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ban lam sai rui de mk lam lai nhe.
\(12.\left(x-1\right):3=4^3-2^3\)
\(12.\left(x-1\right):3=64-8\)
\(12.\left(x-1\right):3=56\)
\(12.\left(x-1\right)=56.3\)
\(12.\left(x-1\right)=168\)
\(x-1=168:12\)
\(x-1=14\)
\(x=15\)
\(\dfrac{24}{x}:\dfrac{8}{3}=\dfrac{3}{5}\)
\(\dfrac{24}{x}=\dfrac{3}{5}.\dfrac{8}{3}\)
\(\dfrac{24}{x}=\dfrac{8}{5}\)
\(\dfrac{24}{x}=\dfrac{24}{15}\)
=>x=5
Vậy x=5
\(x+3\dfrac{1}{2}+x=24\dfrac{1}{4}\)
\(\left(x+x\right)+3\dfrac{1}{2}=24\dfrac{1}{4}\)
\(x.2+\dfrac{7}{2}=\dfrac{97}{4}\)
\(x.2=\dfrac{97}{4}-\dfrac{7}{2}\)
\(x.2=\dfrac{97}{4}-\dfrac{14}{4}\)
\(x.2=\dfrac{83}{4}\)
\(x=\dfrac{83}{4}:2\)
\(x=\dfrac{83}{4}.\dfrac{1}{2}\)
\(x=\dfrac{83}{8}\)
\(x=10\dfrac{3}{8}\)
a) \(\frac{24}{x}:\frac{8}{3}=\frac{3}{5}\) \(\frac{24}{x}=\frac{3}{5}.\frac{8}{3}\) \(\frac{24}{x}=\frac{8}{5}\) \(x=24.5:8\) \(x=15\)
b) Đề bài sai rồi
Giải
a) 24 + (15-x) = 16
<=> 15-x=16-24
<=> 15-x=(-8)
<=> x = 15-(-8) => x = 23
b) 11 - (24-x) =
<=> 24-x= 11 - (-27)
<=> 24 - x = 38
<=> x = 24 - 38 => x = -14
c) |x-16| = 42
<=> x-16=42 và x-16=-42
<=> x = 42 + 16 ; x =-42+16
<=> x = 58 ; x = -26
d) 10-2|x| = (-2) . (-3)
<=> 10 - (-2) . (-3) = 2|x|
<=> 10 - 6 = 2|x|
<=> 2|x| = 4
<=> |x| = 4 : 2 = 2
=> x = 2 hoặc x = (-2)
A/ 24+(15-x)=16
(15-x)= 16- 24
(15-x)= - 18
x = 5 - (-18)
x = 23
B/ 11-(24-x)=(-3)3
(24-x)=11- 27
(24-x)= -16
x = 24 -(- 16)
x= 40
C/
\(\left|x-16\right|=42\hept{\begin{cases}x-16=42\\x-16=-42\end{cases}}\hept{\begin{cases}x=42+16\\x=-42+16\end{cases}}\hept{\begin{cases}x=58\\x=-26\end{cases}}\)
D/ 10-2|x|=(-2)x(-3)
10-2|x|=6
2|x|=10-6
2|x|=4
x=4:2
x=2
a) 15-3(2x-9)=6
=)3(2x-9)=15-6=9
=)6x-27=9 =)6x=9+27=36
=)x=36:6=6
a, 15-3 (2 x -9) = 6
3 (2 x -9) = 9
2 x - 9 = 3
2 x = 12
x = 6
b, 8 (x-3)-2(x-3) = 24
(8-2) (x-3) = 24
6 (x - 3) = 24
x - 3 = 4
x = 7
còn câu c mik ko bik lm thông cảm
Câu 1 : 65.(-19)+19.(-24)= (-65).19+19.(-24)=19.[(-65)+(-24)]=19.(-89)=-1691
Câu 2:a) 7x.(2+x)-7x(x+3)=14
=>7x.(2+x-x-3)=14
=>7x.(-1)=14
=>(-7).x=14
=>x=14:(-7)
=>x=-2
Vậy x=-2
a, \(390-\left(x-7\right)=13^2:12\)
\(390-\left(x-7\right)=\) \(\dfrac{169}{12}\)
\(x-7=390-\dfrac{169}{12}\)
\(x-7=\dfrac{4511}{12}\)
\(x=\dfrac{4511}{12}+7\)
\(x=\dfrac{4595}{12}\)
Vậy ...
b, \(\left(x-35.2^2\right):7=3^3-24\)
\(\left(x-35.4\right):7=27-24\)
\(\left(x-140\right):7=3\)
\(\Leftrightarrow\left(x-140\right)=3.7\)
\(\Leftrightarrow x-140=21\)
\(\Leftrightarrow x=161\)
Vậy .....
c) \(x-6:2-\left(4^2.3-24\right):2:6=3\)
\(x-3-\left(16.3-24\right):2:6=3\)
\(x-3-\left(48-24\right):2:6=3\)
\(x-3-24:2:6=3\)
\(x-3-2=3\)
\(x=3+2+3\)
\(x=8\)
Vậy ......
d) \(4x-5=5+5^2+5^3+.....+5^{99}\)
Đặt :
\(A=5+5^2+.........+5^{99}\)
\(\Leftrightarrow5A=5^2+5^3+..........+5^{100}\)
\(\Leftrightarrow5A-A=\left(5^2+5^3+......+5^{100}\right)-\left(5+5^2+....+5^{99}\right)\)
\(\Leftrightarrow4A=5^{100}-5\)
\(\Leftrightarrow A=\dfrac{5^{100}-5}{4}\)
\(\Leftrightarrow4x+5=\dfrac{5^{100}-5}{4}\)
Đến đây thì sao nữa nhỉ ?
e) \(\left(2x-1\right)^4=625\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=5\\\left(2x-1\right)^4=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy ....
a/
\(-24+\left(x+4\right)^4=10^3\)3
\(\Leftrightarrow-24+x^4+16x^3+96x^2+256x+256=10^3\)
<=>\(x^4+16x^3+96x^2+256x-768=0\)
Giải trên tập số phức ta được
\(x=-\sqrt{32}-4\)
\(x=\sqrt{32}-4\)
\(x=-\sqrt{32}i-4\)
\(x=\sqrt{32}1-4\)=> Phần a kog có giá trị nguyên nào của x thỏa mãn phương trình
b/
2(x+7)-3(6-x)=-24
<=> 2x+14-18+3x=-24
<=>5x=-20
<=>x=-4
Vậy x=-4
c/
\(3x-6x^2=0\)
\(\Leftrightarrow3x\left(1-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\1-2x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)(x = 1/2 kog thỏa mãn yêu cầu)
Vậy x=0
a/\(\left(x+4\right)^4=1000+24\)
\(\Rightarrow x^4+8x^2+4^4-1024=0\)
\(\Rightarrow x^4+8x^2-768\)
\(\Rightarrow x^4-24x+32x-768=0\)
\(\Rightarrow x.\left(x-24\right)+32.\left(x-24\right)\)
\(\Rightarrow\left(x+32\right).\left(x-24\right)\Rightarrow\orbr{\begin{cases}x=-32\\x=24\end{cases}}\)
b/2x+14-18+3x=-24
5x=-24-14+18
x=-20/5=-4
c/3x-6x\(^2\) =0
\(3x.\left(1-2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\rightarrow x=0\\1-2x=0\rightarrow x=\frac{1}{2}\end{cases}}\)
KL bAN tu lam nhe
2\(^{x+3}\) - 23 = 24
2\(^{x+3}\) - 8 = 24
2\(^{x+3}\) = 24 + 8
2\(^{x+3}\) = 32
\(2^{x+3}\) = 25
\(x+3\) = 5
\(x=5-3\)
\(x=2\)
Vậy \(x=2\)