Bài 1:
a) Ta có: -20+35-x=-20
\(\Leftrightarrow15-x=-20\)
\(\Leftrightarrow x=35\)
Vậy: x=35
b) Ta có: \(3x+17=12\)
\(\Leftrightarrow3x=-5\)
hay \(x=-\frac{5}{3}\left(ktm\right)\)
Vậy: x∈∅
c) Ta có: |x-6|=0
⇔x-6=0
hay x=6
Vậy: x=6
d) Ta có: |x|+|y|<2
⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}\left|x\right|=0\\\left|y\right|=1\end{matrix}\right.\\\left\{{}\begin{matrix}\left|x\right|=1\\\left|y\right|=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y\in\left\{1;-1\right\}\end{matrix}\right.\\\left\{{}\begin{matrix}x\in\left\{1;-1\right\}\\y=0\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)∈{(0;1);(0;-1);(-1;0);(-1;0)}
e) Ta có: -13|x|=39
⇔|x|=-3
mà |x|≥0∀x
nên x∈∅
Vậy: x∈∅
f) Ta có: \(-2+\left(-3\right)\left|x\right|=10\)
⇔\(-3\left|x\right|=12\)
⇔|x|=-4
mà |x|≥0∀x
nên x∈∅
Vậy: x∈∅
g) Ta có: 12⋮x
⇔x∈{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}
mà x<0
nên x∈{-1;-2;-3;-4;-6;-12}
Vậy: x∈{-1;-2;-3;-4;-6;-12}
h) Ta có: \(\left(7x-11\right)^3=2^5\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=1000\)
⇔7x-11=10
⇔7x=21
hay x=3(tm)
Vậy: x=3
i) Ta có: \(5^x=125\)
⇔x=3
Vậy: x=3
j) Ta có: \(5^{2x-3}-2\cdot5^2=3\cdot5^2\)
\(\Leftrightarrow\left(5^2\right)^x:5^3-2\cdot5^2-3\cdot5^2=0\)
\(\Leftrightarrow25^x:5^3-50-75=0\)
⇔\(25^x=\frac{125}{5^3}=1\)
hay x=0
Vậy: x=0
m) Ta có: \(\frac{5x+9}{240}=\frac{39}{40}\)
⇔\(5x+9=39\cdot\frac{240}{40}=39\cdot6=234\)
⇔5x=225
hay x=45
Vậy: x=45