1.Tính : A = 2.\(3^4-\left(-2^2\right)\) 2 . Cho A = \(3^{11}+3^{13}+.........+3^{21}+3^{23}\) . A có chia hết cho 5 ko . Tại sao ? 3 . Tính B : \(\dfrac{2017^{2017}-2017^{17}}{2017^{16}.\left(2017^{2001}.2017\right)}\)
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
C = \(\frac{\left(a^{2016}+b^{2016}\right)^{2017}}{\left(c^{2016}+d^{2016}\right)^{2017}}\)= \(\frac{\left(a^{2017}-b^{2017}\right)^{2016}}{\left(c^{2017}-d^{2017}\right)^{2016}}\)
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
cs ng làm đung r
đag định lm
Câu 1. Tính hợp lý giá trị các biểu thức sau :
a. A = ( 689 - 31 ) - ( 269 - 131 )
b. B = \(\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}+1\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}\right)-\left(\frac{1}{2}+\frac{2016}{2017}+\frac{2017}{2018}\right)\times\left(\frac{2016}{2017}+\frac{2017}{2018}+\frac{3}{4}+1\right)\)c. C = \(1-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
Bài 1: Cho \(a^3+b^3+c^3=3abc\) trong đó a,b,c dương
Tính\(A=\frac{a^{2017}}{b^{2017}}+\frac{b^{2017}}{c^{2017}}+\frac{c^{2017}}{a^{2017}}\)
Bài 2: Cho x+y+z=0
Tính \(A=\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Em tham khảo tại đây nhé.
Câu hỏi của Phạm Minh Tuấn - Toán lớp 8 - Học toán với OnlineMath
Ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\)
Do x + y + z = 0 nên \(\left(x+y+z\right)^2=0\)
\(\Rightarrow x^2+y^2+z^2=-2xy-2yz-2zx\)
Vậy thì \(A=\frac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)}\)
\(A=\frac{-2\left(xy+yz+zx\right)}{-4\left(xy+yz+zx\right)-2\left(xy+yz+zx\right)}\)
\(A=\frac{-2\left(xy+yz+zx\right)}{-6\left(xy+yz+zx\right)}=\frac{1}{3}\)
Cho \(\hept{\begin{cases}a\cdot\left(b^{2+c^2}\right)+b\cdot\left(b^2+c^2\right)+c\left(a^2+b^2\right)+2abc=0\\a^{3+}b^3+c^3=1\end{cases}Tính}A=\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}\left(a,b,c#0\right)\)
Tính
a, \(5\dfrac{4}{13}.14\dfrac{3}{41}-5\dfrac{4}{13}.2\dfrac{3}{41}\)
b, \(\dfrac{2^3}{3^3}:\dfrac{16}{27}+\dfrac{2017}{2018}-\dfrac{1}{2}.2017^0\)
c, \(3:\left(\dfrac{-3}{2}\right)^2+\dfrac{1}{9}.\sqrt{36}\)
a, \(5\dfrac{4}{13}.15\dfrac{3}{41}-5\dfrac{4}{13}.2\dfrac{3}{41}\)
\(=\left(15\dfrac{3}{41}-2\dfrac{3}{41}\right).\dfrac{69}{13}=\dfrac{13.69}{13}=69\)
b, \(\dfrac{2^3}{3^3}:\dfrac{16}{27}+\dfrac{2017}{2018}-\dfrac{1}{2}.2017^0\)
\(=\dfrac{8}{27}:\dfrac{16}{27}+\dfrac{2017}{2018}-\dfrac{1}{2}.1=\dfrac{1}{2}+\dfrac{2017}{2018}-\dfrac{1}{2}=\dfrac{2017}{2018}\)
c, \(3:\left(-\dfrac{3}{2}\right)^2+\dfrac{1}{9}.\sqrt{36}=3:\dfrac{9}{4}+\dfrac{1}{9}.6=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2\)
a,
\(5\dfrac{4}{13}.14\dfrac{3}{41}-5\dfrac{4}{13}.2\dfrac{3}{41}=5\dfrac{4}{13}.\left(14\dfrac{3}{41}-2\dfrac{3}{41}\right)\)
=\(5\dfrac{4}{13}.13\)
=\(\dfrac{69}{13}.13\)
=69
Cho a, b, c\(\ne\)0, thỏa mãn:
\(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}-\frac{a^3+b^3+c^3}{abc}=2\)
Tính \(H=\left(\left(a+b\right)^{2017}-c^{2017}\right)\left(\left(b+c\right)^{2017}-a^{2017}\right)\left(\left(c+a\right)^{2017}-b^{2017}\right)\)
Cho \(\hept{\begin{cases}a\cdot\left(b^2+c^2\right)+b\cdot\left(c^2+a^2\right)+c\cdot\left(a^2+b^2\right)+2abc=0\\a^3+b^3+c^3=1\end{cases}}\)Tính A = \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}\)
\(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+2abc=0\)
\(\Rightarrow ab^2+ac^2+bc^2+ba^2+c\left(a+b\right)^2=0\)
\(\Rightarrow ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2=0\)
\(\Rightarrow\left(a+b\right)\left(ab+c^2+ca+cb\right)=0\)
\(\Rightarrow\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
Từ đó a = -b hoặc b = -c hoặc c = -a
Nếu a = -b mà \(a^3+b^3+c^3=1\Rightarrow\left(-b\right)^3+b^3+c^3=1\Rightarrow c^3=1\Rightarrow c=1\)
Khi đó: \(A=\frac{1}{\left(-b\right)^{2017}}+\frac{1}{b^{2017}}+\frac{1}{1^{2017}}=0+1=1\)
Tương tự với các trường hợp b = -c và a = -c, ta tính được A = 1
1.Cho \(a+b+c=1\\ a^2+b^2+c^2=1\\ a^3+b^3+c^3=1\\ \)
Tính \(T=a^{2017}+b^{2017}+c^{2017}\)
2.Chứng minh:
\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Thực hiện phép tính:
1, \(\left(\dfrac{-1}{2}\right)^2.\left|+8\right|-\left(-\dfrac{1}{2}\right)^3:\left|-\dfrac{1}{16}\right|\)
2, \(\left|-0,25\right|-\left(-\dfrac{3}{2}\right)^2:\dfrac{1}{4}+\dfrac{3}{4}.2017^0\)
3, \(\left|\dfrac{2}{3}-\dfrac{5}{6}\right|.\left(3,6:2\dfrac{2}{5}\right)^3\)
4, \(\left|\left(-0,5\right)^2+\dfrac{7}{2}\right|.10-\left(\dfrac{29}{30}-\dfrac{7}{15}\right):\left(-\dfrac{2017}{2018}\right)^0\)
5, \(\dfrac{8}{3}+\left(3-\dfrac{1}{2}\right)^2-\left|\dfrac{-7}{3}\right|\)
`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`