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d) 135(-12 + 247) - 147(135 - 12)= -12.135 + 135.147 - 147.135 + 12.147 = (-12.135 + 12.147) + (135.147 - 147.135) = 12(-135 + 147) + 0 = -12.12 = -144
Bài 1:
a) [3.(-2) - (-8)].(-7) - (-2).(-5) = -7(-6 + 8) - 2.5= -7.2 - 2.5 = -2(7 + 5) = -2.12 = -24
a)(629 +437 )-437 -19
=1066-437-19
=629-19
=610
b)47.86 -47.76 + 2.13.5 -30 +4 .19 .25
=47.(86-76)+(2.5).13-30+(4.25).19
=47.10+10.13-30+100.19
=470+130-30+1900
=600-30+1900
=570+1900
=2470
c) 2928 + (-53) -(-149) + (-247) -2598
=2875+149-247-2598
=3024-247-2598
=2777-2598
=179
32 . 53 - 31 = 53.(31 + 1) - 31 = 53.31 < 53.31 + 32
Bài 2:
a: =>x-1=1 hoặc x-1=-1
=>x=2 hoặc x=0
b: =>x+1=-1
hay x=-2
c: =>(135-7x):9=8
=>135-7x=72
=>7x=63
hay x=9
d: =>(x+7)(x-3)<0
=>-7<x<3
e: \(\Leftrightarrow3^{x-3}=18+9=27\)
=>x-3=3
hay x=6
f: =>4-2x=0
hay x=2
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
Ta có : A = 32 . 53 - 31 = (31 + 1) . 53 - 31 = 31 . 53 + 53 -31 = 53 . 31 +22
Vì 22 < 32 Nên : 53 . 31 + 22 < 53 . 21 +32
Hay : A < B
Vậy A < B .
Ta có: A = 32 . 53 - 31
A = ( 31 + 1 ) . 53 - 31
A = 31 . 53 + 53 - 31
A= 31 . 53 + 22
Vì 31 . 53 + 22 < 53 . 31 + 32 nên A < B
Vậy A < B
Ta có: \(\dfrac{11}{53}+\left(\dfrac{32}{47}+\dfrac{-10}{53}\right)+\dfrac{-64}{94}+\dfrac{1}{-53}+\dfrac{1}{3}\)
\(=\dfrac{11}{53}+\dfrac{32}{47}-\dfrac{10}{53}-\dfrac{32}{47}-\dfrac{1}{53}+\dfrac{1}{3}\)
\(=\dfrac{1}{3}\)
\(147-(2x-53)=32\)
\(2x-53=147-32\)
\(2x-53=115\)
\(2x=115+53\)
\(2x=168\)
\(x=168:2\)
\(x=84.\)Vậy x=84.
_Học tốt_
\(147-\left(2x-53\right)=32\)
\(2x-53=147-32\)
\(2x-53=115\)
\(2x=115+53\)
\(2x=168\)
\(x=168:2\)
\(x=84\)