7-3x=9-x
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NP
0
21 tháng 4 2020
Ta có: (3x - 9)(2x - 7) = (3x - 9)(4x + 3)
<=> 6x2 - 39x + 63 = 12x2 - 27x - 27
<=> 6x2 - 39x + 63 - 12x2 + 27x + 27 = 0
<=> -6x2 - 12x + 90 = 0
<=> -6(x2 + 2x - 15) = 0
<=> x2 + 5x - 3x - 15 = 0
<=> x(x + 5) - 3(x + 5) = 0
<=> (x - 3)(x + 5) = 0
<=> \(\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Vậy pt có tập nghiệm là {3; -5}
21 tháng 4 2020
(3x - 9)(2x -7) = (3x-9)(4x-3)
(3x-9)(2x-7) - (3x-9)(4x-3)=0
(3x-9)(2x-7 - 4x +3)=0
(3x-9)(-2x - 4)=0
3(x-3)(-2)(x+2)=0
(x-3)(x+2)=0
=> x-3=0 hoặc x+2=0
=> x=3 hoặc x=-2
25 tháng 1 2017
a) 9 - 25 = ( 7 - x ) - ( 25 + 7 )
9 - 25 = 7 - x - 25 - 7
=> 9 = 7 - x - 7
9 = 7 - 7 - x
9 = 0 - x
=> x = - 9
b) | x - 2 | + 4 = 10
| x - 2 | = 10 - 4
|x - 2 | = 6
=> x - 2 = ( - 6 ) hoặc x - 2 = 6
=> x = (- 4 ) hoặc x = 8
c) | 3.x + 9 | - 15 = 27
| 3.x + 9 | = 27 + 15
| 3.x + 9 | = 42
=> 3.x + 9 = ( - 42 ) hoặc 2.x+ 9 = 42
3.x = ( - 51 ) hoặc 3.x = 33
x = ( - 17 ) hoặc x = 11
GG
25 tháng 1 2017
a,9-25=(7-x)-(25+7)
-16=7-x-32
-16=(7-32)-x
-16=-25-x
\(\Rightarrow\)-25-x=-16
x =-25-16
x =-41
TG
9 tháng 2 2021
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
9 tháng 2 2021
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
2 tháng 2 2017
\(a,-5.\left(-x+7\right)-3.\left(-x-5\right)=-4.\left(12-x\right)+48\)
\(5x-35-3x+15=-48+4x+48\)
\(2x-10=4x\)
\(2x-4x=10\)
\(-2x=10\)
\(x=-5\)
\(b,\left(-x-7\right)-5.\left(-x-3\right)=12.\left(3-x\right)\)
\(-x-7+5x+15=36-12x\)
\(4x+8=36-12x\)
\(4x+12x=36-8\)
\(16x=28\)
\(x=1,75\)
các câu còn lại tương tự nha
NH
16 tháng 3 2020
a ) \(-5\times\left(-x+7\right)-3\times\left(-x-5\right)=-4\times\left(12-x\right)+48\)
\(\Leftrightarrow5x-35+3x+15=-48+4x+48\)
\(\Leftrightarrow5x-3x+4x=35-15-48+48\)
\(\Leftrightarrow2x=20\)
\(\Leftrightarrow x=10\)
b ) \(-2\times\left(15-3x\right)-4\times\left(-7x+8\right)=-5-9\times\left(-2x+1\right)\)
\(\Leftrightarrow-30+6x-28x-32=-5+18x-9\)
\(\Leftrightarrow6x-28x-18x=30+32-5-9\)
\(\Leftrightarrow-40x=48\)
\(\Leftrightarrow x=-1.2\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
TP
2
12 tháng 12 2023
a) = x.(x-7) + (x-7)^2
= (x-7).(x + x - 7)
= (x-7).(2x-7)
b) = 3x.(9-x)^2 - (9-x)^3
= (9-x)^2.(3x - 9 + x)
= (9-x)^2.(4x-9)
12 tháng 12 2023
a: \(x\left(x-7\right)+\left(7-x\right)^2\)
\(=x\left(x-7\right)+\left(x-7\right)^2\)
\(=\left(x-7\right)\left(x+x-7\right)\)
\(=\left(x-7\right)\left(2x-7\right)\)
b: \(3x\left(x-9\right)^2-\left(9-x\right)^3\)
\(=3x\left(x-9\right)^2+\left(x-9\right)^3\)
\(=\left(x-9\right)^2\left[3x+x-9\right]\)
\(=\left(x-9\right)^2\cdot\left(4x-9\right)\)
-2x=2
x=-1
7-3x=9-x
<=>7-9=-x+3x
<=>-2=2x
=>x=-2:2
=>x= -1