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NT
6 tháng 8 2019
\(\frac{2x+3}{2x+1}-\frac{2x+5}{2x+7}=1-\frac{6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)
\(\Leftrightarrow1+\frac{2}{2x+1}-1-\frac{2}{2x+7}-1=-\frac{6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)
\(\Leftrightarrow\frac{4x+14-4x-2+6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}=1\)
\(\Leftrightarrow\frac{6x^2+9x+3}{\left(2x+1\right)\left(2x+7\right)}=1\)
\(\Leftrightarrow\frac{6x^2+6x+3x+3}{\left(2x+1\right)\left(2x+7\right)}=1\)
\(\Leftrightarrow\frac{6x\left(x+1\right)+3\left(x+1\right)}{\left(2x+1\right)\left(2x+7\right)}=1\)
\(\Leftrightarrow\frac{3\left(2x+1\right)\left(x+1\right)}{\left(2x+1\right)\left(2x+7\right)}=1\)
\(\Leftrightarrow3x+3=2x+7\)
\(\Leftrightarrow x=4\)
J
1
24 tháng 5 2022
\(\Leftrightarrow\dfrac{\left(2x+3\right)\left(2x+7\right)-\left(2x+5\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x+7\right)}=\dfrac{4x^2+16x+7-6x^2-9x+9}{\left(2x+1\right)\left(2x+7\right)}\)
\(\Leftrightarrow-2x^2+7x+16=4x^2+20x+21-4x^2-12x-5\)
\(\Leftrightarrow-2x^2+7x+16=8x+16\)
\(\Leftrightarrow-2x^2-x=0\)
=>x(2x+1)=0
=>x=0(nhận) hoặc x=-1/2(loại)
30 tháng 6 2021
a/ (2x + 1)(4x – 3) – 6x(x + 5) – 2x(x – 7) + 18x
=8x^2-6x+4x-3-6x^2-30x-2x^2+14x+18x
=-3
vậy...
18 tháng 5 2021
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
18 tháng 5 2021
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
15 tháng 3 2020
Điều kiện: \(x \ne -\dfrac{1}{2}\) và \(x \ne -\dfrac{7}{2}\)
\(\begin{array}{l} \dfrac{{\left( {2x + 3} \right)\left( {2x + 7} \right)}}{{\left( {2x + 1} \right)\left( {2x + 7} \right)}} - \dfrac{{\left( {2x + 5} \right)\left( {2x + 1} \right)}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} = \dfrac{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} - \dfrac{{6{x^2} + 9x - 9}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}\\ \Leftrightarrow \dfrac{{4{x^2} + 20x + 21 - 4{x^2} - 12x - 5}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} = \dfrac{{4{x^2} + 16x + 7 - 6{x^2} - 9x + 9}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}\\ \Leftrightarrow \dfrac{{8x + 16}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}} = \dfrac{{ - 2{x^2} + 7x + 16}}{{\left( {2x + 7} \right)\left( {2x + 1} \right)}}\\ \Rightarrow 8x + 16 = - 2{x^2} + 7x + 16 \Leftrightarrow 2{x^2} + x = 0 \Leftrightarrow x\left( {2x + 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0 \text{(nhận)}\\ x = - \dfrac{1}{2} \text{(loại)} \end{array} \right. \end{array}\)
Vậy phương trình có nghiệm duy nhất $x=0$
TP
2
29 tháng 8 2016
TÌM X
a) (3x+2)(2x+9)-(6x+1)(x+2)=7
=> 6x2 + 31x +18 - 6x2 - 13x - 2 - 7 = 0
=> 18x + 9 = 0 => 9(2x + 1) = 0 => 2x + 1 = 0 => x = -1/2
b) (x-2)(x+5)-(x+3)(x+2)=-6
=> x2 + 3x - 10 - x2 - 5x -6 + 6 = 0 => -2x -10 = 0 => -2(x + 5) = 0
=> x + 5 = 0 => x = -5
c) 3(2x-1)(3x-1)-(2x-3)(9x-1)=0
=> 18x2 - 15x +3 - 18x2 + 29x -3 = 0 => 14x = 0 => x = 0
NT
29 tháng 8 2016
a) \(\left(3x+2\right)\left(2x+9\right)-\left(6x+1\right)\left(x+2\right)=7\\\Rightarrow 6x^2+31x+18-6x^2-16x-2-7=0\\ \Rightarrow18x+9=0\Rightarrow9\left(2x+1\right)=0\Rightarrow2x+1=0\Rightarrow x=-\frac{1}{2}\)
b) \(\left(x-2\right)\left(x+5\right)-\left(x+3\right)\left(x+2\right)=-6\\ \Rightarrow x^2+3x-10-x^2-5x-6+6=0\\ \Rightarrow-2x-10=0\\ \Rightarrow-2\left(x+5\right)=0\\ \Rightarrow x+5=0\\ \Rightarrow x=-5\)
c) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\\ \Rightarrow18x^2-15x+3-18x^2+29x-3=0\\ \Rightarrow14x=0\\ \Rightarrow x=0\)
5 tháng 9 2017
(6x+1)(2x-5)=12x2-30x+2x-5=12x2-28x-5
(2x+5)2-2x(2x+8)=4x2+20x+25-4x2-16x=4x+25
(3x-5)(2x-1)-(2x+3)(3x+7)+30x=6x2-3x-10x+5=6x2-13x+5
(X-1)2-(x+1)(x-1)=x2-2x+1-x2+1=-2x+2
(3x+2)(9x2-6x+4)-(3+x)(x-3)=27x3+8+9-x2=27x3-x2+17
VN
0
\(\frac{2x+3}{2x+1}-\frac{2x+5}{2x+7}=\frac{1-6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)
\(\Leftrightarrow\frac{\left(2x+3\right)\left(2x+7\right)-\left(2x+5\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x+7\right)}=\frac{1-6x^2+9x-9}{\left(2x+1\right)\left(2x+7\right)}\)
\(\Rightarrow\left(2x+3\right)\left(2x+7\right)-\left(2x+5\right)\left(2x+1\right)=1-6x^2+9x-9\)
\(\Leftrightarrow4x^2+20x+21-4x^2-12x-5=1-6x^2+9x-9\)
\(\Leftrightarrow8x-16=1-6x^2+9x-9\)
\(\Leftrightarrow8x-16-1+6x^2-9x+9=0\)
\(\Leftrightarrow6x^2-x-8=0\)
Tự làm nốt nha
Trl
-Bạn chuyên toán thcs làm đúng r nhé !~
Học tốt
nhé bạn ~