tìm x : a (1/2+2x)(2x -3)=0
b 1/4 -(2x +1/2)^2 =0
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21 tháng 6 2021
a) (2x + 1)(1 - 2x) + (1 - 2x)2 = 18
= ( 1 - 2x) \(\left[\left(2x+1+1-2x\right)\right]\) = 18
= 2(1 - 2x) - 18 = 0
= 2 - 4x - 18 = 0
= -16 - 4x = 0
= -4x = 16
= x = \(\dfrac{16}{-4}=-4\)
b) 2(x + 1)2 -(x - 3)(x + 3) - (x - 4)2 = 0
= 2 (x2 + 2x + 1) - (x2 - 9) - (x2 - 8x + 16) = 0
= 2x2 + 4x + 2 - x2 + 9 - x2 + 8x - 16 = 0
= 12x - 5 = 0
= 12x = 5
= x = \(\dfrac{5}{12}\)
c) (x - 5)2 - x(x - 4) = 9
= x2 - 10x + 25 - x2 + 4x - 9 = 0
= -6x + 16 = 0
= -6x = -16
= x = \(\dfrac{-16}{-6}=\dfrac{8}{3}\)
d) (x - 5)2 + (x - 4)(1 - x)
= x2 - 10x + 25 + 5x - x2 - 4 = 0
= -5x + 21 = 0
= -5x = -21
= x = \(\dfrac{-21}{-5}=\dfrac{21}{5}\)
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29 tháng 8 2021
\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)
\(c,x^3-2x^2+x=0\\ \Leftrightarrow x\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
29 tháng 8 2021
a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)
Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)
b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)
c) \(x^3-2x^2+x=0\\ \Rightarrow x\left(x^2-2x+1\right)=0\\ \Rightarrow x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy ...
d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
5 tháng 11 2017
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
30 tháng 9 2018
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
18 tháng 7 2023
a)\(\left(x-2\right)^2-\left(2x+3\right)^2=0\Rightarrow\left(x-2+2x+3\right)\left(x-2-2x-3\right)=0\)
\(\Rightarrow\left(3x+1\right)\left(-x-5\right)=0\Rightarrow\left[{}\begin{matrix}3x+1=0\\-x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)
\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
d) \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x^2-1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)\left(x+1\right)+1\right]=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)^2+1\right]=0\)
Do \(\left(x+1\right)^2+1>0\)
\(\Rightarrow x\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
2 tháng 7 2018
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
YN
29 tháng 11 2021
Answer:
\(\left(2x-3\right).\left(x+1\right)-x.\left(2x+3\right)-9=0\)
\(\Rightarrow\left(2x^2+2x-3x-3\right)-2x^2-3x-9=0\)
\(\Rightarrow\left(2x^2-x-3\right)-2x^2-3x-9=0\)
\(\Rightarrow2x^2-x-3-2x^2-3x-9=0\)
\(\Rightarrow\left(2x^2-2x^2\right)-\left(x+3x\right)-\left(3+9\right)=0\)
\(\Rightarrow-4x-12=0\)
\(\Rightarrow x+3=0\)
\(\Rightarrow x=-3\)
\(2x.\left(x-3\right)-x+3=0\) (Sửa đề)
\(\Rightarrow2x.\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right).\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\2x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}}\)
\(2x.\left(x^2-4\right)+6.\left(4-x^2\right)=0\)
\(\Rightarrow2x.\left(x^2-4\right)-6.\left(x^2-4\right)=0\)
\(\Rightarrow2.\left(x-3\right).\left(x+2\right).\left(x-2\right)=0\)
Trường hợp 1: \(x-3=0\Rightarrow x=3\)
Trường hợp 2: \(x+2=0\Rightarrow x=-2\)
Trường hợp 3: \(x-2=0\Rightarrow x=2\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 7 2023
a, (\(x-2\))2 - (2\(x\) + 3)2 = 0
(\(x\) - 2 - 2\(x\) - 3)(\(x\) - 2 + 2\(x\) + 3) = 0
(-\(x\) - 5)(3\(x\) +1) = 0
\(\left[{}\begin{matrix}-x-5=0\\3x+1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\3x=-1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\) { -5;- \(\dfrac{1}{3}\)}
b, 9.(2\(x\) + 1)2 - 4.(\(x\) + 1)2 = 0
{3.(2\(x\) + 1) - 2.(\(x\) +1)}{ 3.(2\(x\) +1) + 2.(\(x\) +1)} = 0
(6\(x\) + 3 - 2\(x\) - 2)(6\(x\) + 3 + 2\(x\) + 2) = 0
(4\(x\) + 1)(8\(x\) + 5) =0
\(\left[{}\begin{matrix}4x+1=0\\8x+5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{5}{8}\end{matrix}\right.\)
S = { - \(\dfrac{5}{8}\); \(\dfrac{-1}{4}\)}
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 7 2023
d, \(x^2\)(\(x\) + 1) - \(x\) (\(x+1\)) + \(x\)(\(x\) -1) = 0
\(x\left(x+1\right)\).(\(x\) - 1) + \(x\)(\(x\) -1) = 0
\(x\)(\(x\) -1)(\(x\) + 1 + 1) = 0
\(x\left(x-1\right)\left(x+2\right)\) = 0
\(\left[{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
S = { -2; 0; 1}
a)
`(1/2+2x)(2x-3)=0`
\(=>\left[{}\begin{matrix}\dfrac{1}{2}+2x=0\\2x-3=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=3\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)
b)
`1/4-(2x+1/2)^2=0`
`=>(2x+1/2)^2=1/4`
\(=>\left[{}\begin{matrix}2x+\dfrac{1}{2}=\dfrac{1}{2}\\2x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\\ =>\left[{}\begin{matrix}2x=0\\2x=-1\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)