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a) \(x+5=20-\left(12-7\right)\)
\(\Rightarrow x+5=20-5\)
\(\Rightarrow x+5=15\)
\(\Rightarrow x=15-5\)
\(\Rightarrow x=10\)
b) \(15-\left(3+2x\right)=2^2\)
\(\Rightarrow3+2x=15-4\)
\(\Rightarrow3+2x=11\)
\(\Rightarrow2x=11-3\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=\dfrac{8}{2}\)
\(\Rightarrow x=4\)
c) \(-11-\left(19-x\right)=50\)
\(\Rightarrow19-x=-11-50\)
\(\Rightarrow19-x=-61\)
\(\Rightarrow x=61+19\)
\(\Rightarrow x=80\)
d) \(159-\left(25-x\right)=43\)
\(\Rightarrow25-x=159-43\)
\(\Rightarrow25-x=116\)
\(\Rightarrow x=25-116\)
\(\Rightarrow x=-91\)
e) \(\left(79-x\right)-43=-\left(17-52\right)\)
\(\Rightarrow\left(79-x\right)-43=52-17\)
\(\Rightarrow79-x-43=35\)
\(\Rightarrow36-x=35\)
\(\Rightarrow x=1\)
f) \(\left(7+x\right)-\left(21-13\right)=32\)
\(\Rightarrow7+x-8=32\)
\(\Rightarrow x-1=32\)
\(\Rightarrow x=32+1\)
\(\Rightarrow x=33\)
g) \(-x+20=-15+8+13\)
\(\Rightarrow-x+20=6\)
\(\Rightarrow x=20-6\)
\(\Rightarrow x=14\)
h) \(-\left(-x+13-142\right)+18=55\)
\(\Rightarrow x-13+142+18=55\)
\(\Rightarrow x+147=55\)
\(\Rightarrow x=55-147\)
\(\Rightarrow x=-92\)
a)
\(x-5=-1\)
\(x=-1+5\)
\(x=4\)
b)
\(x+30=4\)
\(x=4-30\)
\(x=-26\)
c)
\(x-(-24)=3\)
\(x+24=3\)
\(x=3-24\)
\(x=-21\)
d)
\(22-(-x)=12\)
\(22+x=12\)
\(x=12-22\)
\(x=-10\)
e)
\(( x + 5 ) + ( x - 9 ) = x + 2\)
\(x+5+x-9=x+2\)
\(x+x-x=2+9-5\)
\(x=6\)
f)
\(( 27 - x ) + ( 15 + x ) = x - 24\)
\(27-x+15+x=x-24\)
\(-x+x-x=-24-15-27\)
\(-x=-66\)
\(x=66\)
x - 5 = -1 x - (-24) = 3
x = -1 + 5 x + 24 = 3
x = 4 x = 3 - 24
x + 30 = 4 x = - 21
x = 4 - 30 22 - ( -x) = 12
x = - 26 22 + x = 12
x + 5 + ( x - 9) = x + 2 x = 12 - 22
x + 5 + x - 9 = x + 2 x = -10
2x - x = 2 - 5 + 9 ( 27 - x) + ( 15 + x) = x - 24
x = - 3 + 9 27 - x + 15 + x = x - 24
x = 6 27 + 15 = x - 24
x - 24 = 42
x = 42 + 24
x = 66
15 - (13 + x) = x - (23 - 17)
=> 15 - 13 - x = x - 6
=> 2x = 8
=> x = 4
bạn làm kiểu j mà ra 4 thế. cụ thể hơn đi bạn. thanks nhé
a) x + 27 = 9 ⇔ x = 9 − 27 ⇔ x = − 18
b) x – 35 = 15 ⇔ x = 15 + 35 ⇔ x = 50.
c) − 13 − x = 39 ⇔ − x = 39 + 13 ⇔ − x = 52 ⇔ x = − 52.
Ta có:
\(\left(x-1\right)^{15}=\left(x-1\right)^{13}\)
\(\Rightarrow\left(x-1\right)^2\cdot\left(x-1\right)^{13}=\left(x-1\right)^{13}\)
\(\Rightarrow\left(x-1\right)^2=\left(x-1\right)^{13}:\left(x-1\right)^{13}\)
\(\Rightarrow\left(x-1\right)^2=\left(x-1\right)^{13-13}\)
\(\Rightarrow\left(x-1\right)^2=1\)
\(\Rightarrow\left(x-1\right)=1\)
\(\Rightarrow x=1+1\)
\(\Rightarrow x=2\)
Với \(x-1=0\) hay \(x=1\), ta có: \(0^{15}=0^{13}\) (luôn đúng)
Với \(x-1\ne0\) hay \(x\ne1\), ta có: \(\left(x-1\right)^{15}=\left(x-1\right)^{13}\)
\(\Rightarrow\left(x-1\right)^2=1\Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
Vậy các giá trị nguyên \(x\) thỏa mãn là \(0;1;2\).