Tìm X
a) X + 5 = 25
b) X - 12 = 48
c) 66 - X = 16
kb với mik nhé
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a: \(\left(-120\right):15+12\left(2x-1\right)=52\)
=>\(12\left(2x-1\right)-8=52\)
=>\(12\left(2x-1\right)=60\)
=>\(2x-1=\dfrac{60}{12}=5\)
=>2x=5+1=6
=>\(x=\dfrac{6}{2}=3\)
c: \(x+4⋮x+1\)
=>\(x+1+3⋮x+1\)
=>\(3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
d: \(2x+7⋮x+2\)
=>\(2x+4+3⋮x+2\)
=>\(3⋮x+2\)
=>\(x+2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{-1;-3;1;-5\right\}\)
e: \(3x⋮x-1\)
=>\(3x-3+3⋮x-1\)
=>\(3⋮x-1\)
=>\(x-1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{2;0;4;-2\right\}\)
a) Ta có: \(\left(-12\right)-\left|13-x\right|=-21\)
\(\Leftrightarrow-\left|x-13\right|-12=-21\)
\(\Leftrightarrow-\left|x-13\right|=-9\)
\(\Leftrightarrow\left|x-13\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=9\\x-13=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=22\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{22;4\right\}\)
b) Ta có: \(8\le\left|x\right|< 9\)
\(\Leftrightarrow\left|x\right|=8\)
\(\Leftrightarrow x\in\left\{8;-8\right\}\)
Vậy: \(x\in\left\{8;-8\right\}\)
c) Ta có: \(x-\left(-25+x\right)=13-x\)
\(\Leftrightarrow x+25-x-13+x=0\)
\(\Leftrightarrow x+12=0\)
hay x=-12
Vậy: x=-12
d) Ta có: \(\left(15-30\right)+x=x-\left(27-\left|-8\right|\right)\)
\(\Leftrightarrow x-15=x-\left(27-8\right)\)
\(\Leftrightarrow x-15-x+19=0\)
\(\Leftrightarrow-4=0\)(vô lý)
Vậy: \(x\in\varnothing\)
\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
Câu a) xem lại đề giùm nhé em
b) \(\left(x-1\right)^3=9^3\)
\(x-1=9\)
\(x=10\)
Vậy \(x=10\)
c) \(\left(x-1\right)^2=25\)
\(x-1=5\) hoặc \(x-1=-5\)
* \(x-1=5\)
\(x=6\)
* \(x-1=-5\)
\(x=-4\)
Vậy \(x=-4\); \(x=6\)
d) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy \(x=2\)
e) Sửa đề: \(\left(2x+4\right)^3=64\)
\(\left(2x+4\right)^3=4^3\)
\(2x+4=4\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)
a: \(3x\left(x-3\right)+4x-12=0\)
=>\(3x\left(x-3\right)+\left(4x-12\right)=0\)
=>\(3x\left(x-3\right)+4\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(3x+4\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b: Sửa đề:\(\left(x+1\right)\left(x^2-x+1\right)-x^3+2x=17\)
\(\Leftrightarrow x^3+1-x^3+2x=17\)
=>2x+1=17
=>2x=17-1=16
=>\(x=\dfrac{16}{2}=8\)
c: \(\left(x-3\right)\left(x+5\right)+\left(x-1\right)^2-6x^4y^2:3x^2y^2=15x\)
=>\(x^2+2x-15+x^2-2x+1-2x^2=15x\)
=>\(15x=-14\)
=>\(x=-\dfrac{14}{15}\)
a. 8 x 37 x 5 x 25 = (8 x 25) x 37 x 5
= 200 x 5 x 37 = 1000 x 37 = 37000
b. 48 x 34 + 6 x 66 x 8 = 48 x 34 + 48 x 66 = 48 x ( 34 + 66) = 48 x 100 = 4800
c. 17 x (13 - 5) + 17 x 5 = 17 x 8 + 17 x 5 = 17 x (8 + 5) = 17 x 13 = 221
d. 78 x 100 - 100 = 78 x 100 - 100 x 1 = 100(78 - 1) = 100 x 77 = 7700
e. 6 x 17 x 8 + 3 x 37 x 16 + 4 x 46 x 12
= 48 x 17 + 48 x 37 + 48 x 46
= 48 x (17 + 37 + 46)
= 48 x 100 = 4800
f. 47 x 43 + 47 x 73 - 8 x 47 x 2 = 47 x 43 + 47 x 73 - 47 x 16 = 47 x (43 + 73 - 16) = 47 x 100 = 4700
8×37×5×25=296×5×25=1480×25=37000
48×34+6×66×8=1632+396×8=1632+3168=4800
15×(13-7)+17×5=15×6+17×5=90+85=175
78×100-100=7800-100=7700
a)\(\dfrac{24}{36}\)=\(\dfrac{8}{12}\)
b)\(\dfrac{14}{56}\)=\(\dfrac{1}{4}\)
c)\(\dfrac{9}{24}\)=\(\dfrac{21}{56}\)
Chúc bạn học tốt!
a = 20
b = 60
c = 50
x+5=25
x=25-5
x=20