K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TP
1
6 tháng 2 2017
a/ (x+5)(3x+2)^2=x^2(x+5)
(x+5)(9x^2+12x+4)=x^2(x+5)
9x^3+12x^2+4x+45x^2+60x+20=x^3+5x^2
9x^3-x^3+12x^2+45x^2-5x^2+4x+60x=-20
8x^3+52x^2+64x+20=0
........................
HX
2
31 tháng 3 2018
x + 3x + 4x + 3x + 1 = 0
⇒x + x + 2x + 2x + 2x + 2x + x + 1 = 0
⇒x x + 1 + 2x x + 1 + 2x x + 1 + x + 1 = 0 ⇒ x + 1 x + x + x + x + x + 1 = 0 ⇒ x + 1 x x + 1 + x x + 1 + x + 1 = 0 ⇒ x + 1 x + 1 x + x + 1 = 0 ⇒ x + 1 x + x + 1 = 0 ⇒ x + 1 = 0 vix̀ + x + 1 ≠ 0 ⇒x + 1 = 0 ⇒x = −1 vậy pt có No ......... 3 2x − 3 − 6 x − 3 = 5 4x + 3 − 17 ⇔ 30 10 2x − 3 − 30 5 x − 3 = 30 6 4x + 3 − 30 17.30 ⇔20x − 30 − 5x + 15 = 24x + 18 − 510 ⇔20x − 5x − 24x = 18 − 510 + 30 − 15
⇔− 9x = −477 ⇔x = 53
vậy pt có No........
31 tháng 3 2018
\(x^4+3x^3+4x^2+3x+1=0\)
\(\Rightarrow x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)
\(\Rightarrow x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^3+x^2+x^2+x+x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left[x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\right]=0\)
\(\Rightarrow\left(x+1\right)\left(x+1\right)\left(x^2+x+1\right)=0\)
\(\Rightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)
\(\Rightarrow\left(x+1\right)^2=0\left(vìx^2+x+1\ne0\right)\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
vậy pt có No .........
\(\frac{2x-3}{3}-\frac{x-3}{6}=\frac{4x+3}{5}-17\)
\(\Leftrightarrow\frac{10\left(2x-3\right)}{30}-\frac{5\left(x-3\right)}{30}=\frac{6\left(4x+3\right)}{30}-\frac{17.30}{30}\)
\(\Leftrightarrow20x-30-5x+15=24x+18-510\)
\(\Leftrightarrow20x-5x-24x=18-510+30-15\)
\(\Leftrightarrow-9x=-477\)
\(\Leftrightarrow x=53\)
vậy pt có No........
3 tháng 5 2017
a. (3x-4)2=9(x-1)(x+1)
<=> 9x2-24x+16=9x2-9
<=> -24x=-25
<=> x=\(\dfrac{25}{24}\)
Vậy S=\(\left\{\dfrac{25}{24}\right\}\)
b. (4x-5)2-4(x-2)2=0
<=> (4x-5)2-(2x-4)2=0
<=> (4x-5-2x+4)(4x-5+2x-4)=0
<=> (2x-1)(6x-9)=0
<=> \(\left[{}\begin{matrix}2x-1=0\\6x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy S=\(\left\{\dfrac{1}{2};\dfrac{3}{2}\right\}\)
3 tháng 5 2017
c. |x2-x|= -2x
Ta có: |x2-x|=x2-x khi x2-x\(\ge0\) hay x\(\ge1\)
=> x2-x= -2x
<=> x2-x+2x=0
<=> x2+x=0
<=> x(x+1)=0
<=> \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) (không thỏa mãn điều kiện x\(\ge1\))
Lại có: |x2-x|= x-x2 khi x2-x<0 hay x<1
=> x-x2= -2x
<=> x-x2+2x=0
<=> 3x-x2=0
<=> x(3-x)=0
x=0 (thỏa mãn điều kiện x<1)
hoặc: 3-x=0<=> x=3 (không thỏa mãn điều kiện x<1)
Vậy S=\(\left\{0\right\}\)
d. \(\dfrac{x+3}{x-3}+\dfrac{48x^3}{9-x^2}=\dfrac{x-3}{x+3}\)
ĐKXĐ: \(x\ne\pm3\)
Ta có:\(\dfrac{x+3}{x-3}+\dfrac{48x^3}{9-x^2}=\dfrac{x-3}{x+3}\)
<=> \(\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{48x^3}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}\)
=> x2+6x+9-48x3=x2-6x+9
<=> 12x-48x3=0
<=> 12x(1-4x2)=0
<=> 12x(1-2x)(1+2x)=0
<=> \(\left[{}\begin{matrix}x=0\\1-2x=0\\1+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\) (thỏa mãn ĐKXĐ)
Vậy S=\(\left\{0;\pm0,5\right\}\)
TD
4 tháng 5 2017
a ) ( 3x - 4 )2 = 9 (x-1)(x+1)
\(\Leftrightarrow\) 9x2 - 24x + 16 = 9 ( x2 - 1 )
\(\Leftrightarrow\) 9x2 - 24x + 16 = 9x2 - 9
\(\Leftrightarrow\) 9x2 - 24x - 9x2 = - 9 - 16
\(\Leftrightarrow\) -24x = -24
\(\Leftrightarrow\) x = 1
Vậy phương trình có nghiệm x = 1 .
21 tháng 8 2020
a/\(\left(4x-1\right)\left(x+5\right)=x^2-25\Leftrightarrow4x^2+20x-x-5=x^2-25\Leftrightarrow3x^2+19x+20\)\(\Leftrightarrow\left[{}\begin{matrix}\frac{-4}{3}\\-5\end{matrix}\right.\)
b/
\(2x^3-6x^2=x^2-3x\Leftrightarrow2x^3-6x^2-x^2+3x=0\Leftrightarrow2x^2\left(x-3\right)-x\left(x-3\right)=0\Leftrightarrow\left(2x^2-x\right)\left(x-3\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}2x^2-x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}\\3\\0\end{matrix}\right.\)
c/\(x\left(x+3\right)^3-\frac{x}{4}\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2x-\frac{x}{4}\right]=0\Leftrightarrow\left(x+3\right)\left[\left(x^2+6x+9\right)x-\frac{x}{4}\right]=0\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\frac{x}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^3+6x^2+\frac{35}{4}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{5}{2}\\x=-\frac{7}{2}\end{matrix}\right.\)
d/\(\left(x-1\right)^2=\left(2x+5\right)^2\Leftrightarrow\left(x-1\right)^2-\left(2x+5\right)^2=0\Leftrightarrow\left(x-1+2x+5\right)\left(x-1-2x-5\right)=0\Leftrightarrow\left(3x+4\right)\left(-x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}3x+4=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{-4}{3}\\0\\-6\end{matrix}\right.\)
23 tháng 2 2020
a) Ta có: \(\left(2x+3\right)^2-\left(5+x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3+5+x\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-3\\3x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{2}\\x=\frac{-8}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-3}{2};\frac{-8}{3}\right\}\)
b) Ta có: \(\left(2x+5\right)^2-\left(2x-5\right)^2=6x+8\)
\(\Leftrightarrow\left(2x+5+2x-5\right)\left(2x+5-2x+5\right)-6x-8=0\)
\(\Leftrightarrow40x-6x-8=0\)
\(\Leftrightarrow34x=8\)
\(\Leftrightarrow x=\frac{8}{34}=\frac{4}{17}\)
Vậy: \(x=\frac{4}{17}\)
c) Ta có: \(\left(4x+3\right)^2=4\left(x-1\right)^2\)
\(\Leftrightarrow16x^2+24x+9=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)
\(\Leftrightarrow12x^2+32x+5=0\)
\(\Leftrightarrow12x^2+2x+30x+5=0\)
\(\Leftrightarrow2x\left(6x+1\right)+5\left(6x+1\right)=0\)
\(\Leftrightarrow\left(6x+1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+1=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=-1\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)
d) Ta có: \(\left(7x-1\right)\left(3x-2\right)-49x^2+14x=1\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2\right)-\left(49x^2-14x+1\right)=0\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2\right)-\left(7x-1\right)^2=0\)
\(\Leftrightarrow\left(7x-1\right)\left[3x-2-\left(7x-1\right)\right]=0\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2-7x+1\right)=0\)
\(\Leftrightarrow\left(7x-1\right)\left(-4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-1=0\\-4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=1\\-4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{7}\\x=\frac{-1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{7};\frac{-1}{4}\right\}\)
7 tháng 5 2020
a)
\(\left(5x+3\right)\cdot\left(x^2+4\right)\cdot\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x+3=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{5}\\x=4\end{matrix}\right.\)
b)
\(\left(4x-1\right)\cdot\left(x-3\right)-\left(x-2\right)\cdot\left(5x+2\right)=0\\ \Leftrightarrow4x^2-12x-x+3-5x^2-2x+10x+4=0\\ \Leftrightarrow-x^2-5x+7=0\\ \Rightarrow x=\left[{}\begin{matrix}-\frac{5+\sqrt{53}}{2}\\-\frac{5-\sqrt{53}}{2}\end{matrix}\right.\)
c)
\(\left(x+3\right)\cdot\left(x-5\right)+\left(x+3\right)\cdot\left(3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(x-5+3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(4x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)
d)
\(\left(x+6\right)\cdot\left(3x-1\right)+x^2-36=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x^2-36\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x+6\right)\cdot\left(x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1+x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(4x-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)
e)
\(0.75x\cdot\left(x+5\right)=\left(x+5\right)\cdot\left(3-1.25x\right)\\ \Leftrightarrow0.75x\cdot\left(x+5\right)-\left(x+5\right)\cdot\left(3-1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(0.75x-3+1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(2x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.\)