X^2(2x-5)-4x+10=0
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LV
1
13 tháng 3 2022
\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)
2 tháng 4 2022
a: (3x-2)(4x+5)=0
=>3x-2=0 hoặc 4x+5=0
=>x=2/3 hoặc x=-5/4
b: (2,3x-6,9)(0,1x+2)=0
=>2,3x-6,9=0 hoặc 0,1x+2=0
=>x=3 hoặc x=-20
c: =>(x-3)(2x+5)=0
=>x-3=0 hoặc 2x+5=0
=>x=3 hoặc x=-5/2
PA
30 tháng 11 2016
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
10 tháng 12 2021
Câu 9:
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
10 tháng 12 2021
\(9,\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow x^2+5x-x-5=0\\ \Leftrightarrow\left(x+5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ 12,\Leftrightarrow\left(x+1\right)^2-36=0\\ \Leftrightarrow\left(x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\\ 13,\Leftrightarrow x^3-25x-x^3-8=17\\ \Leftrightarrow-25x=25\Leftrightarrow x=-1\\ 14,\Leftrightarrow x\left(2x^2+8x-3x-12\right)=0\\ \Leftrightarrow x\left(x+4\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{3}{2}\end{matrix}\right.\)
JY
1
UT
19 tháng 4 2021
1, \(4x-10=0\\ \Leftrightarrow x=\dfrac{5}{2}\)
vậy tập no S=\(\left\{\dfrac{5}{2}\right\}\)
2, \(2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\) \(x=0\) hoặc \(2x-1=0\) hoặc \(x+3=0\)
\(\Leftrightarrow\) \(x=0\) hoặc \(x=\dfrac{1}{2}\) hoặc \(x=-3\)
vậy tập no S=\(\left\{0,\dfrac{1}{2},-3\right\}\)
3, \(x-5=3-x\\ \Leftrightarrow2x=8\\ \Leftrightarrow x=4\)
vậy tập no S=\(\left\{4\right\}\)
4,\(\left(-10x+5\right)\left(2x-8\right)=0\)
\(\Leftrightarrow\) \(-10x+5=0\) hoặc \(2x-8=0\)
\(\Leftrightarrow\) \(x=\dfrac{1}{2}\) hoặc \(x=4\)
vậy tập no S=\(\left\{\dfrac{1}{2},4\right\}\)
6 tháng 6 2018
Bài 1. a) 4x - 3 = 0
⇔ x = \(\dfrac{3}{4}\)
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b) - x + 2 = 6
⇔ x = - 4
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c) -5 + 4x = 10
⇔ 4x = 15
⇔ x = \(\dfrac{15}{4}\)
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d) 4x - 5 = 6
⇔ 4x = 11
⇔ x = \(\dfrac{11}{4}\)
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h) 1 - 2x = 3
⇔ -2x = 2
⇔ x = -1
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Bài 2. a) ( x - 2)( 4 + 3x ) = 0
⇔ x = 2 hoặc x = \(\dfrac{-4}{3}\)
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b) ( 4x - 1)3x = 0
⇔ x = 0 hoặc x = \(\dfrac{1}{4}\)
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c) ( x - 5)( 1 + 2x) = 0
⇔ x = 5 hoặc x = \(\dfrac{-1}{2}\)
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d) 3x( x + 2) = 0
⇔ x = 0 hoặc x = -2
KL.....
6 tháng 6 2018
Bài 3.a) 3( x - 4) - 2( x - 1) ≥ 0
⇔ x - 10 ≥ 0
⇔ x ≥ 10
b) 3 - 2( 2x + 3) ≤ 9x - 4
⇔ - 4x - 3 ≤ 9x - 4
⇔ 13x ≥1
⇔ x ≥ \(\dfrac{1}{13}\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 7 2023
a) \(x^2+4x+10\)
\(=x^2+4x+4+6\)
\(=\left(x+2\right)^2+6\)
Mà: \(\left(x+2\right)^2+6>0\forall x\)
\(\Rightarrow x^2+4x+10>0\forall x\)
b) \(x^2-6x+15\)
\(=x^2-6x+9+6\)
\(=\left(x-3\right)^2+6\)
Mà: \(\left(x-3\right)^2+6>0\forall x\)
\(\Rightarrow x^2-6x+16>0\forall x\)
c) \(-x^2+2x-5\)
\(=-\left(x^2-2x+5\right)\)
\(=-\left(x^2-2x+1+4\right)\)
\(=-\left(x-1\right)^2-4\)
Mà: \(-\left(x-1\right)^2-4< 0\forall x\)
\(\Rightarrow-x^2+2x-5< 0\forall x\)
M
1 tháng 4 2020
e, 3x(2-x) =15(x-2)
\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy..
f, (x+5)(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)
Vậy..
g, x(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
,h, (2x -4)(x-2)=0
\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
i, (x+1/5)(2x-3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)
k, x²-4x=0
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
m, 4x²-1=0
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
n, x²-6x+9=0
\(\Leftrightarrow x^2-2.x.3+3^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)
<=> x=3
l, (3x-5)²-(x+4)²=0
\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)
\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy ..
o, 7x(x+2)-5(x+2)=0
\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)
Vậy....
p, 3x(2x-5)-4x+10=0
\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)
\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...
q, (2-2x)-x²+1=0
\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)
\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy ....
r, x(1-3x)=5(1-3x)
\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)
s, 2x-3/4+x+1/6=3
\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)
x^2*(2x-5)-2(2x-5)=0
(2x-5)*(x^2-2)=0
2x-5=0
x^2-2=0
x=5/2
x^2=2 x=4
x2.(2x-5) - 4x + 10 = 0
x2.(2x-5) - 2.(2x-5) = 0
(2x-5).(x2 - 2) = 0
\(\left(2x-5\right).\left(x-\sqrt{2}\right).\left(x+\sqrt{2}\right)=0.\)
=> 2x - 5 => 2x = 5 => x = 5/2
\(x-\sqrt{2}=0\Rightarrow x=\sqrt{2}\)
\(x+\sqrt{2}=0\Rightarrow x=-\sqrt{2}\)
KL:...