( x - 140 ) : 7 = 32 - 23 . 3
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a) (x - 140) : 7 = 33 - 23 . 3
(x - 140) : 7 = 27 - 8 . 3 = 27 - 24 = 3
x - 140 = 3 x 7 = 21
x = 21 + 140 = 161
b) x3 . x2 = 28 : 23
x5 = 25
=> x = 2
c) (x + 2) . ( x - 4) = 0
x = -2 hoặc 4
d) 3x-3 - 32 = 2 . 32 =
3x-3 - 9 = 2 . 9 = 18
3x-3 = 18 + 9 = 27
3x-3 = 33
=> x - 3 = 3
x = 3 + 3 = 6
a: \(280-\left(x-140\right):35=270\)
=>(x-140):35=280-270=10
=>x-140=350
=>x=350+140
=>x=490
b: \(\left(190-2x\right):35-32=16\)
=>\(\left(190-2x\right):35=32+16=48\)
=>\(190-2x=35\cdot48=1680\)
=>2x=190-1680=-1490
=>x=-745
c: \(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=2^3\cdot5\)
=>\(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=8\cdot5=40\)
=>41-(2x-5)=720:40=18
=>2x-5=41-18=23
=>2x=28
=>x=14
d: \(\left(x:23+45\right)\cdot37-22=2^4\cdot105\)
=>\(\left(\frac{x}{23}+45\right)\cdot37=16\cdot105+22=1702\)
=>\(\frac{x}{23}+45=46\)
=>\(\frac{x}{23}=1\)
=>x=23
e: \(\left(3x-4\right)\left(x-1\right)^3=0\)
=>\(\left[\begin{array}{l}3x-4=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac43\\ x=1\end{array}\right.\)
f: \(2^{2x-1}:4=8^3\)
=>\(2^{2x-1-2}=2^9\)
=>2x-3=9
=>2x=12
=>x=6
g: \(x^{17}=x\)
=>\(x^{17}-x=0\)
=>\(x\left(x^{16}-1\right)=0\)
=>\(\left[\begin{array}{l}x=0\\ x^{16}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{16}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)
h: \(\left(x-5\right)^4=\left(x-5\right)^6\)
=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)
=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)
=>\(\left(x-5\right)^4\cdot\left(x-4\right)\left(x-6\right)=0\)
=>x∈{4;5;6}
i: \(\left(x+2\right)^5=2^{10}\)
=>\(\left(x+2\right)^5=\left(2^2\right)^5=4^5\)
=>x+2=4
=>x=2
k: 1+2+3+...+x=78
=>\(\frac{x\left(x+1\right)}{2}=78\)
=>x(x+1)=156
=>\(x^2+x-156=0\)
=>(x+13)(x-12)=0
=>x=-13(loại) hoặc x=12(nhận)
l: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)
=>3x=16+14=30
=>\(x=\frac{30}{3}=10\)
n: \(5^{x}:5^2=125\)
=>\(5^{x-2}=5^3\)
=>x-2=3
=>x=5
m: \(\left(x+1\right)^2=\left(x+1\right)^0\)
=>\(\left(x+1\right)^2=1\)
=>\(\left[\begin{array}{l}x+1=1\\ x+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-2\end{array}\right.\)
o: Số số hạng của dãy số 2;4;..;52 là:
(52-2):2+1=50:2+1=25+1=26(số)
Tổng của dãy số 2;4;...;52 là:
\(\left(52+2\right)\cdot\frac{26}{2}=54\cdot13=702\)
(2+x)+(4+x)+...+(52+x)=780
=>26x+702=780
=>26x=78
=>x=3
p: \(70=2\cdot5\cdot7;80=2^4\cdot5\)
=>ƯCLN(70;80)=\(2\cdot5=10\)
70⋮x; 80⋮x
=>x∈ƯC(70;80)
=>x∈Ư(10)
mà x>8
nên x=10
q: \(12=2^2\cdot3;25=5^2;30=2\cdot3\cdot5\)
=>BCNN(12;25;30)=\(2^2\cdot3\cdot5^2=300\)
x⋮12; x⋮25; x⋮30
=>x∈BC(12;25;30)
=>x∈B(300)
mà 0<x<500
nên x=300
1, <=>33+x=45
<=> x = 12
Vậy x=12
2, <=>x+73 =102
<=> x =29
3,x=2
4, -23+x=25<=>x=2
6, x-8=20<=>x=28
a: \(\frac{23}{45}\cdot\frac{15}{16}\cdot\frac{32}{23}+0,3\cdot x=5\frac35\cdot7\frac{7}{12}\cdot\frac{2}{27}\cdot\frac{5}{13}\)
=>\(\frac{12}{16}\cdot\frac{15}{45}+0,3x=\frac{28}{5}\cdot\frac{91}{12}\cdot\frac{2}{27}\cdot\frac{5}{13}\)
=>\(0,3x+\frac{1}{12}=\frac{91}{13}\cdot\frac{2}{12}\cdot\frac{28}{27}=7\cdot\frac16\cdot\frac{28}{27}=\frac{7\cdot14}{3\cdot27}=\frac{98}{81}\)
=>\(0,3x=\frac{98}{81}-\frac{1}{12}=\frac{365}{324}\)
=>\(x=\frac{365}{324}:0,3=\frac{365}{324\cdot0,3}=\frac{365}{97,2}=\frac{1825}{486}\)
b: \(\left\lbrack\left(35\frac57+2\frac34\right)-5\frac57+\frac14\right\rbrack:11+x=3\)
=>\(\left\lbrack35+\frac57+2+\frac34-5-\frac57+\frac14\right\rbrack:11+x=3\)
=>(32+1):11+x=3
=>33/11+x=3
=>x+3=3
=>x=0

\(\left(x-140\right):7=3^2-2^3.3\)
\(\left(x-140\right):7=3\)
\(\left(x-140\right)=3.7\)
\(\left(x-140\right)=21\)
\(x=21+140\)
\(x=161\)