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7 tháng 11 2021
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
7 tháng 11 2021
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
NH
1 tháng 9 2020
\(\text{a)}\Rightarrow x-1-x-1-x+2=5\)
\(\Rightarrow-x=5\)
\(\Rightarrow x=-5\)
\(\text{Vậy x=-5}\)
\(\text{b)}\left(2x-1\right)^2-\left(2x+3\right)^2=7\)
\(\Rightarrow\left(4x^2-4x+1\right)-\left(4x^2+12x+9\right)=7\)
\(\Rightarrow4x^2-4x+1-4x^2-12x-9=7\)
\(\Rightarrow-16x-8=7\)
\(\Rightarrow-16x=15\)
\(\Rightarrow x=\frac{-15}{16}\)
\(\text{Vậy }x=\frac{-15}{16}\)
\(\text{c)}\Rightarrow16x^2-9-\left(16x^2-8x+1\right)=8\)
\(\Rightarrow-9+8x-1=8\)
\(\Rightarrow8x=18\)
\(\Rightarrow x=\frac{18}{8}=\frac{9}{4}\)
\(\text{Vậy }x=\frac{9}{4}\)
\(\text{Phần d số rất lẻ, có thể bạn chép sai đề nên mình ko chữa nha~}\)
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, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)
20 tháng 7 2018
Tìm x, biết:
1) 2x ( x - 5) - x ( 2x - 4 ) = 15
<=> 2x2 - 10x - 2x2 + 4x - 15 = 0
<=> -6x - 15 = 0
<=> -6x = 15
<=> x = -15/6
2) ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6
<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0
<=> -4x = -16
<=> x = 4
3) 4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x
<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0
<=> x + 4 = 0
<=> x = -4
4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5
<=> 2x2 + x + 6x + 3 - 2x2 - 4x + 5 = 0
<=> 3x + 8 = 0
<=> 3x = -8
<=> x = -8/3
5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0
<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0
.......
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0
<=> -2x + 40 = 0
<=> -2x = -40
<=> x = 20
Còn lại tương tự ....
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
p) \(\left(9-x\right)\left(x^2+2x-3\right)\)
\(=9\left(x^2+2x-3\right)-x\left(x^2+2x-3\right)\)
\(=9x^2+18x-27-x^3-2x^2+3x\)
\(=-x^3+7x^2+21x-27\)
n) \(\left(-x+3\right)\left(x^2+x+1\right)\)
\(=-x\left(x^2+x+1\right)+3\left(x^2+x+1\right)\)
\(=-x^3-x^2-x+3x^2+3x+3\)
\(=-x^2+2x^2+2x+3\)
o) \(\left(-6x+\dfrac{1}{2}\right)\left(x^2-4x+2\right)\)
\(=-6x\left(x^2-4x+2\right)+\dfrac{1}{2}\left(x^2-4x+2\right)\)
\(=-6x^3+24x^2-12x+\dfrac{1}{2}x^2-2x+1\)
\(=-6x^3+\dfrac{49}{2}x^2-14x+1\)
q) \(\left(6x+1\right)\left(x^2-2x-3\right)\)
\(=6x\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)
\(=6x^3-12x^2-18x+x^2-2x-3\)
\(=6x^3-11x^2-20x-3\)
r) \(\left(2x+1\right)\left(-x^2-3x+1\right)\)
\(=2x\left(-x^2-3x+1\right)+\left(-x^2-3x+1\right)\)
\(=-2x^3-6x^2+2x-x^2-3x+1\)
\(=-2x^3-7x^2-x+1\)
u) \(\left(2x-3\right)\left(-x^2+x+6\right)\)
\(=2x\left(-x^2+x+6\right)-3\left(-x^2+x+6\right)\)
\(=-2x^3+2x^2+12x+3x^2-3x-18\)
\(=-2x^3+5x^2+9x-18\)
s) \(\left(-4x+5\right)\left(x^2+3x-2\right)\)
\(=-4x\left(x^2+3x-2\right)+5\left(x^2+3x-2\right)\)
\(=-4x^3-12x^2+8x+5x^2+15x-10\)
\(=-4x^3-7x^2+23x-10\)
v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4x^3\right)\)
\(=-\dfrac{1}{2}x\left(2x+6-4x^3\right)+3\left(2x+6-4x^3\right)\)
\(=-x^2-3+2x^4+6x+18-12x^3\)
\(=2x^4-12x^3-x^2+6x+15\)
30 tháng 7 2023
p: (-x+9)(x^2+2x-3)
=-x^3-2x^2+3x+9x^2+18x-27
=-x^3+7x^2+21x-27
n: (-x+3)(x^2+x+1)
=-x^3-x^2-x+3x^2+3x+3
=-x^3+2x^2+2x+3
o: (-6x+1/2)(x^2-4x+2)
=-6x^3+24x^2-12x+1/2x^2-2x+1
=-64x^3+49/2x^2-14x+1
q: (6x+1)(x^2-2x-3)
=6x^3-12x^2-18x+x^2-2x-3
=6x^3-11x^2-20x-3
r: (2x+1)(-x^2-3x+1)
=-2x^3-6x^2+2x-x^2-3x+1
=-2x^3-7x^2-x+1
u: =-2x^3+2x^2+12x+3x^2-3x-18
=-2x^3+5x^2+9x-18
s: =-4x^3-12x^2+8x+5x^2+15x-10
=-4x^3-7x^2+23x-10
29 tháng 7 2021
\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ =\dfrac{11x+x-18}{2x-3}\\ =\dfrac{12x-18}{2x-3}\\ =\dfrac{6\left(2x-3\right)}{2x-3}\\ =6\)
\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\left(ĐKXĐ:x\ne\dfrac{3}{2};x\ne\dfrac{-3}{2}\right)\\ =\dfrac{2x+12}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x+5}{2\left(2x-3\right)}\\ =\dfrac{4x+24}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{\left(2x+5\right)\left(2x+3\right)}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x+24+4x^2+6x+10x+15}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x^2+20x+39}{2\left(2x-3\right)\left(2x+3\right)}\)
\(\dfrac{x}{2x+1}+\dfrac{-1}{4x^2-1}+\dfrac{2-x}{2x-1}\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne\dfrac{-1}{2}\right)\\ =\dfrac{x\left(2x-1\right)-1+\left(2-x\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{2x^2-x-1+4x+2-2x^2-x}{\left(2x-1\right)\left(2x+1\right)}\\ =\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{1}{2x-1}\)
25 tháng 2 2022
1: \(\Leftrightarrow\left(x-4\right)^2+14=-9\left(x-4\right)\)
\(\Leftrightarrow x^2-8x+16+14+9x-36=0\)
\(\Leftrightarrow x^2+x-6=0\)
=>(x+3)(x-2)=0
=>x=-3(nhận) hoặc x=2(nhận)
2: \(\Leftrightarrow\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9=0\)
\(\Leftrightarrow16x^2-8x+2x-1-4x^2-2x-12x^2+9=0\)
=>-8x+8=0
hay x=1(nhận)
c: \(\dfrac{1}{2\left(x-3\right)}-\dfrac{3x-5}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{2}\)
\(\Leftrightarrow x-1-2\left(3x-5\right)=\left(x-3\right)\left(x-1\right)\)
\(\Leftrightarrow x^2-4x+3=x-1-6x+10=-5x+9\)
\(\Leftrightarrow x^2+x-6=0\)
=>(x+3)(x-2)=0
=>x=-3(nhận) hoặc x=2(nhận)
14 tháng 10 2018
a) \(x^2-4x=0\)
\(x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
b) \(4x^2-9=0\)
\(\left(2x\right)^2-3^2=0\)
\(\left(2x+3\right)\left(2x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}\)
c) \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\left(2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)
d) \(x\left(2x+9\right)-4x-18=0\)
\(x\left(2x+9\right)-2\left(2x+9\right)=0\)
\(\left(2x+9\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)
e) \(\left(2x-1\right)^2-\left(x+2\right)^2=0\)
\(\left(2x-1-x-2\right)\left(2x-1+x+2\right)=0\)
\(\left(x-3\right)\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{3}\end{cases}}}\)
T
14 tháng 10 2018
\(x^2-4x=0\)
\(x.\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4=0\Leftrightarrow x=4\end{cases}}\)
\(4x^2-9=0\)
\(2^2x^2-9=0\)
\(\left(2x\right)^2-9=0\)
\(\left(2x\right)^2-3^2=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x\right)^2=\left(-3\right)^2\\\left(2x\right)^2=3^2\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\2x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}}\)
\(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\cdot\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+3\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)
\(x\left(2x+9\right)-4x-18=0\)
\(x\left(2x+9\right)-\left(4x+18\right)=0\)
\(x\left(2x+9\right)-\left(2\cdot2x+2\cdot9\right)=0\)
\(x\left(2x+9\right)-2.\left(2x+9\right)=0\)
\(\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-9\\x=0+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)
\(\left(2x-1\right)^2-\left(x+2\right)^2=0\)
\(\Rightarrow\left(2x-1\right)^2=\left(x+2\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=x+2\\2x-1=-x+2\end{cases}\Rightarrow\orbr{\begin{cases}2x=3+x\\2x=-x+3\end{cases}\Rightarrow\orbr{\begin{cases}2x-x=3\\2x+x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}}}\)
\(\)
AH
Akai Haruma
Giáo viên
27 tháng 2 2019
1.
PT \(\Leftrightarrow (x+2)(x-3)(x-4)(x+6)=16x^2\)
\(\Leftrightarrow [(x+2)(x+6)][(x-3)(x-4)]=16x^2\)
\(\Leftrightarrow (x^2+8x+12)(x^2-7x+12)=16x^2\)
\(\Leftrightarrow (a+8x)(a-7x)=16x^2\) (đặt \(x^2+12=a\) )
\(\Leftrightarrow a^2+ax-72x^2=0\)
\(\Leftrightarrow (a-8x)(a+9x)=0\Rightarrow \left[\begin{matrix} a-8x=0\\ a+9x=0\end{matrix}\right.\)
Nếu \(a-8x=0\Leftrightarrow x^2+12-8x=0\Leftrightarrow (x-2)(x-6)=0\Rightarrow \left[\begin{matrix} x=2\\ x=6\end{matrix}\right.\)
Nếu \(a+9x=0\Leftrightarrow x^2+12+9x=0\Leftrightarrow x=\frac{-9\pm \sqrt{33}}{2}\)
Vậy...........
AH
Akai Haruma
Giáo viên
27 tháng 2 2019
2.
PT \(\Leftrightarrow [(4x+7)(2x+1)][(4x+5)(x+1)]=9\)
\(\Leftrightarrow (8x^2+18x+7)(4x^2+9x+5)=9\)
\(\Leftrightarrow (2a+7)(a+5)=9\) (đặt \(a=4x^2+9x\) )
\(\Leftrightarrow 2a^2+17a+26=0\)
\(\Leftrightarrow (a+2)(2a+13)=0 \)\(\Rightarrow \left[\begin{matrix} a+2=0\\ 2a+13=0\end{matrix}\right.\)
Nếu \(a+2=0\Leftrightarrow 4x^2+9x+2=0\Leftrightarrow (4x+1)(x+2)=0\)
\(\Rightarrow \left[\begin{matrix} x=\frac{-1}{4}\\ x=-2\end{matrix}\right.\)
Nếu \(2a+13=0\Leftrightarrow 8x^2+18x+13=0\) (pt này dễ thấy vô nghiệm)
Vậy.........
\(\left(2x-3\right)^2-\left(x+5\right)\left(4x-1\right)=-9\)
\(\Leftrightarrow4x^2-12x+9-4x^2+x-20x+5=-9\)
\(\Leftrightarrow-31x+14=-9\)
\(\Leftrightarrow-31x=-23\)
\(\Leftrightarrow x=\dfrac{-23}{-31}\)
\(\Leftrightarrow x=\dfrac{23}{31}\)
\(\Leftrightarrow4x^2-12x+9-4x^2+x-20x-5=-9\)
\(\Leftrightarrow31x=13\Leftrightarrow x=\dfrac{13}{31}\)