tìm x biết:x^2+2x-10=0 mình cần gấp ah
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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
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)
PT
5
XO
17 tháng 8 2020
Ta có : x2 - 2x + 10 = 0
=> x2 - 2x + 1 = -9
=> (x - 1)2 = -9
=> \(x\in\varnothing\)
17 tháng 8 2020
\(x^2-2x+10=0\)
\(\Leftrightarrow x^2-2x+1+9=0\)
\(\Leftrightarrow\left(x-1\right)^2+9=0\)
Mà \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\9>0\end{cases}}\)
=> Phương trình vô nghiệm
4 tháng 10 2019
Ta có: \(x+2\sqrt{2}.x^2+2x^3=0\)
\(\Leftrightarrow x\left(1+2\sqrt{2}.x+2x^2\right)=0\)
\(\Leftrightarrow x\left[1^2+2.x\sqrt{2}.1+\left(x\sqrt{2}\right)^2\right]=0\)
\(\Leftrightarrow x\left(1+x\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+x\sqrt{2}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{\sqrt{2}}\end{cases}}\)
Vậy\(x\in\left\{0;\frac{-1}{\sqrt{2}}\right\}\)
4 tháng 10 2019
\(x+2\sqrt{2}x^2+2x^3=0\)
\(x\left(1+2\sqrt{2}x+2x^2\right)=0\)
\(x\left(2\sqrt{2}x+1\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2\sqrt{2}x+1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2x\sqrt{2}}\end{cases}}\)
H
1
AH
Akai Haruma
Giáo viên
10 tháng 8 2021
Lời giải:
$x+(x+1)+(x+2)+....+(x+21)=231$
$\underbrace{x+x+....+x}_{22}+(1+2+3+...+21)=231$
$22x+231=231$
$22x=0$
$x=0$
12 tháng 6 2018
a) \(\left(\frac{1}{7}x-\frac{2}{3}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x-\frac{2}{3}=0\\-\frac{1}{5}x+\frac{3}{5}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x=\frac{2}{3}\\-\frac{1}{5}x=-\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{14}{3}\\x=3\end{cases}}\)
b)\(\frac{1}{10}x-\frac{4}{5}x+1=0\)
\(\Leftrightarrow x.\left(\frac{1}{10}-\frac{4}{5}\right)+1=0\)
\(\Rightarrow-\frac{7}{10}x=-1\)
\(\Rightarrow x=\frac{10}{7}\)
c)\(\left(2x-\frac{1}{3}\right).\left(5x+\frac{2}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\5x+\frac{2}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\5x=-\frac{2}{7}\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{2}{35}\end{cases}}\)
A
12 tháng 6 2018
a, (1/7 . x - 2/3) . (-1/5 . x + 3/5) = 0
Suy ra : 1/7 .x -2/3 = 0 hoặc -1/5 .x + 3/5 =0
Vậy : 1/7 .x = 2/3 hoặc -1/5 .x = 3/5
x =2/3 : 1/7 hoặc x = 3/5 : (-1/5)
x = 14/3 hoặc x = -3
b, 1/10 .x - 4/5 .x + 1 =0
x . (1/10 - 4/5) + 1 = 0
x . (-7/10) + 1 = 0
x . -7/10 =0 +1 = 1
x = 1 : (-7/10)
x = -10/7
c, (2x - 1/3 ) . (5x +2/7) = 0
Suy ra : 2x - 1/3 = 0 hoặc 5x + 2/7 = 0
Vậy : 2x = 1/3 hoặc 5x = 2/7
x = 1/3 : 2 hoặc x = 2/7 : 5
x = 1/6 hoặc x = 2/35
DN
1
TN
4 tháng 8 2016
Đặt t = 2x^2 +x pt trở thành
t^2 - 4t + 3=0
=>t^2 -t -3t +3 =0
=>t( t - 1) -3( t - 1)=0
=>(t - 3)(t - 1 )=0
*)Với t-3=0 <=> 2x^2 + x -3=0
=>2x^2 +3x -2x - 3 =0
=>x(2x + 3) - (2x + 3)=0
=>(x - 1)(2x + 3)=0 <=>x=1 hoặc x=-3/2
*)Với t-1=0 <=> 2x^2 + x -1=0
=>2x^2 - x + 2x -1=0
=>x(2x - 1) + (2x - 1) =0
=>(x + 1)(2x - 1)=0 <=> x=-1 hoặc x=1/2
JL
1
15 tháng 9 2021
\(\Leftrightarrow9x\left(x+2\right)+9y\left(y-\dfrac{2}{3}\right)=10\\ \Leftrightarrow9x^2+18x+9y^2-6y-10=0\\ \Leftrightarrow\left(9x^2+18x+9\right)+\left(9y^2-6y+1\right)=0\\ \Leftrightarrow9\left(x+1\right)^2+\left(3y-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
18 tháng 7 2023
a. \(8x\left(x-2007\right)-2x+4034=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy x=2017 hoặc x=1/4
b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)
\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy x=0 hoặc x=-4
c.\(4-x=2\left(x-4\right)^2\)
\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)
\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy x=4 hoặc x=7/2
d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)
\(\Rightarrow\left(x-2\right)\left(x^2+3\right)=0\)
Nxet: (x2+3)>0 với mọi x
=> x-2=0 <=>x=2
Vậy x=2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 7 2023
a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0
4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0
4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0
4\(x^2\) - 8029\(x\) + 2017 = 0
4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))2) - (\(\dfrac{8029}{4}\))2 + 2017 = 0
4.(\(x\) + \(\dfrac{8029}{8}\))2 = (\(\dfrac{8029}{4}\))2 - 2017
\(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)
thấy sai sai bạn ạ
\(x^2+2x-10=0\)
\(\Leftrightarrow x^2+2x+1-9=0\)
\(\Leftrightarrow\left(x+1\right)^2-9=0\\\)
\(\Leftrightarrow\left(x+1\right)^2=9\)
\(\Leftrightarrow\left(x+1\right)^2=\pm\sqrt{9}\)
\(\Leftrightarrow\left(x+1\right)^2=\left(\pm3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3-1\\x=-3-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy S={2;-4}