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Bài 7:
x/1=z/2 nên x/6=z/12
=>x/6=y/9=z/12
=>x/2=y/3=z/4
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{27}{9}=3\)
=>x=6; y=9; z=12
Bài 2:
1: =>5x+1=6/7 hoặc 5x+1=-6/7
=>5x=-1/7 hoặc 5x=-13/7
=>x=-1/35 hoặc x=-13/35
2: =>x-1=4
=>x=5
3: =>3x-1=3
=>3x=4
=>x=4/3
4: \(\Leftrightarrow\dfrac{5}{x+3}=\dfrac{-5}{6}+\dfrac{1}{2}=\dfrac{-5+3}{6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
=>x+3=-15
=>x=-18
7: \(\Leftrightarrow2^{2x+1}+2^{2x+6}=264\)
=>2^2x+1*(1+2^5)=264
=>2^2x+1=8
=>2x+1=3
=>x=1
9: =>x^4=8x
=>x^4-8x=0
=>x=2
Bài 2:
a: \(3B=3+3^2+3^3+...+3^{90}\)
\(\Leftrightarrow2B=3^{90}-1\)
hay \(B=\dfrac{3^{90}-1}{2}\)
b: \(B=\left(1+3+3^2+3^3+3^4+3^5\right)+3^6\left(1+3+3^2+3^3+3^4+3^5\right)+...+3^{84}\left(1+3+3^2+3^3+3^4+3^5\right)\)
\(=384\cdot\left(1+3^6+...+3^{84}\right)⋮52\)
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
~ Học tốt ~
Bài 1:
1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)
\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)
\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)
\(=3^2=9\)
2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)
\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)
\(=2^7:2^3:\dfrac{1}{2^4}\)
\(=2^4.2^4=256\)
3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}.1:2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)
\(=\dfrac{43}{48}\)
4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)
\(=-3-1+\dfrac{1}{8}\)
\(=-4+\dfrac{1}{8}\\ \)
\(=-\dfrac{31}{8}\)
5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)
Chúc bạn học tốt 
e, Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=k\left(k\in Z\right)\)
\(\Leftrightarrow x=4k,y=5k\) (1)
Theo bài ra ta có: xy = 80
Từ (1) \(\Rightarrow4k.5k=80\Rightarrow20.k^2=80\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
+ Với k = 2 \(\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
+ Với k = -2 \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-10\end{matrix}\right.\)
Vậy \(\left(x,y\right)\in\left\{\left(8,10\right);\left(-8,-10\right)\right\}\)
a) \(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5-6}=\dfrac{-16}{4}=-4\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-4\\\dfrac{y}{5}=-4\\\dfrac{z}{-2}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-12\\y=-20\\z=8\end{matrix}\right.\)
Mình làm cách 1 theo cách này bạn xem được không nhé :
Đặt \(A=-\dfrac{5}{70}-\dfrac{5}{700}-\dfrac{5}{7000}-\dfrac{5}{70000}-\dfrac{5}{700000}\)
\(\Rightarrow10A=-\dfrac{5}{7}-\dfrac{5}{70}-\dfrac{5}{700}-\dfrac{5}{7000}-\dfrac{5}{70000}\)
\(\Rightarrow10A-A=9A=-\dfrac{5}{7}+\dfrac{5}{700000}\)
\(9A=\dfrac{-500000}{700000}+\dfrac{5}{700000}=\dfrac{-450000}{700000}=\dfrac{-9}{14}\)
\(\Rightarrow A=\dfrac{-9}{14}:9=\dfrac{-1}{14}\)
Mình không biết làm bài 1 thông cảm nha
\(2,\)
\(x^5:x^3=\sqrt{4}\)
\(\Rightarrow x^5:x^3=2\)
\(\Rightarrow x^2=2\)
\(\Rightarrow x^2=\sqrt{2^2}=\sqrt{\left(-2\right)^2}\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
\(3,\)
\(a,\) \(8^7-2^{18}\)
\(=\left(2^3\right)^7-2^{18}\)
\(=2^{21}-2^{18}\)
\(=2^{18}.\left(2^3-1\right)\)
\(=2^{18}.\left(8-1\right)=2^{18}.7\)
Vì \(7⋮7\)
\(\Rightarrow2^{18}.7⋮7\)
Vậy \(8^7-2^{18}\) chia hết cho 7
\(b,\)
\(10^6+5^7\)
\(=\left(5.2\right)^6+5^7\)
\(=5^6.2^6+5^7\)
\(=5^6.\left(2^6+5\right)\)
\(=5^6.\left(64+5\right)=5^6.69\)
Vì \(69⋮69\)
\(\Rightarrow5^6.69⋮69\)
\(\Rightarrow10^6+5^7\) chia hết cho 69
\(c,14^6-49^3\)
\(=\left(7.2\right)^6-\left(7^2\right)^3\)
\(=7^6.2^6-7^6\)
\(=7^6.\left(2^6-1\right)\)
\(=7^6.\left(64-1\right)=7^6.63\)
Vì \(63⋮63\)
\(\Rightarrow7^6.63⋮63\)
Vậy \(14^6-49^3⋮63\)
\(d,14^9-49^2\)
\(=\left(7.2\right)^9-\left(7^2\right)^2\)
\(=7^9.2^9-7^4\)
\(=7^4.\left(7^5-2^9\right)\)
Xét : \(7^5-2^9\)
\(=\left(7^2\right)\left(7^2\right).7-\left(2^4\right)\left(2^4\right).2\)
\(=\overline{...9}.\overline{...9}.\overline{...7}-\overline{...6}.\overline{...6}.\overline{...2}\)
\(=\overline{...7}-\overline{...2}=\overline{...5}\)
\(\overline{...5}⋮5\)
Vì \(7\) không chia hết cho 3
\(\Rightarrow7^5\) không chia hết cho 3
mà \(7^5\) không phải là số chính phương
⇒ \(7^5\) chia 3 dư 1 \(\left(1\right)\)
Tương tự \(\Rightarrow2^9\) chia 3 dư 1 \(\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\)\(\Rightarrow7^5-2^9⋮3\)
Vì 5;3 là hai số nguyên tố cùng nhau
\(\Rightarrow7^5-2^9⋮\left(5.3\right)=15\)