Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1, bạn xem lại đề
2, 15(x-3) + 8x-21 = 12(x+1) +120
<=> 23x - 66 = 12x + 132
<=> 11x = 198 <=> x = 198/11
3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4
<=> 30x + 10 - 95 = 18x -12
<=> 12x = 73 <=> x = 73/12
\(P=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^{128}-1\right)< 3^{128}-1=Q\)
\(P=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\\ 2P=\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\\ 2P=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1\\ P=\dfrac{3^{128}-1}{2}< Q=3^{218}-1\)
\(A=\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right).....\left(51^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)....\left(52^4+\frac{1}{4}\right)}\)
\(=\frac{\left(1+1+\frac{1}{2}\right)\left(1-1+\frac{1}{2}\right)....\left(11^2-11+\frac{1}{2}\right)}{\left(2+2^2+\frac{1}{2}\right)\left(2^2-2+\frac{1}{2}\right)....\left(12^2-12+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)....\left(11.12+\frac{1}{2}\right)}{\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)....\left(12.13+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{12.13+\frac{1}{2}}\)
\(=\frac{1}{313}\)
Chúc bạn học tốt !!!
a) \(A=123\left(123+154\right)+77^2\)
\(A=123^2+\left(123.154\right)+77^2=\left(123+77\right)^2=200^2=400\)
b) \(B=3^{24}-\left(2^{47}+1\right)\left(9^6-1\right)\)
\(B=3^{24}-\left(3^{12}-1\right)\left(3^{12}+1\right)\)
\(B=3^{24}-3^{24}+1=1\)
c) \(C=85^2+75^2+65^2+55^2-45^2-35^2-25^2-15^2\)
\(C=\left(85^2-15^2\right)+\left(75^2-25^2\right)+\left(65^2-35^2\right)+\left(55^2-45^2\right)\)
\(C=\left(85+15\right)\left(85-15\right)+\left(75+25\right)\left(75-25\right)+\left(65+35\right)\left(65-35\right)\left(55+45\right)\left(55-45\right)\)
\(C=100\left(60+50+40+30+20+10\right)\)
\(C=100.210=21000\)
Lời giải:
a)
\(A=75^2+50.75+25^2=75^2+2.25.75+25^2\)
\(=(75+25)^2=100^2=10000\)
b) \(123^2+23^2-46.123=123^2-2.23.123+23^2\)
\(=(123-23)^2=100^2=10000\)
c) \((3^4-1)(3^4+1)-9^4\)
\(=[(3^4)^2-1^2]-9^4=(9^4-1)-9^4=-1\)
Nếu đúng đề thì là bằng 9 hoặc 16
Còn nếu bạn ghi thiếu số 8 + 11 thì bằng 96
k cho mk nha