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a ) 76 + 75 - 74
= 74 ( 72 + 7 - 1 )
= 74. 55 chia hết cho 55
b ) 165 + 215
= ( 24 ) 5 + 215
= 220 + 215
= 215 ( 25 + 1 )
= 215 . 33 chia hết cho 33
c ) 817 - 279 - 913
= ( 34 )7 - ( 33 )9 - ( 32 )13
= 328 - 327 - 326
= 326 ( 32 - 3 - 1 )
= 326 . 5
= 322 . 34 . 5
= 322 . 81 . 5
= 322 . 405 chia hết cho 405
Bài 2:
\(A=\frac{8^5(-5)^8+(-2)^5.10^9}{2^{16}.5^7+20^8}\) \(=\frac{(2^3)^5(-5)^8+(-2)^5.2^9.5^9}{2^{16}.5^7+(2^2.5)^8}\)
\(=\frac{2^{15}.5^8-2^5.2^9.5^9}{2^{16}.5^7+2^{16}.5^8}\)
\(=\frac{2^{14}.5^8(2-5)}{2^{16}.5^7(1+5)}\)
\(=\frac{5(-3)}{2^2.6}=\frac{-5}{8}\)
Bài 3:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)
Thay vào:
\(\frac{5a+3b}{5a-3b}=\frac{5bt+3b}{5bt-3b}=\frac{b(5t+3)}{b(5t-3)}=\frac{5t+3}{5t-3}\)
\(\frac{5c+3d}{5c-3d}=\frac{5dt+3d}{5dt-3d}=\frac{d(5t+3)}{d(5t-3)}=\frac{5t+3}{5t-3}\)
Do đó: \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) (đpcm)
Bài 4:
Ta có:
\(A=3+3^2+3^3+3^4+...+3^{100}\)
\(=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+....+(3^{97}+3^{98}+3^{99}+3^{100})\)
\(=3(1+3+3^2+3^3)+3^5(1+3+3^2+3^3)+...+3^{97}(1+3+3^2+3^3)\)
\(=3.40+3^5.40+....+3^{97}.40\)
\(=120(1+3^4+....+3^{96})\vdots 120\)
Ta có đpcm.
\(a.\)
\(8^7-2^{18}\)
\(=\left(2^3\right)^7-2^{18}\)
\(=2^{21}-2^{18}\)
\(=2^{18}.2^3-2^{18}\)
\(=2^{18}\left(2^3-1\right)\)
\(=2^{18}.7\)
\(=2^{17}.7.2⋮14\)
Vậy \(8^7-2^{18}⋮14\)
\(b.\)
\(5^5-5^4+5^3\)
\(=5^3\left(5^2-5+1\right)\)
\(=5^3.21\)
\(=5^3.7.3⋮7\)
Vậy \(5^5-5^4+5^3⋮7\)
\(c.\)
\(7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)\)
\(=7^4.55\)
\(=7^4.5.11⋮11\)
Vậy \(7^6+7^5-7^4⋮11\)
Ta có : \(\frac{a+5}{a-5}=\frac{b+6}{b-6}\Rightarrow\frac{b-6}{a-5}=\frac{b+6}{a+5}\)
Áp dụng t/c dãy tỉ số bằng nhau :
\(\frac{b-6}{a-5}=\frac{b+6}{a+5}=\frac{\left(b+6\right)-\left(b-6\right)}{\left(a+5\right)-\left(a-5\right)}=\frac{12}{10}=\frac{6}{5}\)
\(\Rightarrow5\left(b-6\right)=6\left(a-5\right)\Leftrightarrow5b-30=6a-30\Leftrightarrow5b=6a\Leftrightarrow\frac{a}{b}=\frac{5}{6}\)
a) \(5^5-5^4+5^3\)
\(=5^3.5^2-5^3.5+5^3\)
=\(5^3.\left(5^2-5+1\right)\)
\(=5^3.21\)
vì \(21⋮7\Rightarrow5^3.21⋮7\)
\(\Rightarrow5^5-5^4+5^3⋮7\)
a)
55-54+53
=53(52-5+1)
=53.21
=53.3.7
vì trong tích 53.3.7 có chứa 1 thừa số chia hết cho 7
=> 53.3.7 chia hết cho 7 hay 55-54+53 chia hết cho 7
b)
76+75-74
=74(72+7-1)
=74.55
=74.5.11
vì trong tích 74.5.11 có 1 thừa số chia hết cho 11 nên
74.5.11 chia hết cho 11 hay 76+75-74 chia hết cho 11
a/ Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c};c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=k^3\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)
Áp dụng tính chất của tỉ lệ thức ta có:\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=k^3\)
Mặt khác: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\frac{a+b+c}{b+c+d}=k\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=k^3\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\frac{a+b+c}{b+c+d}\right)^3\left(=k^3\right)\)
a) 55 - 54 + 53
= 53.(52 - 5 + 1)
= 53.(25 - 5 + 1)
= 53.21
= 53.3.7 chia hết cho 7 (đpcm)
b) 76 + 75 - 74
= 74.(72 + 7 - 1)
= 74.(49 + 7 - 1)
= 74.(56 - 1)
= 74.55
= 74.5.11 chia hết cho 11 (đpcm)
Giảng phần a thôi,phần b chị làm tương tự!
a)\(5^3.\left(5^2-5+1\right)\)
\(=5^3.\left(25-4\right)\)
\(=5^3.7.3\)chia hết cho 7.
Chúc chị học tốt^^
a)Ta có:\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55\)
=>\(7^6+7^5-7^4⋮55\)
b)\(A=1+5+5^2+...+5^{50}\)
\(5A=5\left(1+5+5^2+...+5^{50}\right)=5+5^2+5^3+...+5^{51}\)
\(5A-A=5+5^2+5^3+...+5^{51}-\left(1+5+5^2+...+5^{50}\right)\)
\(4A=5^{51}-1\)
\(\Rightarrow A=\dfrac{5^{51}-1}{4}\)
a) \(7^6+7^5+7^4=7^4\left(7^2+7+1\right)\)
= \(7^4.55\)
Vậy: \(7^6+7^5+7^4\) chia hết cho 55.
b) A= \(1+5+5^2+5^3+5^4+.....+5^{50}\)
5A= 5+\(5^2+5^3+5^4+5^{51}\)
5A-A= 5+\(5^2+5^3+5^4+......+5^{51}\)\(-\left(1+5^2+5^3+5^4+......+5^{51}\right)\)
4A= 5+\(5^2+5^3+5^4+......+5^{51}\)\(-1-5-5^2-5^3-5^4-.......-5^{50}\)
= \(5^{51}-1\)
Vậy A= \(\left(5^{51}-1\right):4\)
Tick mk nha!
Bài 1:
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^4=0\)
=>2x(2x-1)(2x-2)=0
hay \(x\in\left\{0;\dfrac{1}{2};1\right\}\)
Bài 3:
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Leftrightarrow\dfrac{a-5+10}{a-5}=\dfrac{b-6+12}{b-6}\)
\(\Leftrightarrow\dfrac{10}{a-5}=\dfrac{12}{b-6}\)
\(\Leftrightarrow\dfrac{a-5}{5}=\dfrac{b-6}{6}\)
\(\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{6}\)
hay a/b=5/6
a
\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3\cdot21⋮7\)
b
\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4\cdot55⋮11\)
a)\(5^5-5^4+5^3\)
\(=5^3\left(5^2-5+1\right)\)
\(=5^3\times21⋮7\)
b) \(7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)\)
\(=7^4\times55⋮11\)
A=5^3(5^2-5+1)
=5^3*21 chia hết cho 7
=5^5 -5^4+5^3=5^3.5^2 -5^3.5+5^3
=5^3(5^2-5+1)=5^3.21
Vì 21 chia hết cho 7 =>5^3.21 chia hết cho 7
Vậy 5^5 -5^4+5^3 chia hết cho 7