$
$\Leftrightarrow x^2-2x-24=0$
$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$
$\Leftrightarrow x=-4$ hoặc $x=6$
HP
3
LB
5 tháng 2 2018
1) \(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
5 tháng 1 2020
Bài 2:
\(3x^2+10x-8=5x^2-2x+10\)
\(\Leftrightarrow 2x^2-12x+18=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow (x-3)^2=0\Rightarrow x=3\)