2x(x+1)=x+9
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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: \(S=\left\{\dfrac{7-\sqrt{1057}}{12};\dfrac{7+\sqrt{1057}}{12}\right\}\)
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
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: \(S=\left\{-\dfrac{1}{2}\right\}\)
16 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
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
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
MH
14 tháng 1 2016
1. a. 9.(x-2)+1=2-2x
=> 9x-18+1=2-2x
=> 9x+2x = 2+18-1
=> 11x = 19
=> x=19/11
b. 9|x| = 45
=> |x| = 5
=> x=5 hoặc x=-5
c. 9:|x|=3
=> |x| = 3
=> x=3 hoặc x=-3
d. 5.(1-x)+2x=9-2x
=> 5-5x+2x=9-2x
=> -5x+2x+2x=9-5
=> -x=4
=> x=-4
DC
24 tháng 4 2021
\(\dfrac{x+1}{x-3}\) + \(\dfrac{x-1}{x+3}\) - \(\dfrac{2x-2x^2}{9-x^2}\)
= \(\dfrac{x+1}{x-3}\)+ \(\dfrac{x-1}{x+3}\) + \(\dfrac{2x-2x^2}{x^2-9}\)
= \(\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\) + \(\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\) + \(\dfrac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{\left(x+1\right)\left(x+3\right)+\left(x-1\right)\left(x-3\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x^2+3x+x+3+\left(x^2-3x-x+3\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x^2+4x+3+x^2-3x-x+3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)
18 tháng 3 2020
\(a,3\left(2x-1\right)-2\left(1-x\right)=x+9\)
\(6x-3-2+2x=x+9\)
\(8x-5=x+9\)
\(8x-5-x-9=0\)
\(7x-14=0\)
\(7x=14\)
\(x=2\)
\(-3\left(2x-1\right)-2\left(1-x\right)=x+9\left(1-x\right)\)
\(-6x+3-2+2x=x+9-9x\)
\(-4x+1=-8x-9\)
\(-4x+1+8x+9=0\)
\(4x+10=0\)
\(4x=10\)
\(x=\frac{10}{4}=\frac{5}{2}\)
\(c,\left(1-x\right)\left(2x-1\right)-2\left(2-x\right)\left(2+x\right)=x=9\)
SAI ĐỀ
18 tháng 3 2020
a) 3(2x - 1) - 2(1 - x) = x + 9
<=> 6x - 3 - 2 + 2x = x + 9
<=> 6x + 2x - x = 9 + 3 + 2
<=> 7x = 14
<=> x = 14/7 = 2
vậy giải phương trình ta đc x = 2
b) -3(2x - 1) - 2(1 - x) = x + 9(1 - x)
<=> -6x + 3 - 2 + 2x = x + 9 - 9x
<=> -6x + 2x + 9x - x = 9 - 3 + 2
<=> 4x = 8
<=> x = 8/4 = 2
c) (1 - x)(2x - 1) - 2(2 - x)(2 + x) = x + 9
<=> 2x - 1 - 2x2 + x - 8 + 2x2 = x + 9
<=> 2x + x - x = 9 +1 +8
<=> 2x = 18
<=> x = 9
2*x|x+1|=x9=2*x+2=x+9=x=7 .Vậy x bằng 7