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24 tháng 2 2020
a) Ta có: \(\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x+5\right)\left(2x-1\right)-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2-x+10x-5-\left(2x^2+2x-3x-3\right)=0\)
\(\Leftrightarrow2x^2+9x-5-2x^2+x+3=0\)
\(\Leftrightarrow10x-2=0\)
hay 10x=2
\(\Leftrightarrow x=\frac{1}{5}\)
Vậy: \(x=\frac{1}{5}\)
b) Ta có: \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+9x+x+9=x^2+5x+3x+15\)
\(\Leftrightarrow x^2+10x+9-x^2-8x-15=0\)
\(\Leftrightarrow2x-6=0\)
hay 2x=6
\(\Leftrightarrow x=3\)
Vậy: x=3
c) Ta có: \(\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)\)
\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)
\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)
\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)
\(\Leftrightarrow33x-1=0\)
\(\Leftrightarrow33x=1\)
hay \(x=\frac{1}{33}\)
Vậy: \(x=\frac{1}{33}\)
d) Ta có: \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)\)
\(\Leftrightarrow3x^2+5x-6x-10=2x^2+2x-4x-4\)
\(\Leftrightarrow3x^2-x-10=2x^2-2x-4\)
\(\Leftrightarrow3x^2-x-10-2x^2+2x+4=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
đ) Ta có: \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)
e) Ta có: \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)
Vì \(3\ne0\)
nên \(\left[{}\begin{matrix}x-4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4\right\}\)
AH
Akai Haruma
Giáo viên
24 tháng 2 2020
a) $(x+5)(2x-1)=(2x-3)(x+1)$
$\Leftrightarrow 2x^2+9x-5=2x^2-x-3$
$\Leftrightarrow 10x=2\Rightarrow x=\frac{1}{5}$
b)
$(x+1)(x+9)=(x+3)(x+5)$
$\Leftrightarrow x^2+10x+9=x^2+8x+15$
$\Leftrightarrow 2x=6\Rightarrow x=3$
c)
$(3x+5)(2x+1)=(6x-2)(x-3)$
$\Leftrightarrow 6x^2+13x+5=6x^2-20x+6$
$\Leftrightarrow 33x=1\Rightarrow x=\frac{1}{33}$
NK
10 tháng 12 2019
(2x-1)(3-2x)+(2x+3)2=(6x-4x2-3+2x)+(4x2+12x+9)
= 6x-4x2-3+2x+4x2+12x+9
=20x+6
(2x-1)2-x(4x-2)= (4x2-4x+1)-(4x2-2x)
= 4x2-4x+1-4x2+2x
=-2x +1
(x+4)(x-3)-(x-5)2 =(x2-3x+4x-12)-(x2-10x+25)
=x2-3x+4x-12-x2+10x-25
=11x-37
(2-3x)(x+4)+(2x-5)2= (2x+8-3x2-12x)+(4x2-20x+25)
= 2x+8-3x2-12x+4x2-20x+25
= x2-30x+32
(x-1)(2x+3)-(x+2)2 =(2x2+3x-2x-3)-(x2+4x+4)
=2x2+3x-2x-3-x2-4x-4
=x2-3x-7
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 ....
BL
3 tháng 3 2020
Bài 1:
1. \(x-8=3-2\left(x+4\right)\)
\(x-8=3-2x-8\)
\(3x=3\Rightarrow x=1\)
2. \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(2x+6-3x+3=2\)
\(-x+9=2\Rightarrow x=7\)
3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(4x-20-3x+1=x-19\)
\(0x=0\Rightarrow x=0\)
4. \(7-\left(x-2\right)=5\left(2x-3\right)\)
\(7-x+2=10x-15\)
\(-11x=-24\Rightarrow x=\frac{24}{11}\)
5. \(32-4\left(0,5y-5\right)=3y+2\)
\(32-2y+20=3y+2\)
\(-5y=-50\Rightarrow y=10\)
6. \(3\left(x-1\right)-x=2x-3\)
\(3x-3-x=2x-3\)
\(0x=0\Rightarrow x=0\)
Bài 2:
1. \(\frac{2-x}{3}=\frac{3-2x}{5}\)
\(\frac{\left(2-x\right)5}{15}-\frac{\left(3-2x\right)3}{15}=0\)
\(\frac{10-5x-9+6x}{15}=0\)
\(x+1=0\Rightarrow x=-1\)
2. \(\frac{3-4x}{4}=\frac{x+2}{5}\)
\(\frac{5\left(3-4x\right)}{20}-\frac{4\left(x+2\right)}{20}=0\)
\(\frac{15-20x-4x-8}{20}=0\)
\(7-24x=0\)
\(24x=7\Rightarrow x=\frac{7}{24}\)
16 tháng 6 2023
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
TT
1 tháng 5 2018
a \(2x+2>4\\ \Leftrightarrow2\left(x+1\right)>4\\ \Leftrightarrow x+1>2\\ \Leftrightarrow x>1\)
b \(3x+2>-5\\ \Leftrightarrow3x>-7\\ \Leftrightarrow x>\dfrac{-7}{3}\)
c \(10-2x>2\\ \Leftrightarrow2\left(5-x\right)>2\\ \Leftrightarrow5-x>1\\ \Leftrightarrow-x>-4\\ \Leftrightarrow x< 4\)
d \(1-2x< 3\\ \Leftrightarrow-2x< 2\\ \Leftrightarrow2x>2\\ \Leftrightarrow x>1\)
KT
4 tháng 5 2018
a)2x+2>4
<=> 2x>4-2
<=>2x>2
<=>x>1
Vậy...
b)3x+2>-5
<=>3x>-5-2
<=>3x>-7
<=>x>\(\dfrac{-7}{3}\)
Vậy...
c)10-2x>2
<=>-2x>-10+2
<=>-2x>-8
<=>x<4
Vậy...
d)1-2x<3
<=>-2x<3-1
<=>-2x<2
<=>x>-1
Vậy...
e)10x+3-5\(\le\)14x+12
<=>10x-2\(\le\)14x+12
<=>10x-14x\(\le\)2+12
<=>-4x\(\le\)14
<=>x\(\ge\)\(\dfrac{-7}{2}\)
Vậy...
f)(3x-1)<2x+4
<=> 3x-2x<1+4
<=>x<5
Vậy...
12 tháng 9 2017
bài này bạn nhân lần lượt ra, cuối cùng hết giá trị của x, cò lại số tự nhiên. vậy là đã cm được biểu thức k phụ thuộc vào giá trị của biến rồi đó.
VD:
\(\left(x-3\right)\left(x^2+3x+9\right)-x^3+7\)
\(=x^3+3x^2+9x-3x^2-9x-27-x^3+7\)
\(=-20\)
HL
12 tháng 7 2017
A = \(\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)\)
A = \(x^2-6x+9-4x^2+1=-3x^2-6x+10\)
B = \(\left(2x-3\right)^2-\left(x-1\right)\left(2x+1\right)\)
B = \(4x^2-12x+9-2x^2-x+2x+1\)
B = \(2x^2-11x+10\)
C = \(4x\left(x-3\right)^2-\left(4-2x\right)^2\)
C = \(4x\left(x^2-6x+9\right)-16+16x-4x^2\)
C = \(4x^3-24x^2+36x-16+16x-4x^2\)
C = \(4x^3-28x^2+52x-16\)
D = \(3x\left(x-1\right)\left(x-2\right)-x\left(2x-1\right)^2\)
D = \(\left(3x^2-3x\right)\left(x-2\right)-x\left(2x-1\right)^2\)
D = \(3x^3-6x^2-3x^2+6x-x\left(4x^2-4x+1\right)\)
D = \(3x^3-9x^2+6x-4x^3+4x^2-x\)
D = \(-x^3-5x^2+5x\)
8 tháng 6 2019
Tìm x:
1. 3x (2x + 3) - (2x + 5).(3x - 2) = 8
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=0 \)
\(\Leftrightarrow-2x+10=0\Leftrightarrow x=5\)
Vậy x = 5
2. 4x (x -1) - 3(x2 - 5) -x2 = (x - 3) - (x + 4)
\(\Leftrightarrow4x^2-4x-3x^2+15-x^2=x-3-x-4\)
\(\Leftrightarrow-4x+15=-7\)
\(\Leftrightarrow-4x=-22\Leftrightarrow x=\frac{11}{2}\)
Vậy x = \(\frac{11}{2}\)
3. 2 (3x -1) (2x +5) - 6 (2x - 1) (x + 2) = -6
\(\Leftrightarrow2\left(6x^2+15x-2x-5\right)-6\left(2x^2+4x-x-2\right)=-6\)
\(\Leftrightarrow12x^2+30x-4x-10-12x^2-24x+6x+12=-6\)
\(\Leftrightarrow8x=-8\Leftrightarrow x=-1\)
Vậy x = -1
4. 3 ( 2x - 1) (3x - 1) - (2x - 3) (9x - 1) - 3 = -3
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-18x^2+2x+27x-3-3=-3\)
\(\Leftrightarrow18x^2-6x-9x+3-18x^2+2x+27x-6=-3\)
\(\Leftrightarrow14x=0\Leftrightarrow x=0\)
Vậy x = 0
5. (3x - 1) (2x + 7) - ( x + 1) (6x - 5) = (x + 2) - (x - 5)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5=7\)
\(\Leftrightarrow18x=9\Leftrightarrow x=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
6. 3xy (x + y) - (x + y) (x2 + y2 + 2xy) + y3 = 27
\(\Leftrightarrow3x^2y+3xy^2-\left(x+y\right)^3+y^3=27\)
\(\Leftrightarrow3x^2y+3xy^2-x^3-y^3-3x^2y-3xy^2+y^3=27\)
\(\Leftrightarrow-x^3=27\)
\(\Leftrightarrow x=-3\)
Vậy x = -3
7. 3x (8x - 4) - 6x (4x - 3) = 30
\(\Leftrightarrow24x^2-12x-24x^2+12x=30\)
\(\Leftrightarrow0=30\) ( vô lý)
Vậy pt vô nghiệm
8. 3x (5 - 2x) + 2x (3x - 5) = 20
\(\Leftrightarrow15x-6x^2+6x^2-10x=20\)
\(\Leftrightarrow5x=20\Leftrightarrow x=4\)
Vậy x = 4
\(\frac{4}{x^2+2x-3}-\frac{2x-5}{x+3}-\frac{2x}{x-1}ĐK:x\ne1;-3\)
\(=\frac{4}{\left(x+3\right)\left(x-1\right)}-\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{4-2x^2+2x+5x-5-2x^2-6x}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{-1-4x^2+x}{\left(x-1\right)\left(x+3\right)}\)