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AH
Akai Haruma
Giáo viên
11 tháng 8 2021
1.
$(x-2)(x-5)=(x-3)(x-4)$
$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)
Vậy pt vô nghiệm.
2.
$(x-7)(x+7)+x^2-2=2(x^2+5)$
$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$
$\Leftrightarrow -51=10$ (vô lý)
Vậy pt vô nghiệm.
AH
Akai Haruma
Giáo viên
11 tháng 8 2021
3.
$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$
$\Leftrightarrow 4x+10=-8$
$\Leftrightarrow 4x=-18$
$\Leftrightarrow x=-4,5$
4.
$(x+1)^2=(x+3)(x-2)$
$\Leftrightarrow x^2+2x+1=x^2+x-6$
$\Leftrightarrow x=-7$
12 tháng 9 2021
1: Ta có: \(x^2-2x+5-\left(x-7\right)\left(x+2\right)\)
\(=x^2-2x+5-x^2-2x+7x-14\)
\(=3x-9\)
2: Ta có: \(-5x\left(x-5\right)+\left(x-3\right)\left(x^2-7\right)\)
\(=-5x^2+25x+x^3-7x-3x^2+21\)
\(=x^3-8x^2+18x+21\)
3: Ta có: \(x\left(x^2-x-2\right)-\left(x+5\right)\left(x-1\right)\)
\(=x^3-x^2-2x-x^2-4x+5\)
\(=x^3-2x^2-6x+5\)
HH
8 tháng 7 2018
1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)
ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
8 tháng 9 2023
2:
a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8
=>x^2-x-12-x^2+4x+5=8
=>3x-7=8
=>3x=15
=>x=5
b: =>3x^2+3x-2x-2-3x^2-21x=13
=>-20x=15
=>x=-3/4
c: =>x^2-25-x^2-2x=9
=>-2x=25+9=34
=>x=-17
d: =>x^3-1-x^3+3x=1
=>3x-1=1
=>3x=2
=>x=2/3
H9
HT.Phong (9A5)
CTVHS
30 tháng 8 2023
Bài 1:
a) \(x\left(x+1\right)+x\left(x-1\right)-2x^2\)
\(=x^2+x+x^2-x-2x^2\)
\(=2x^2-2x^2\)
\(=0\)
b) \(\left(x+2\right)\left(x^2-x+1\right)-\left(x-2\right)\left(x^2+x+1\right)\)
\(=x^3-x^2+x+2x^2-2x+2-x^3-x^2-x+2x^2+2x+2\)
\(=\left(x^3-x^3\right)+\left(-x^2+2x^2-x^2+2x^2\right)+\left(x-2x-x+2x\right)+\left(2+2\right)\)
\(=2x^2+4\)
c) \(\left(3-x\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)
\(=\left(x-3\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)
\(=\left[\left(x-3\right)+\left(x+7\right)\right]^2\)
\(=\left(x-3+x+7\right)^2\)
\(=\left(2x+4\right)^2\)
2 tháng 10 2020
Sorry mình nhầm câu a
a) (2x - 1)2 + (x + 3)2 - 5(x + 7)(x - 7) = 0
b) (x + 2)(x2 - 2x + 4) - x(x2 + 2) = 15
c) (x + 3)3 - x(3x + 1)2 + (2x - 1)(4x2 - 2x + 1) = 28
d) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
Giải:
a) (2x - 1)2 + (x + 3)2 - 5(x + 7)(x - 7) = 0
\(\Leftrightarrow\) 4x2 - 4x + 1 + x2 + 6x + 9 - 5(x2 - 49) = 0
\(\Leftrightarrow\) 4x2 - 4x + 1 + x2 + 6x + 9 - 5x2 + 245 = 0
\(\Leftrightarrow\) 2x + 255 = 0
\(\Leftrightarrow\) 2x = - 255
\(\Leftrightarrow\) x = - 255 : 2
\(\Leftrightarrow\) x = \(-\frac{255}{2}\)
Vậy x = \(-\frac{255}{2}\)
b) (x + 2)(x2 - 2x + 4) - x(x2 + 2) = 15
\(\Leftrightarrow\) x3 + 8 - x3 - 2x = 15
\(\Leftrightarrow\) 8 - 2x = 15
\(\Leftrightarrow\) 2x = 8 - 1
\(\Leftrightarrow\) 2x = - 7
\(\Leftrightarrow\) x = - 7 : 2
\(\Leftrightarrow\) x = \(-\frac{7}{2}\)
Vậy x = \(-\frac{7}{2}\)
c) (x + 3)3 - x(3x + 1)2 + (2x - 1)(4x2 - 2x + 1) = 28
\(\Leftrightarrow\) x3 + 6x2 + 27x + 27 - x(9x2 + 6x + 1) + 8x3 - 1 = 28
\(\Leftrightarrow\) x3 + 6x2 + 27x + 27 - 9x3 - 6x2 - x + 8x3 - 1 = 28
\(\Leftrightarrow\) 26x + 26 = 28
\(\Leftrightarrow\) 26x = 28 - 26
\(\Leftrightarrow\) 26x = 2
\(\Leftrightarrow\) x = 2 : 26
\(\Leftrightarrow\) x = \(\frac{1}{13}\)
Vậy x = \(\frac{1}{13}\)
d) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
\(\Leftrightarrow\) x6 - 2x2 + 1 - (x6 - 1) = 0
\(\Leftrightarrow\) x6 - 2x2 + 1 - x6 + 1 = 0
\(\Leftrightarrow\) -2x2 + 2 = 0
\(\Leftrightarrow\) -2x2 = - 2
\(\Leftrightarrow\) x2 = - 2 : (- 2)
\(\Leftrightarrow\) x2 = 1
\(\Leftrightarrow\) x = 1 hoặc x = - 1
Vậy x \(\in\) {1; - 1}
26 tháng 7 2022
1: \(A=\left(-x+5\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)\)
\(=-x^2+2x+5x-10+x^2-49=7x-59\)
\(B=\left(3x+1\right)^2-\left(3x-2\right)\left(3x+2\right)\)
\(=9x^2+6x+1-9x^2+4=6x+5\)
=>7x-59=6x+5
=>x=64
2: \(A=\left(5x-1\right)\left(x+1\right)-2\left(x-3\right)^2\)
\(=5x^2+5x-x-1-2x^2+12x-9\)
\(=3x^2+16x-10\)
\(B=\left(x+2\right)\left(3x-1\right)-\left(x+4\right)^2+x^2-x\)
\(=3x^2-x+6x-2-x^2-8x-16+x^2-x\)
\(=3x^2-4x-18\)
=>16x-10=-4x-18
=>20x=-8
hay x=-2/5
\(\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-7}{x+2}\)
ĐKXĐ : x ≠ 1 , x ≠ -2
pt <=> \(\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{x+2}{\left(x-1\right)\left(x+2\right)}+\frac{7\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=0\)
<=> \(\frac{3-x-2+7x-7}{\left(x-1\right)\left(x+2\right)}=0\)
<=> \(\frac{6x-6}{\left(x-1\right)\left(x+2\right)}=0\)
=> 6x - 6 = 0
<=> x = 1 ( ktm )
Vậy phương trình vô nghiệm