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các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha
a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)
-\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)
-6 < \(x\) < -4
vì \(x\) \(\in\) Z nên \(x\) = -5
a)4.(x+3)-(2x-12)=x-(-11+4)
4x+12-2x+12=x+11-4
2x+24=x+7
2x-x=-24+7
x=-17
b)-(4x-13)+(5x-4)=-3-(-15+7)
-4x+13+5x-4=-3+15-7
(-4x+5x)+13-4=12-7
x+9=5
x=-4
c)(5x-3)-(-2x+4)=6x-12
5x-3+2x-4=6x-12
5x+2x-6x=3+4-12
x=-5
d)(15x+20)-(9x-3)=5x-(-12)
15x+20-9x+3=5x+12
15x-9x-5x=-20-3+12
x=-11
e,(7x+14)+(3x-8)=-(-9x+3)
7x+14+3x-8=9x-3
7x+3x-9x=-14+8-3
x=-9
a) \(\left(9^4.8+9^4.5\right):\left(9^2.\left(10-1\right)\right)\)
=\(9^4.13:9^3=13.9=117\)
b) 100-(75-25)=100-50=50
Bài 1 Tìm x biết:
a)65-(29-x)=32
65 -29+x=31
x=31-65+29
x=-5
b)(x+5)-(x+23)=x-34
x+5 -x +23 = x-34
(x-x)+ (23+5)=x-34
0+28=x-34
28=x-34
28+34=x
62=x
=>x=62
c)(16-x)+(x-38)=x+44
16-x+x-38=x+44
-x+x-x=44-16+38
-x=36
=>x=-36
d)-12+3(-x+7)=-18
3(-x+7)=-18+12
3(-x+7)=-6
-x+7=-6:3
-x+7=-2
-x=-2-7
-x=-9
=>x=9
Baif 2
d)|7-x|=10
=> \(\left[{}\begin{matrix}7-x=10\\7-x=-10\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=7-10\\x=-10-7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=-3\\x=-17\end{matrix}\right.\)
e)(x-6).(7-2x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}x-6=0\\7-2x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+6\\2x=7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=7:2\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=3,5\end{matrix}\right.\)
f)(9-x).(2x+8)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}9-x=0\\2x+8=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+9\\2x=-8\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
g)x(-x+8).(-3x-18)=0
\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\-x+8=0\\-3x-18=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=0+8\\-3x=0+18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=8\\-3x=18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=18:\left(-3\right)\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=-6\end{matrix}\right.\)
h)(-x+8).(x-54).(-24-x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)
\(\text{A)17×(x-8)=-54}\)
\(x-8=-54:17\)
\(x-8=\frac{-54}{17}\)
\(x=\frac{-54}{17}+8\)
\(x=\frac{82}{17}\)
\(\text{B)(12-2×x)÷4=-12}\)
\(12-2.x=-12:4\)
\(12-2.x=-3\)
\(2.x=12+3\)
\(2.x=15\)
\(\Rightarrow x=\frac{15}{2}\)
\(\text{C)9x+(-7)x=(-56)}\)
\(x.\left[9+\left(-7\right)\right]=-56\)
\(x.2=-56\)
\(x=-56:2\)
\(x=-28\)
\(\text{D)(3x -9)×(x+12)=0}\)
\(\Rightarrow\orbr{\begin{cases}3x-9=0\\x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=9\\x=-12\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-12\end{cases}}}\)
\(\Rightarrow x\in\left\{3;-12\right\}\)
\(\text{E) (x+4)^3=-8}\)
\(\left(x+4\right)^3=2^3\)
\(\Rightarrow x+4=2\)
\(x=2-4\)
\(x=-2\)
\(\text{F)(x)-7=11}\)
mk ko hiểu đề bài
Th1 , \(x-7=11\)
\(x=11+7\)
\(x=18\)
TH2, \(|x|-7=11\)
\(|x|=11+7\)
\(|x|=18\)
\(\Rightarrow x\in\left\{\pm18\right\}\)
học tốt
E) (x + 4)3 = -8
=> (x+4)3 = (-2)3
=> x + 4 = -2
=> x = -2 - 4
=> x = -6
Vậy x = -6
F) Cái chỗ (x) có phải là giá trị tuyệt đối không ạ ? Mình sửa đề bài như sau :
|x| - 7 = 11
\(\Rightarrow\orbr{\begin{cases}x-7=11\\x-7=-11\end{cases}\Rightarrow\orbr{\begin{cases}x=11+7\\-11+7\end{cases}\Rightarrow}\orbr{\begin{cases}x=18\\x=-4\end{cases}}}\)
Vậy x = 18 hoặc x= -4