(5x - 30 - 2(x+6)=9
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TL
1
12 tháng 5 2023
c: =>(x+2)(x+3)(x-5)(x-6)=180
=>(x^2-3x-10)(x^2-3x-18)=180
=>(x^2-3x)^2-28(x^2-3x)=0
=>x(x-3)(x-7)(x+4)=0
=>\(x\in\left\{0;3;7;-4\right\}\)
c: =>(x-3)(x+2)(2x+1)(3x-1)=0
=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
10 tháng 11 2018
2) \(\left(x+3\right)\left(5x-15\right)+2x-6=0\)
\(\Leftrightarrow\left(x+3\right).5.\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[5\left(x+3\right)+2\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x+17\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x+17=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{17}{5}\end{matrix}\right.\)
câu 1 phải là \(\left(x-5\right)\left(x+9\right)+6x=30\)
\(\Leftrightarrow\left(x-5\right)\left(x+9\right)+6x-30=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+9\right)+6\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+9+6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-15\end{matrix}\right.\)
TN
18 tháng 7 2017
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Vậy:....
\(b,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25^2+9=30\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy :....
\(c,\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)\(\Leftrightarrow x^3+27-x\left(x^2-4\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=15-27=-12\)
\(\Leftrightarrow x=-3\)
vậy : .....
NT
1
20 tháng 7 2015
1 , <=> 25x^2 + 10x + 1 - ( 25x^2 - 9) = 30
<=> 25x^2 + 10x + 1 - 25x^2 + 9 = 30
<=> 10x + 10 = 30
<=> 10 ( x + 1) = 30
<=> x + 1 = 3
<=> x = 2
2, ( x + 3)(x^2 - 3x + 9 ) - x(x+2)(x-2) = 15
<=> x^3 - 27 - x(x^2 - 4) = 15
<=> x^3 - 27 - x^3 + 4x = 15
<=> 4x -27 = 15
<=> 4x = 15 + 27
<=> 4x =42
<=> x = 42/4 = 21/2
******************
18 tháng 9 2020
(x+2)^2= 9
=> (x+2)^2= 3^2=(-3)^2
TH1: x+2=3
=> x=3-2=1
TH2: x+2=-3
=> x=(-3)-2=-5
18 tháng 9 2020
Bài làm :
\(a,\left(x+2\right)^2-9=0\)
\(\Leftrightarrow\left(x+2\right)^2=9\)
\(\Leftrightarrow\left(x+2\right)^2=3^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=3\\x+2=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy x = 1 hoặc x = -5 .
\(b,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-\left(25x^2-3^2\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow\left(25x^2-25x^2\right)+10x=30-9-1\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)
Vậy x = 2 .
\(c,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)
\(\Leftrightarrow x^3+x^2+x-x^2-x-1+\left(x^2+2x\right)\left(2-x\right)=5\)
\(\Leftrightarrow x^3-1+2x^2-x^3+4x-2x^2=5\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(2x^2-2x^2\right)+4x=5+1\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy x = 3/2 .
Học tốt nhé
NT
1
18 tháng 11 2018
1) 2(3x+5)-6=9
2(3x+5) =9+6
2(3x+5) =15
3x+5 = 15:2
3x+5 = \(\frac{15}{2}\)
3x = \(\frac{15}{2}-5\)
3x =\(\frac{5}{2}\)
x = \(\frac{5}{2}:3\)
x = \(\frac{5}{6}\)
2) 5x+3(4+2x)=25
5x+12+6x=25
5x+6x =-12+25
11x =13
x =13:11
x =\(\frac{13}{11}\)
3) 3(4x+1)+2(x-1)=105
12x+3+2x-2=105
12x+2x = -3+2+105
14x =104
x = 104:14
x = \(\frac{52}{7}\)
4) \(30-[2\left(x-3\right)-2]=14\)
\(2\left(x-3\right)-2\)=\(30-14\)
\(2\left(x-3\right)-2=16\)
\(2\left(x-3\right)=16+2\)
\(2\left(x-3\right)=18\)
\(x-3=18:2\)
\(x-3=9\)
\(x=9+3\)
\(x=12\)
k mình nha !!!!!!
5x - 30 - 2x - 12 - 9 = 0
3x - 51 = 0
x = 17