Tìm x, biết: a) x2+14x+48=0
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1 tháng 8 2016
b. 1500(x-7)=0
x-7=0
x=7
c. (2x-4)(48-12x)=0
2x-4=0 hoặc 48-12x=0
x=2 hoặc x=4
d. (x+12)(x-1)=0
x+12=0 hoặc x-1=0
x=-12 hoặc x=1
bài 2 :
a . 128-3(x+4)=23
3(x+4)=105
x+4=35
x=31
b. [(14X+26).3+55]:5=35
(14x+26).3+55=175
(14x+26).3=120
14x+26=40
14x=14
x=1
d. 720:[41-(2X-5)]=23.5
41-(2x-5)=720:(23.5)
41-(2x-5)=144/23
2x-5=799/23
2x=914/23
x=457/23
23 tháng 7 2017
b, 1500.(x – 7) = 0
<=>1500x-10500=0
<=>1500x=10500
<=>x=7
Vậy x=7
c,(2.x – 4).(48 – 12.x) = 0
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-4=0\\48-12x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\12x=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy x=2 hoặc x=4
d, (x + 12).(x – 1) =0
\(\Leftrightarrow\left\{{}\begin{matrix}x+12=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-12\\x=1\end{matrix}\right.\)
Vậy x=-12 hoặc x=1
Bài 2:
a) 128- 3(x+ 4) = 23
\(\Leftrightarrow\)128-(3x+12)=23
\(\Leftrightarrow\)128-3x-12=23
\(\Leftrightarrow\)116-3x=23
\(\Leftrightarrow\)3x=116-23
\(\Leftrightarrow\)3x=93
\(\Leftrightarrow\)x=31
Vậy x=31
b) [(14x+ 26). 3+ 55]: 5= 35
\(\Leftrightarrow\)(14x+ 26). 3+ 55=175
\(\Leftrightarrow\)42x+78+55=175
\(\Leftrightarrow\)42x+133=175
\(\Leftrightarrow\)42x=175-133
\(\Leftrightarrow\)42x=42
\(\Leftrightarrow\)x=1
Vậy x=1
d, 720: [41- (2x- 5)]= 23. 5
\(\Leftrightarrow\)720: 41- (2x- 5)=115
\(\Leftrightarrow\)41-(2x- 5)=720:115
\(\Leftrightarrow\)41-(2x- 5)=\(\dfrac{144}{23}\)
\(\Leftrightarrow\)2x-5=\(\dfrac{799}{23}\)
\(\Leftrightarrow\)2x=\(\dfrac{914}{23}\)
\(\Leftrightarrow\)x=\(\dfrac{457}{23}\)
Vậy x=\(\dfrac{457}{23}\)
3 tháng 7 2021
a) Ta có: \(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow-\left(x-2\right)-2\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)\left[1+2\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
b) Ta có: \(8x^3-72x=0\)
\(\Leftrightarrow8x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy: S={0;3;-3}
c) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
d) Ta có: \(2x^3+3x^2+3+2x=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
hay \(x=-\dfrac{3}{2}\)
e) Ta có: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
Vậy: S={0;1;-2}
f) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
Vậy: S={0;2;12}
L
24 tháng 8 2024
a)
\(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9=3^2\)
\(\Rightarrow x+2=\pm3\)
\(\Rightarrow x=-5;1\)
b)
\(25x^2-10x+1=0\)
\(\left(5x\right)^2-2\cdot5x+1^2=0\)
\(\Rightarrow\left(5x+1\right)^2=0\)
\(\Rightarrow5x+1=0\)
\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)
c)
\(x^2+14x+49=0\)
\(\Rightarrow x^2+2\cdot7x+7^2=0\)
\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)
\(\Rightarrow x=-7\)
d)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)
\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)
\(\Rightarrow2x+255=0\)
\(\Rightarrow2x=-255\)
\(\Rightarrow x=\dfrac{-255}{2}\)
LD
1
17 tháng 9 2018
a, |1 - 9x| - 10 = 0
=> |1 - 9x| = 10
=> 1 - 9x = 10 hoặc 1 - 9x = -10
=> 9x = -9 hoặc 9x = 11
=>x = -1 hoặc x = 11/9
vậy_
b, |3 - 14x| - 13 = 0
=> |3 - 14x| = 13
=> 3 - 14x = 13 hoặc 3 - 14x = -13
=> 14x = -10 hoặc 14x = 16
=> x = -10/14 hoặc x = 16/14
các phần sau tương tự
\(x^2+14x+48=0\)
\(\Leftrightarrow x^2+6x+8x+48=0\)
\(\Leftrightarrow x\left(x+6\right)+8\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+6=0\\x+8=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-6\\x=-8\end{array}\right.\)