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a) \(\left(3^2-2\right).\left(x-12+35\right)\)\(=\)\(5^2+279:5\)
\(7.\left(x-12+35\right)=80,8\)
\(x-12+35=80,8:7\)
\(x-12+35=\frac{404}{35}\)
\(x-12=\frac{404}{35}-35\)
\(x-12=\frac{-821}{35}\)
\(x=\frac{-821}{35}+12\)
\(x=\frac{-401}{35}\)
b) \(260:\left(x+4\right)\)\(=\)\(5\left(2^3+5\right)-3\left(3^2+2^2\right)\)
\(260:\left(x+4\right)=26\)
\(x+4=260:26\)
\(x+4=10\)
\(x=10-4\)
\(x=6\)
c) \(7^{x-3}=343\)
\(7^x:7^3=343\)
\(7^x=343.7^3\)
\(7^x=117649\)
vì \(117649=7^6\Rightarrow x=6\)
d) \(\left(x-3\right)^{2017}=\left(x-3\right)^{2016}\)
\(\Rightarrow\hept{\begin{cases}x-3=1\\x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=3\end{cases}}}\)
e) \(\left(2^3+3\right).\left(x-5\right)+14\)\(=\)\(5^2+124:2^2\)
\(\left(2^3+3\right).\left(x-5\right)+14=56\)
\(\left(2^3+3\right).\left(x-5\right)=56-14\)
\(\left(2^3+3\right).\left(x-5\right)=42\)
\(x-5=42:\left(2^3+3\right)\)
\(x-5=\frac{42}{11}\)
\(x=\frac{42}{11}+5\)
\(x=\frac{97}{11}\)
Lời giải:
$(2^2+3)(x-5)+14-5^2+124=2^2$
$7(x-5)+113=4$
$7(x-5)=4-113=-109$
$x-5=-109:7=\frac{-109}{7}$
$x=5+\frac{-109}{7}=\frac{-74}{7}$
a) 7(x-5)+14=56
<=>7(x-5)=42
<=>x-5=6
<=>x=11
b)9(x+1)-3=15
<=>9(x+1)=18
<=>x+1=2
<=>x=1
c)12(x+5)-36=48
<=>12(x+5)=84
<=>x+5=7
<=>x=2
d)5(x+14)=145
<=>x+14=29
<=>x=15
13/ 75 :(x+2)=53-(34-1).23
15/ (23+ 3)(x-5)+14=52 + 124 : 23
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( 2^2 + 1 ) ( x + 14 ) = 5^2 * 4 + ( 2^5 + 3^2 + 7^2 ) :2
Ta có :
52 . 4 + ( 25 + 32 + 72 ) : 2
= 25 . 4 + ( 32 + 9 + 49 ) : 2
= 100 + 90 : 2
= 100 + 45
= 145
⇒ ( 22 + 1 ) . ( x + 14 ) = 145
⇒ ( 4 + 1 ) . ( x + 14 ) = 145
⇒ 5 . ( x + 14) = 145
⇒ x + 14 = 145 : 5
⇒ x + 14 = 29
⇒ x = 29 - 14
⇒ x = 15
Vậy x = 15
Ta có: \(\left(2^2+1\right)\left(x+14\right)=5^2\cdot4+\left(2^5+3^2+7^2\right):2\)
\(\Leftrightarrow5\cdot\left(x+14\right)=100+45=145\)
\(\Leftrightarrow x+14=29\)
hay x=15
(22+3).(x-5)+14=52+124:22
(4+3).(x-5)+14=25+124:4
7.(x-5)+14=25+31
7.(x-5)+14=56
7.(x-5)=42
x-5=6
x=11
Ta có: \(\left(2^2+3\right)\left(x-5\right)+14=5^2+124:2^2\)
\(\Leftrightarrow7\left(x-5\right)=25+31-14=42\)
\(\Leftrightarrow\left(x-5\right)=6\)
hay \(x=11\)