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a)[(6x-72)/2-84]/28=5628
(6x-72)/2-84=5628*28
3x-36-84=157584
3x=157584+36+84
3x=157704
x=157704/3
x=52568
b)14(x-3)-138=8*9
14(x-3)=72+138
x-3=210/14
x-3=15
x=15+3
x=18
c)2*3x-2*92=4*33
2*(3x-34)=2*2*33
3x=54+81
3x=135
nên x\(\in\phi\)
a)\(\left(\frac{1}{2}-\frac{1}{3}\right).6^x+6^{x+2}=6^{15}+6^{18}\)
\(\frac{1}{6}.6^x+6^{x+2}=6^{15}\left(1+6^3\right)\)
\(\frac{1}{6}.6^x\left(1+6^3\right)=6^{15}.217\)
\(6^{x-1}.217=6^{15}.217\)
\(6^{x-1}=6^{15}\)
\(x-1=15\)
\(x=16\)
b) \(\left(\frac{1}{2}-\frac{1}{6}\right).3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
\(\frac{1}{3}.3^x.4\left(3^4-1\right)=3^{13}.4\left(3^3-1\right)\)
\(3^x.4.\left(3^3-1\right)=3^{13}.4.\left(3^3-1\right)\)
\(3^x=3^{13}\)
\(x=13\)
\(\left(\frac{1}{2}-\frac{1}{6}\right).\left(3^x.3^4\right)-4.3^x=3^{16}-4.3^{13}\)
=> \(\frac{1}{3}.3^x.3^4-4.3^x=3^{16}-4.3^{13}\)
=> \(3^x.3^4-4.3^x=\left(3^{16}-4.3^{13}\right):\frac{1}{3}\)
=> \(3^x.3^4-4.3^x=-386339074,3\)
=> \(3^x.\left(3^4-4\right)=-386339074,3\)
=> \(3^x.77=-386339074,3\)
=> \(3^x=-386339074,3:77\)
=> \(3^x=-5017390,575\)
=> x = ... chắc tự ngồi tính đc
a) Ta có: \(\frac{1}{9}\cdot27^n=3^n\)
\(\Leftrightarrow\frac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)
\(\Leftrightarrow3^{3n}=3^{n+2}\)
\(\Rightarrow3n=n+2\)
\(\Rightarrow n=1\)
b) Ta có: \(3^2.3^4.3^n=3^7\)
\(\Rightarrow3^n=3\)
\(\Rightarrow n=1\)
c) Ta có: \(2^{-1}.2^n+4.2^n=9.2^5\)
\(\Leftrightarrow2^n\cdot\frac{9}{2}=9.2^5\)
\(\Rightarrow2^n=2^6\)
\(\Rightarrow n=6\)
d) Ta có: \(32^{-n}.16^n=2048\)
\(\Leftrightarrow\frac{1}{2^{5n}}\cdot2^{4n}=2^{11}\)
\(\Leftrightarrow2^{4n}=2^{5n+11}\)
\(\Rightarrow4n=5n+11\)
\(\Rightarrow n=-11\)
a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\dfrac{1}{1}+0+0+...+0-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
a) 11.2+12.3+13.4+....+199.10011.2+12.3+13.4+....+199.100
= 11−12+12−13+13−14+....+199−110011−12+12−13+13−14+....+199−1100
=11+0+0+...+0−110011+0+0+...+0−1100
=1−11001−1100
= 99100
a) 32 . 3n = 35
=> 3n = 35 : 32
=> 3n = 33
=> n = 3
b) (22 : 4) . 2n = 4
=> (4 : 4) . 2n = 4
=> 2n = 4
=> 2n = 22
=> n = 2
c) \(\frac{1}{9}.3^4.3^n=3^7\)
\(\Rightarrow3^{-2}.3^4.3^n=3^7\)
\(\Rightarrow3^{-2+4+n}=3^7\)
\(\Rightarrow3^{2+n}=3^7\)
\(\Rightarrow2+n=7\)
\(\Rightarrow n=5\)
d) \(\frac{1}{9}.27^n=3^n\)
\(\Rightarrow3^{-2}.3^{3n}=n\)
\(\Rightarrow3^{-2+3n}=n\)
\(\Rightarrow-2+3n=n\)
\(\Rightarrow2n=2\)
\(\Rightarrow n=1\)
mk ko viết lại đề
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)
\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)
Vậy A= \(\frac{1}{2}\)
Mình làm câu đầu tiên.Câu thứ 2 tương tự nhé bạn.
3^x ( 3^2 + 4x3)=7x3^6
3^x = 7x3^6: 21 = 3^5
=> x=5
Từ đề bài ta suy ra:
\(x+1=6\)
\(y-2=9\)(y này là thay thế cho x thứ 2 để gọi cho dễ)
Vậy x=5; y(x2)=11