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NN
1
GH
5 tháng 3 2018
a) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy .................
b) \(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(2x+1-x+3\right)\left(2x+1+x-3\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...............
c) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
P/s: tới đây bn tự giải tiếp nha
TT
6
6 tháng 4 2017
(1)
\(\left\{{}\begin{matrix}\left|x^2-1\right|\ge0\\\left|x-1\right|\ge0\end{matrix}\right.\) => để (1) có nghiệm \(\left\{{}\begin{matrix}x^2-1=0\\x-1=0\end{matrix}\right.\)=> x=1
(2)
tương tự \(\left\{{}\begin{matrix}x^2-1=0\\2x-2=0\\x-1=0\end{matrix}\right.\) => 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/\(\left(x-1\right)^2=\left(2x+5\right)^2\Leftrightarrow\left(x-1\right)^2-\left(2x+5\right)^2=0\Leftrightarrow\left(x-1+2x+5\right)\left(x-1-2x-5\right)=0\Leftrightarrow\left(3x+4\right)\left(-x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}3x+4=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{-4}{3}\\0\\-6\end{matrix}\right.\)
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 .
TT
3
29 tháng 3 2017
a) 2x3+5x2-3x=0
<=> 2x3+6x2-x2-3x=0
<=> 2x2(x+3)-x(x+3)=0
<=> (x+3)(2x2-x)=0
<=> (x+3)x(2x-1)=0
\(\Rightarrow\left\{{}\begin{matrix}x+3=0\\x=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
c) x3+1=x(x+1)
<=> (x+1)(x2+1-x)-x(x+1)=0
<=> (x+1)(x2-2x+1)=0
<=> (x+1)(x-1)2=0
\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy ...
a. (x-1) (x2 +x+1)= x3+x
=>x3 -1=x3 +x
=> x3 -1-x3 =x=>x=-1
b)(3x+2)2 - (2x+3)2=0
(3x+2-2x-3)(3x+2+2x+3)=0
=>(x-1)(5x+5)=0
=>x-1=0 hoặc 5x+5 =0
+nếu x-1=0 thì x=1
+nếu 5x+5 =0 thì 5x=-5 =>x=-1