tìm x biết a/ |x+15|+1=3x
b/|2x-5|+x=2
ai giải đc 3 tick
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LV
1
13 tháng 3 2022
\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)
24 tháng 7 2016
a) a^x-15 = 17
a^x=32
a^x=2^5
\(\Rightarrow\)x=5
Vậy x=5
b/ (7.x-11)^3 = 2^5.5^2+200
\(\Leftrightarrow\)(7.x-11)^3=800+200
\(\Leftrightarrow\)(7.x-11)^3= 1000
\(\Leftrightarrow\)(7x-11)^3= 10^3
\(\Leftrightarrow\) 7x-11=10
\(\Leftrightarrow\)7x=21
\(\Leftrightarrow\) x=3
Vậy x=3
VA
24 tháng 7 2016
a) ax - 15 = 17
ax = 17 + 15
ax = 32
ax = 25
=> x = 5
Vậy x = 5
b) (7 . x - 11)3 = 25 . 52 + 200
(7 . x - 11)3 = 32 . 25 + 200
(7 . x - 11)3 = 1000
(7 . x - 11)3 = 103
=> 7 . x - 11 = 10
7 . x = 10 + 11
7 . x = 21
x = 21 : 7
x = 3
Vậy x = 3
Ủng hộ mk nha !!! *_*
HN
15 tháng 8 2019
a) \(2x\left(x-3\right)+6\left(3-x\right)=0\)
\(\Leftrightarrow2\left[x\left(x-3\right)+3\left(3-x\right)\right]=0\)
\(\Leftrightarrow x\left(x-3\right)+3\left(3-x\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Rightarrow x=3\)
b) \(3x\left(2x-5\right)-15\left(5-2x\right)=0\)
\(\Leftrightarrow3\left[x\left(2x-5\right)-5\left(5-2x\right)\right]=0\)
\(\Leftrightarrow x\left(2x-5\right)-5\left(5-2x\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{2}\end{cases}}\)
6 tháng 2 2016
1.4x - 5(-3+x)=7
4x - 5(x-3) =7
4x - 5x + 15=7
-1x + 15=7
-1x =-8
=> x =8
2.5(x-3) - 2(x+6)=9
5x - 15 -2x -12=9
5x - 2x -15 - 12=9
5x - 2x=9 + 12 + 15
5x - 2x= 36
3x = 36
=> x = 12
3.4(x-1) - 3(x-2)=15
4x - 4 - 3x + 6=15
4x - 3x =15 - 6 + 4
4x - 3x = 13
=> x = 13
Nhớ mink nhoa pn
4 tháng 7 2016
x10=1x x10=x (2x-15)5=(2x-15)3
=>x=1(suy đoán) =>x=1(suy đoán) Hình như đề sai (suy đoán)
4 tháng 7 2016
a)x10=1x=>x10=1=>x=1
b)x10=x=>x10-x=0=>x(x9-1)=0
TH1.x=0
TH2.x9-1=0
=>x9=1=>x=1
c)(2x-15)5=(2x-15)3
như câu b
29 tháng 3 2020
a) Mình chưa thấy x ở đâu nha bạn
b) Ta có : ( 2.x - 15) 5 = ( 2.x - 15 )3
<=. ( 2.x - 15 )5 - ( 2.x - 15 )3 = 0
<=> ( 2.x -15 )3 . [ ( 2.x - 15 )2 - 1 ] = 0
<=> \(\orbr{\begin{cases}\left(2.x-15\right)^3=0\\\left[\left(2.x-15\right)^2-1\right]=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2.x-15=0\\2.x-15=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}2.x=15\\2.x=16\end{cases}\Leftrightarrow}\orbr{\begin{cases}\frac{15}{2}\left(L\right)\\x=8\end{cases}}}\)
Vậy x = 8
DA
2
TG
16 tháng 1 2021
a) \(x^2+2x=\left(x-2\right).3x\)
\(\Leftrightarrow x^2+2x=3x^2-6x\)
\(\Leftrightarrow x^2+2x-3x^2+6x=0\)
\(\Leftrightarrow-2x^2+8x=0\)
\(\Leftrightarrow-2x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy S = {0;4}
b) \(x^3+x^2-x-1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\mp1\end{matrix}\right.\)
Vậy: S = {-1; 1}
c) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]=40\)
\(\Leftrightarrow\left(x^2+5x+x+5\right)\left(x^2+4x+2x+8\right)=40\)
\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt x2 + 6x + 5 = t
\(\Leftrightarrow t.\left(t+3\right)=40\)
\(\Leftrightarrow t^2+3t=40\)
\(\Leftrightarrow t^2+2.t.\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{169}{4}\)
\(\Leftrightarrow\left(t+\dfrac{3}{2}\right)^2=\dfrac{169}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}t+\dfrac{3}{2}=\dfrac{13}{2}\\t+\dfrac{3}{2}=-\dfrac{13}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{13}{2}-\dfrac{3}{2}=\dfrac{10}{2}=5\\t=-\dfrac{13}{2}-\dfrac{3}{2}=-\dfrac{16}{2}=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x+5=5\\x^2+6x+5=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\x^2+6x+13=0\end{matrix}\right.\)
Mà: \(x^2+6x+13=x^2+2.x.3+9+4=\left(x+3\right)^2+4\ne0\)
=> x2 + 6x = 0
<=> x. (x + 6) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy S = {0; -6}
16 tháng 1 2021
a) Ta có: \(x^2+2x=\left(x-2\right)\cdot3x\)
\(\Leftrightarrow x\left(x+2\right)-3x\left(x-2\right)=0\)
\(\Leftrightarrow x\left[\left(x+2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow x\left(x+2-3x+6\right)=0\)
\(\Leftrightarrow x\left(-2x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy: S={0;4}
b) Ta có: \(x^3+x^2-x-1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(x-1\right)\cdot\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy: S={-1;1}
c) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-40=0\)
\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)
\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+40-40=0\)
\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)=0\)
\(\Leftrightarrow\left(x^2+6x\right)\left(x^2+6x+13\right)=0\)
\(\Leftrightarrow x\left(x+6\right)\left(x^2+6x+13\right)=0\)
mà \(x^2+6x+13>0\forall x\)
nên \(x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy: S={0;-6}
a. |x+15|+1=3x
=> |x+15|=3x-1
+) x+15=3x-1
=> x-3x=-1-15
=> -2x=-16
=> x=8
+) x+15=-(3x-1)
=> x+15=-3x+1
=> x+3x=1-15
=> 4x=-14
=> x=-7/2
Vậy x E {-7/2; 8}
b. |2x-5|+x=2
=> |2x-5|=2-x
+) 2x-5=2-x
=> 2x+x=2+5
=> 3x=7
=> x=7/3
+) 2x-5=-(2-x)
=> 2x-5=x-2
=> 2x-x=-2+5
=> x=3
Vậy x E {7/3; 3}.
a.|x+15|+1=3x
=>|x+15|=3x-1
+)x+15=3x-1
=>x-3x=-1-15
=>-2x=-16
=>x=8
+)x+15=-3x+1
=>x+3x=1-15
=>4x=-14
=>x=-7/2
Vậy:x=8 và x=-7/2
b.|2x-5|+x=2
=>|2x-5|=2-x
+)2x-5=2-x
=>2x+x=2+5
=>3x=7
=>x=7/3
+)2x-5=-2-x
=>2x+x=-2+5
=>3x=3
=>x=1
Vậy:x=7/3 và x=1.