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8) Ta có: \(x+\dfrac{3}{2}=-\dfrac{5}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{3}-\dfrac{3}{2}=\dfrac{-10}{6}-\dfrac{9}{6}\)
hay \(x=-\dfrac{19}{6}\)
Vậy: \(x=-\dfrac{19}{6}\)
10) Ta có: \(\left|x-\dfrac{1}{2}\right|+75\%=\dfrac{9}{10}\)
\(\Leftrightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{3}{20}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{20}\\x-\dfrac{1}{2}=-\dfrac{3}{20}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{20}\\x=\dfrac{7}{20}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{13}{20};\dfrac{7}{20}\right\}\)
11) Ta có: \(x+\dfrac{2}{3}=-\dfrac{1}{2}\)
nên \(x=-\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{-3}{6}-\dfrac{4}{6}\)
hay \(x=-\dfrac{7}{6}\)
Vậy: \(S=\left\{-\dfrac{7}{6}\right\}\)
Bài 2:
Cả mảnh vải dài:
\(25:\dfrac{3}{5}=\dfrac{125}{3}\left(m\right)\)
Ta có : \(A=\frac{5}{3.9}+\frac{5}{9.15}+\frac{5}{5.21}+.....+\frac{5}{63.69}\)
\(\Rightarrow A=\frac{5}{6}\left(\frac{6}{3.9}+\frac{6}{9.15}+\frac{6}{15.21}+......+\frac{6}{63.69}\right)\)
\(\Rightarrow A=\frac{5}{6}\left(\frac{1}{3}-\frac{1}{9}+\frac{1}{9}-\frac{1}{15}+.....+\frac{1}{63}-\frac{1}{69}\right)\)
\(\Rightarrow A=\frac{5}{6}\left(\frac{1}{3}-\frac{1}{69}\right)\)
\(\Rightarrow A=\frac{5}{6}.\frac{22}{69}=\frac{55}{207}\)
a: Ta có: \(\dfrac{n\left(n+1\right)}{2}=231\)
\(\Leftrightarrow n^2+n-462=0\)
\(\Leftrightarrow n^2+22n-21n-462=0\)
\(\Leftrightarrow\left(n+22\right)\left(n-21\right)=0\)
\(\Leftrightarrow n=21\)
Bài 1 :
\(a,6^3.6^7.6^5=6^{3+7+5}=6^{15}\)
\(b,17^9:17^5:17^2=17^{9-5-2}=17^2\)
\(c,=\left(3^3\right)^3.3^3=3^9.3^3=3^{9+3}=3^{12}\)
\(d,=\left(2^4\right)^3.\left(2^6\right)^5=2^{12}.2^{30}=2^{12+30}=2^{42}\)
Bài 2 :
\(a,11^{60}:11^{58}=11^{60-58}=11^2=121\)
\(b,8^{10}:8^5:8^4=8^{10-5-4}=8^1=8\)
\(c,=\left(5^2\right)^9:\left(5^3\right)^5=5^{18}:5^{15}=5^{18-15}=5^3=125\)
\(d,=\left(2^4\right)^5:\left(2^2\right)^6:\left(2^3\right)^2=2^{20}:2^{12}:2^6=2^{20-12-6}=2^2=4\)
\(e,=10^5.\left(10^2\right)^5.\left(10^3\right)^2=10^5.10^{10}.10^6=10^{5+10+6}=10^{21}\)
Bài 3:
a)\(58.75+58.50-58.25\)
=\(58.\left(75+50-25\right)\)
=\(58.100\)
=\(5800\)
b)\(27.39+27.63-2.27\)
=\(27.\left(39+63-2\right)\)
=\(27.100\)
=\(2700\)
c)\(156.25+5.156+156.14+36.156\)
=\(156.\left(25+5+14+36\right)\)
=\(156.80\)
=\(12480\)
d)\(12.35+35.182-35.94\)
=\(35.\left(12+182-94\right)\)
=\(35.100\)
=\(3500\)
e)\(48.19+48.115+67.104\)
=\(48.\left(19+115\right)+67.104\)
=\(48.134+67.104\)
=\(48.67+48.67+67.104\)
=\(67.\left(48+48+104\right)\)
=\(67.200\)
=\(13400\)
f)\(128.72+128.67+128.72+11.72\)
=\(128.\left(72+67\right)+72.\left(128+11\right)\)
=\(128.139+72.139\)
=\(139.\left(72+128\right)\)
=\(139.200\)
=\(27800\)
\(A=3+3^2+3^3+...+3^{60}\)
\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{57}+3^{58}+3^{59}+3^{60}\right)\)
\(\Rightarrow A=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{57}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=\left(3+3^5+...+3^{57}\right)\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=40\left(3+3^5+...+3^{57}\right)⋮40\)
https://toancap2.net/14-de-thi-hoc-sinh-gioi-toan-lop-6-co-dap-an/
A=1/1-1/2+1/3-1/4+...+1/99-1/100
B=1/51+1/52+...1/100
tính 31A/30B
a)2\(^3\)-5\(^3\):5\(^2\)+12.2\(^2\) c)2.[(7-3\(^3\):3\(^2\)):2\(^2\)+99]-100 e)(3\(^5\).3\(^7\)):3\(^{10}\)+5.2\(^4\)-7\(^3\):7
)=8-5\(^1\)+12.4 =2.[(7-3):4+99]-100 =3\(^{12}\):3\(^{10}\)+5.32-49
=8-5+12.4 =2.[4:4+99]-100:3 =9+160-49
=8-5+48 =2.100-100 =120
=51 =200-100=100
Câu g với câu i mik ko bt nên bạn cố gắng tự làm hai câu đó nha
Hok tốt!
Những câu kia bạn trên làm rồi nên mk làm câu khác :
g) \(\left(6^{2007}-6^{2006}\right):6^{2006}=\left(6^{2006}\cdot6-6^{2006}\right):6^{2006}=\left(6-1\right)\cdot6^{2006}:6^{2006}=5\)
i) \(\left(7^{2005}+7^{2004}\right):7^{2004}=\left(7^{2004}\cdot7+7^{2004}\right):7^{2004}=7^{2004}\left(7+1\right):7^{2004}=8\)