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26 tháng 10 2021
a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)
\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)
\(\Leftrightarrow-11x^2+23x=0\)
\(\Leftrightarrow x\left(-11x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)
25 tháng 1 2023
a: =>\(\left(\dfrac{2x+1}{9}+1\right)+\left(\dfrac{2x+2}{8}+1\right)+...+\left(\dfrac{2x+9}{1}+1\right)=0\)
=>2x+10=0
=>x=-5
b: \(\Leftrightarrow\left(\dfrac{x-1}{2015}-1\right)+\left(\dfrac{x-2}{2014}-1\right)+...+\left(\dfrac{x-2014}{2}-1\right)+\left(x-2016\right)=0\)
=>x-2016=0
=>x=2016
29 tháng 10 2023
c: \(\left(2x+3\right)^2+\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\)
\(=4x^2+12x+9+4x^2-12x+9-\left(4x^2-9\right)\)
\(=8x^2+18-4x^2+9=4x^2+27\)
d: \(\left(x-1\right)\cdot\left(x^2+x+1\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\)
\(=\left(x-1\right)\left(x^2+x\cdot1+1^2\right)-\left(2x+3\right)\left[\left(2x\right)^2-2x\cdot3+3^2\right]\)
\(=x^3-1-8x^3-27=-7x^3-28\)
e: \(\left(x+1\right)^3-\left(x-1\right)^3-6x^2\)
\(=x^3+3x^2+3x+1-6x^2-\left(x^3-3x^2+3x-1\right)\)
\(=x^3-3x^2+3x+1-x^3+3x^2-3x+1\)
=2
NM
2
2 tháng 4 2021
a)ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
Ta có: \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)
Suy ra: \(x^2-1+x-2x+1=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1}
2 tháng 4 2021
b) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(\dfrac{5}{x-3}-\dfrac{2x-3}{x+3}=\dfrac{2x\left(1-x\right)}{x^2-9}\)
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(2x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(5x+15-2x^2+6x+3x-9-2x+2x^2=0\)
\(\Leftrightarrow12x+6=0\)
\(\Leftrightarrow12x=-6\)
hay \(x=-\dfrac{1}{2}\)(thỏa ĐK)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
H
4 tháng 8 2019
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
DC
24 tháng 4 2021
\(\dfrac{x+1}{x-3}\) + \(\dfrac{x-1}{x+3}\) - \(\dfrac{2x-2x^2}{9-x^2}\)
= \(\dfrac{x+1}{x-3}\)+ \(\dfrac{x-1}{x+3}\) + \(\dfrac{2x-2x^2}{x^2-9}\)
= \(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\) + \(\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\) + \(\dfrac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{\left(x+1\right)\left(x+3\right)+\left(x-1\right)\left(x-3\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x^2+3x+x+3+\left(x^2-3x-x+3\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x^2+4x+3+x^2-3x-x+3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)
HB
1
19 tháng 2 2021
Ta có: \(\left(2x+3\right)\left(2x+1\right)-\left(2x+5\right)\left(2x+7\right)=1-\left(6x^2+9x-9\right)\)
\(\Leftrightarrow4x^2+2x+6x+3-\left(4x^2+14x+10x+35\right)=1-6x^2-9x+9\)
\(\Leftrightarrow4x^2+8x+3-4x^2-24x-35-1+6x^2+9x-9=0\)
\(\Leftrightarrow6x^2-7x-42=0\)
\(\Delta=49-4\cdot6\cdot\left(-42\right)=1057\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{1057}}{12}\\x_2=\dfrac{7+\sqrt{1057}}{12}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{7-\sqrt{1057}}{12};\dfrac{7+\sqrt{1057}}{12}\right\}\)
2 tháng 7 2023
1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
23 tháng 10 2021
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
25 tháng 8 2020
a) ( 3x + 2 )( x - 1 ) - ( x + 2 )( 3x + 1 ) = 7
<=> 3x2 - x - 2 - ( 3x2 + 7x + 2 ) = 7
<=> 3x2 - x - 2 - 3x2 - 7x - 2 = 7
<=> -8x - 4 = 7
<=> -8x = 11
<=> x = -11/8
b) ( 6x + 5 )( 2x + 3 ) - ( 4x + 3 )( 3x - 2 ) = 8
<=> 12x2 + 28x + 15 - ( 12x2 + x - 6 ) = 8
<=> 12x2 + 28x + 15 - 12x2 - x + 6 = 8
<=> 27x + 21 = 8
<=> 27x = -13
<=> x = -13/27
c) 2x( x + 3 ) - ( x + 1 )( 2x + 1 ) - 5 = 9
<=> 2x2 + 6x - ( 2x2 + 3x + 1 ) - 5 = 9
<=> 2x2 + 6x - 2x2 - 3x - 1 - 5 = 9
<=> 3x - 6 = 9
<=> 3x = 15
<=> x = 5
d) ( 5x + 3 )( 4x - 7 ) - ( 10x + 9 )( 2x - 3 ) = 10
<=> 20x2 - 23x - 21 - ( 20x2 - 12x - 27 ) = 10
<=> 20x2 - 23x - 21 - 20x2 + 12x + 27 = 10
<=> -11x + 6 = 10
<=> -11x = 4
<=> x = -4/11
A
25 tháng 8 2020
a, \(\left(3x+2\right)\left(x-1\right)-\left(x+2\right)\left(3x+1\right)=7\Leftrightarrow-8x-4=7\Leftrightarrow x=-\frac{11}{8}\)
b, \(\left(6x+5\right)\left(2x+3\right)-\left(4x+3\right)\left(3x-2\right)=8\Leftrightarrow27x+21=8\Leftrightarrow x=-\frac{13}{27}\)
c, \(2x\left(x+3\right)-\left(x+1\right)\left(2x+1\right)-5=9\Leftrightarrow3x-6=9\Leftrightarrow x=5\)
d, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)
\(\left(x+3\right)=\left(6x-3\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=6x-3\\x+3=3-6x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=0\end{cases}}\)
Vậy x = {6/5; 0}