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22 tháng 4 2020
d, (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
Đặt x2 + 4x + 8 = t ta được:
t2 + 3xt + 2x2 = 0
\(\Leftrightarrow\) t2 + xt + 2xt + 2x2 = 0
\(\Leftrightarrow\) t(t + x) + 2x(t + x) = 0
\(\Leftrightarrow\) (t + x)(t + 2x) = 0
Thay t = x2 + 4x + 8 ta được:
(x2 + 4x + 8 + x)(x2 + 4x + 8 + 2x) = 0
\(\Leftrightarrow\) (x2 + 5x + 8)[x(x + 4) + 2(x + 4)] = 0
\(\Leftrightarrow\) (x2 + 5x + \(\frac{25}{4}\) + \(\frac{7}{4}\))(x + 4)(x + 2) = 0
\(\Leftrightarrow\) [(x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\)](x + 4)(x + 2) = 0
Vì (x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\) > 0 với mọi x
\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
Vậy S = {-4; -2}
Mình giúp bn phần khó thôi!
Chúc bn học tốt!!
A
22 tháng 4 2020
c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)
⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇒x2+x+1+2x2-5=4x-4
⇔3x2-3x=0
⇔3x(x-1)=0
⇔x=0 (TMĐK) hoặc x=1 (loại)
Vậy tập nghiệm của phương trình đã cho là:S={0}
12 tháng 10 2020
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
19 tháng 8 2018
Bài 4 : Tìm x biết:
a, 4x2 - 49 = 0
\(\Leftrightarrow\) (2x)2 - 72 = 0
\(\Leftrightarrow\) (2x - 7)(2x + 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-7=0\\2x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b, x2 + 36 = 12x
\(\Leftrightarrow\) x2 + 36 - 12x = 0
\(\Leftrightarrow\) x2 - 2.x.6 + 62 = 0
\(\Leftrightarrow\) (x - 6)2 = 0
\(\Leftrightarrow\) x = 6
19 tháng 8 2018
e, (x - 2)2 - 16 = 0
\(\Leftrightarrow\) (x - 2)2 - 42 = 0
\(\Leftrightarrow\) (x - 2 - 4)(x - 2 + 4) = 0
\(\Leftrightarrow\) (x - 6)(x + 2) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
f, x2 - 5x -14 = 0
\(\Leftrightarrow\) x2 + 2x - 7x -14 = 0
\(\Leftrightarrow\) x(x + 2) - 7(x + 2) = 0
\(\Leftrightarrow\) (x + 2)(x - 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)
21 tháng 9 2017
a) \(x^3-4x^2-9x+36=0\Leftrightarrow x^3-7x^2+12x+3x^2-21x+36=0\) \(x\left(x^2-7x+12\right)+3\left(x^2-7x+12\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-7x+12\right)=0\) \(\Leftrightarrow\left(x+3\right)\left(x^2-7x+12\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x-4x+12\right)=0\) \(\Leftrightarrow\left(x+3\right)\left(x\left(x-3\right)-4\left(x-3\right)\right)=0\Leftrightarrow\left(x+3\right)\left(x-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\x-4=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=4\\x=3\end{matrix}\right.\) vậy \(x=-3;x=4;x=3\)
b) \(5x^2-4\left(x^2-2x+1\right)-5=0\) \(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\Leftrightarrow x^2-x+9x-9=0\)
\(\Leftrightarrow x\left(x-1\right)+9\left(x-1\right)\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+9=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\) vậy \(x=-9;x=1\)
c) đề có sai o bn
d) \(x^3-3x+2=0\Leftrightarrow x^3+x^2-2x-x^2-x+2=0\)
\(\Leftrightarrow x\left(x^2+x-2\right)-\left(x^2+x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x^2+x-2\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x+2x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x\left(x-1\right)+2\left(x-1\right)\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-1\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+2=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-2\\x=1\end{matrix}\right.\)
vậy \(x=1;x=-2\)
LH
21 tháng 9 2017
1. \(x^3-4x^2-9x+36=0\)
\(\Rightarrow x^2.\left(x-4\right)-9\left(x-4\right)=0\)
\(\Rightarrow\left(x^2-9\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-9=0\Rightarrow x\in\left\{3;-3\right\}\\x-4=0\Rightarrow x=4\end{matrix}\right.\)
Vậy ..........
2. \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Rightarrow5x^2-4\left(x^2-1\right)-5=0\)
\(\Rightarrow5x^2-4x^2+4-5=0\)
\(\Rightarrow x^2-1=0\)
\(\Rightarrow x^2=1\)
\(\Rightarrow x=\pm1\)
Vậy .......
3. \(x^3-3x+2=0\)
\(\Rightarrow x^3-4x+x+2=0\)
\(\Rightarrow x.\left(x^2-4\right)+x+2=0\)
\(\Rightarrow x.\left(x-2\right).\left(x+2\right)+x+2=0\)
\(\Rightarrow\left(x+2\right).\left(x^2-2x+1\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x-1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+2\right)=0\\\left(x-1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy .......
Q
18 tháng 7 2017
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
AH
23 tháng 10 2018
Thời gian có hạn copy cái này hộ mình vào google xem nha :
https://lazi.vn/quiz/d/16491/nhac-edm-la-loai-nhac-the-loai-gi
Vào xem xong các bạn nhận được 1 thẻ cào mệnh giá 100k nhận thưởng bằng cách nhắn tin vs mình và 1 phần thưởng bí mật là chiếc áo đá bóng,....
Có 500 giải nhanh nha đã có 401 người nhận rồi
OKz
MB
8
MC
16 tháng 7 2018
\(a)9.x^2=25\)
\(\Leftrightarrow x^2=\dfrac{25}{9}\)
\(\Leftrightarrow x^2=\left(\pm\dfrac{5}{9}\right)^2\)
\(\Leftrightarrow x=\pm\dfrac{5}{9}\)
\(\Leftrightarrow x\in\left\{\dfrac{5}{9};-\dfrac{5}{9}\right\}\)
16 tháng 7 2018
b, \(x^3-\dfrac{1}{4}x=0\)
⇔ \(x\left(x^2-\dfrac{1}{4}\right)=0\)
⇔ \(x\left(x+\dfrac{1}{2}\right)\left(x-\dfrac{1}{2}\right)=0\)
⇔\(\left[{}\begin{matrix}x=0\\x+\dfrac{1}{2}\\x-\dfrac{1}{2}\end{matrix}\right.=0\) ⇔ \(\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy xϵ \(\left\{0,\pm\dfrac{1}{2}\right\}\)
b) Ta có: \(\dfrac{x-2}{4}=\dfrac{2x+1}{3}\)
\(\Leftrightarrow3\left(x-2\right)=4\left(2x+1\right)\)
\(\Leftrightarrow3x-6=8x+4\)
\(\Leftrightarrow3x-8x=4+6\)
\(\Leftrightarrow-5x=10\)
hay x=-2
Vậy: x=-2
a) x3-9x2-4x-36=0
⇔ x2(x-9)-4(x-9)=0
⇔ (x-9)(x2-4)=0
⇒ Xảy ra 2 trường hợp:
- TH1: x-9=0 ⇔ x=9
- TH2: x2-4=0 ⇔ x=2 hoặc x=-2
Vậy x=9 hoặc x=2 hoặc x=-2.