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a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
a) Ta có: \(\left(2x-5\right)^3=216\)
\(\Leftrightarrow2x-5=6\)
\(\Leftrightarrow2x=11\)
hay \(x=\dfrac{11}{2}\)
b) Ta có: \(2x-3⋮x+4\)
\(\Leftrightarrow-11⋮x+4\)
\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)
hay \(x\in\left\{-3;-5;7;-15\right\}\)
Alo, sugeni two wai phem. Si ga no, you woo be the me that nas te, ai gi da
1) Ta có: \(\left(-86-x\right)-\left(3+2x\right)=-4-15\)
\(\Leftrightarrow-86-x-3-2x+4+15=0\)
\(\Leftrightarrow-3x-70=0\)
\(\Leftrightarrow-3x=70\)
hay \(x=-\dfrac{70}{3}\)
Vậy: \(x=-\dfrac{70}{3}\)
2) Ta có: \(18+\left(-x\right)-\left(40-28\right)=-32-\left(-18\right)\)
\(\Leftrightarrow18-x-40+28+32-18=0\)
\(\Leftrightarrow-x+20=0\)
\(\Leftrightarrow-x=-20\)
hay x=20
Vậy: x=20
3) Ta có: \(-27-\left(-31+x\right)-25=-5-17\)
\(\Leftrightarrow-27+31-x-25+5+17=0\)
\(\Leftrightarrow-x+1=0\)
\(\Leftrightarrow-x=-1\)
hay x=1
Vậy: x=1
4) Ta có: \(-9-14-x+42-38=-5+13\)
\(\Leftrightarrow-x-19=8\)
\(\Leftrightarrow-x=27\)
hay x=-27
Vậy: x=-27
32x + 3x + 3 = 759
3x.3x + 3x + 3 = 759
3x.3x + 3x = 759 - 3
3x.3x + 3x = 756
3x(3x + 1) = 756
3x(3x + 1) = 27.28
=> 3x = 27
3x = 33
=> x = 3
(x+3).(2x-18)=0
=>x+3=0 hoặc 2x-18 =0
+)x+3=0
x=-3
+)2x-18=0
2x =18
x =9
Vậy x \(\in\)\(_{\left\{-3;9\right\}}\)
a ) x = 1 ; x = 0
b ) x = 1 ; x = 0
c ) x không có ; x không tồn tại
d ) x không có ; x không tồn tại
\(a,x-\dfrac{1}{18}=\dfrac{-4}{15}\\ \Rightarrow x=\dfrac{-4}{15}+\dfrac{1}{18}\\ \Rightarrow x=\dfrac{-19}{90}\)
\(b,2x+\dfrac{3}{4}=x-\dfrac{1}{3}\\ \Leftrightarrow2x-x=-\dfrac{1}{3}-\dfrac{3}{4}\\ \Leftrightarrow x=\dfrac{-4-9}{12}\\ \Leftrightarrow x=-\dfrac{13}{12}\)
\(c,\dfrac{x-3}{5}=\dfrac{5}{x-3}\\ \Leftrightarrow\left(x-3\right)^2=25\\ \Leftrightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
1e) Để \(\frac{2x-1}{x-3}\) nguyên thì \(2x-1⋮x-3\)
\(\Leftrightarrow2x-6+5⋮x-3\)
\(\Leftrightarrow2\left(x-3\right)+5⋮x-3\)
Do \(2\left(x-3\right)⋮x-3\) \(\Rightarrow5⋮x-3\)
\(\Rightarrow x-3\in\left\{-5;-1;1;5\right\}\)
\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)
Vậy:...................
(18-x) - (3-2x) = -5
18 - x - 3 + 2x = -5
2x - x = -5 - 18 + 3
x= -20