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a, \(3^4\div3^2-\left[120-\left(2^6.2+5^2.2\right)\right]\)
\(=3^2-\left\{120-\text{[}2.\left(2^6+5^2\right)\text{]}\right\}\)
\(=3^2-\left(120-2\cdot89\right)\)
\(=9--58=9+58=67\)
1. \(a,3^4:3^2-\left[120-(2^6\cdot2+5^2\cdot2)\right]\)
\(=3^2-\left[120-\left\{(2^6+5^2)\cdot2\right\}\right]\)
\(=3^2-\left[120-\left\{(64+25)\cdot2\right\}\right]\)
\(=9-\left[120-89\cdot2\right]\)
\(=9-\left[120-178\right]=9-(-58)=67\)
b, Tương tự như bài a
2.a,\(4^x\cdot5+4^2\cdot2=2^3\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+16\cdot2=8\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+32=56+56\)
\(\Leftrightarrow4^x\cdot5+32=112\)
\(\Leftrightarrow4^x\cdot5=80\)
\(\Leftrightarrow4^x=16\Leftrightarrow4^x=4^2\Leftrightarrow x=2\)
\(b,24:(2x-1)^3-2=1\)
\(\Leftrightarrow24:(2x-1)^3=3\)
\(\Leftrightarrow(2x-1)^3=8\)
\(\Leftrightarrow(2x-1)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
Làm nốt là xong thôi
A, 5.(-8) + (-2).(-3)
=(-40)+6
=(-34)
B,4.(-5)mũ2 + 2.(-15)
=4.25+2.(-15)
=100+(-30)
=70
C,(25-39):(-2-5)
=(-14):(-7)
=2
D,(-5)mũ2 + 3 . (-2)mũ3 - (-1)mũ2016
=25+3.(-8)-1
=25+(-24)-1
=1-1
=0
E,(-5) . (-8+12) - (-3) . (-12+2)
=(-5) . 4+3.(-10)
=(-20)+(-30)
=(-50)
F,(-3)2 + (-2)9 : (-2)7 -(-2017)0
=9+ (-2)2 -1
=9+4-1
=13-1
=12
5 . ( -8 ) + (-2 ) . (-3 )
= [ 5 . ( -2 ) ] + [ (-8 ) . ( -3 ) ]
= -10 + 24
= 14
4 . ( -5 )2 + 2 . ( -15 )
= 4 . 225 + 2 . ( -15 )
= 900 + ( -30 )
= 870
a,\(3^3.3^2-3^5+5^8.1-5^{12}:5^4\)
=\(3^5-3^5+5^8-5^8\)
=0
9!-8!-7! :\(8^2\)
=8!.9-8!-8!.8
=8!.(9-1-8)
=8!.0
=0
mời bạn đừng thi thì tốt hơn 
Bài 1:
\(2B=2^2+2^3+2^4+2^5+...+2^{101}\\ \Rightarrow2B-B=2^{101}-2\\ \Leftrightarrow B=2^{101}-2\)
\(3C=3+3^2+3^3+3^4+...+3^{2004}\\ \Rightarrow3C-C=3^{2004}-3\\ \Leftrightarrow2C=3^{2004}-3\\ \Leftrightarrow C=\frac{3^{2004}-3}{2}\)
Mấy câu sau tương tự nhân 4 và 5 nhé bạn!
Bài 2: Giải theo lớp 6 nhé! :) Mình nghĩ đề bài cần a nguyên nữa nhé nếu không giải theo lớp 8,9 mất rồi! :)
\(a,2a+27⋮2a+1\\ \Leftrightarrow2a+1+26⋮2a+1\\ \Rightarrow26⋮2a+1\left(vì2a+1⋮2a+1\right)\\ \Rightarrow2a+1\inƯ_{\left(26\right)}mà2a+1lẻnên:\\ 2a+1\in\left\{1;-1;13;-13\right\}\\ \Leftrightarrow a\in\left\{0;-1;6;-7\right\}\\ Vậy...\)
Mấy bài sau tương tự nhé! :)
Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)
Bài 1:
a/ \(5^2.3-16:2^2\)\(=25.3-16:4=75-4=71\)
b/\(3.5^2-40:2^3=3.25-40:8=75-5=70\)
c/\(4.5^3-81.3^2=4.125-81.9=500-729=\)\(-229\)
d/\(2^3.15-\left[115-\left(12-5\right)^2\right]\)\(=8.115-\left[115-7^2\right]\)\(=920+115+49=1084\)
Chúc bạn học tốt!!!
