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\(a)\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{-11.13^7}{11^5.13^8}=\frac{-1}{11^4.13}\) (Bạn xem thử xem có sai đề không nhé)
\(c)\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3+1\right)}{2^9.3^{10}}=\frac{2.4}{3}=\frac{8}{3}\)
\(d)\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.4+9\right)}=\frac{8}{20+9}=\frac{8}{29}\)
\(a)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{\left(-11\right)\cdot13^7}{11^5\cdot13^8}=\frac{-1}{11^4\cdot13}=\frac{-1}{14641\cdot13}=\frac{-1}{190333}\)
\(c)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\cdot2}{2^9\cdot3^{10}}=\frac{2\cdot2}{3}=\frac{4}{3}\)
A và B dễ
Bài 2:
sai đề bài vì ngay từ cái phép tính đầu đã ko theo quy luật rồi
\(A=\frac{-3}{5}-\frac{2}{5}+2\)
\(A=-1+2=1\)
\(B=\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}=\frac{1}{4}\)
nÀ NÍ sao lại = đây là dấu trừ hay cộng 1/4
Bài làm
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)
\(1-\frac{1}{x+2}=\frac{2015}{2016}\)
\(\frac{1}{x+2}=\frac{1}{2016}\)
\(\Rightarrow x+2=2016\)
\(x=2014\)
(x2-1)2=9
=> x2-1 = 3
x2 = 3+1
x2 = 4
=> x2 = 4 = 22 ( x2=22 )
<=> x = 2
12:{390:[5.102-(53+x.72)]} = 4
390:[5.102-(53+x.72)] = 12:4
390:[5.102-(53+x.72)] = 3
5.102-(53+x.72) = 390 : 3
5.102-(53+x.72) = 130
=> 500-(125+x+49)=130
125+x+49 = 500-130
125+x+49 = 370
125+x = 370-49
125+x = 321
x = 321-125
x = 106
53(3x+2):13=103:(135:134)
53(3x+2):13=103:13
53(3x+2):13= 1000/13
125(3x+2):13 = 1000/13
125(3x+2) = 1000/13 . 13
125(3x+2) = 1000
3x+2 = 1000:125
3x+2 = 8
3x = 8-2
3x = 6
x = 6:3
x = 2
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
\(a,\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=5^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=9\)
\(\Leftrightarrow19x+50=126\)
\(\Leftrightarrow19x=76\Leftrightarrow x=4\)
b) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 30 ) = 1240
x + x + 1 + x + 2 + ... + x + 30 = 1240
( x + x + ... + x ) + ( 1 + 2 + ... + 30 ) = 1240
Số số hạng là : ( 30 - 1 ) : 1 + 1 = 30 ( số )
Tổng là : ( 30 + 1 ) . 30 : 2 = 465
=> 31x + 465 = 1240
=> 31x = 775
=> x = 25
Vậy........
TL:
\(\left\{x^2-\left[6^2-\left(8^2-9\cdot7\right)^3-7\cdot5\right]^3-5\cdot3\right\}^3=1\\ \Rightarrow x^2-\left[36-\left(64-63\right)^3-35\right]^3-15=1\\ \Rightarrow x^2-\left[36-35-1^3\right]^3=16\\ \Rightarrow x^2-0^3=16\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)Vậy \(x=\pm4\)
^HT^
TL:
TL:
{x2−[62−(82−9·7)3−7·5]3−5·3}3=1 \\ ⇒x2−[36−(64−63)3−35]3−15=1 \\ ⇒x2−[36−35−13]3=16 \\ ⇒x2−03=16 \\ ⇒x2=16 \\ ⇒x=±4Vậy x=±4
^HT^