Tính:
Q=\(\dfrac{2.tanπ/6 - sinπ/4.cosπ/6 + 3.cotπ/4}{2.sin3π/4 + 6.cos2π/3 - 5.tan5π/6}\)
Tính A= \(\dfrac{1}{2}\)- \(\dfrac{2}{3}\)+\(\dfrac{3}{4}\)-\(\dfrac{4}{5}\)+\(\dfrac{5}{6}\)-\(\dfrac{6}{7}\)-\(\dfrac{6}{5}\)+\(\dfrac{4}{5}\)-\(\dfrac{3}{4}\)+\(\dfrac{2}{3}\)-\(\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}+\dfrac{4}{5}-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\left(-\dfrac{4}{5}+\dfrac{4}{5}\right)+\left(\dfrac{5}{6}-\dfrac{6}{7}-\dfrac{6}{5}\right)\)
\(=0+0+0+0-\dfrac{257}{210}\)
\(=\dfrac{257}{210}\)
Có ai biết câu này không, làm giúp mình với
Tính (theo mẫu).
Mẫu: \(2+\dfrac{1}{6}=\dfrac{12}{6}+\dfrac{1}{6}=\dfrac{13}{6};1-\dfrac{1}{4}=\dfrac{4}{4}-\dfrac{1}{4}=\dfrac{3}{4}\) |
a) \(1+\dfrac{4}{9}\) b) \(5+\dfrac{1}{2}\) c) \(3-\dfrac{5}{6}\) d) \(\dfrac{31}{7}-2\)
a) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)
b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)
c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)
d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)
Tính giá trị biểu thức :
\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+\left(\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{3}{6}+\dfrac{4}{6}+\dfrac{5}{6}\right)-\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{4}{7}+\dfrac{5}{7}+\dfrac{6}{7}\right)+...+\left(100+...+\dfrac{99}{100}\right)\)
Tìm x,biết
\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^x\)
\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)
\(=\dfrac{4.4^5.6.6^5}{3.3^5.2.2^5}\)
\(=\dfrac{4^6.6^6}{3^6.2^6}\)
\(=\dfrac{2^6.2^6.2^6.3^6}{3^6.2^6}\)
\(=2^{12}=2^{3^4}=8^4=8^x\)
Vậy x = 4
45+45+45+4535+35+35.65+65+65+65+65+6525+2545+45+45+4535+35+35.65+65+65+65+65+6525+25
=4.45.6.653.35.2.25=4.45.6.653.35.2.25
=46.6636.26=46.6636.26
=26.26.26.3636.26=26.26.26.3636.26
=212=234=84=8x=212=234=84=8x
Vậy x = 4
Tìm x biết
\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5++3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^{\left|2x+6\right|}\)
Lời giải:
\(\text{VT}=\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)
\(=\frac{4.4^5}{3.3^5}.\frac{6.6^5}{2.2^5}=\frac{4^6.6^6}{3^6.2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)
Do đó: \(8^{|2x+6|}=2^{12}\Leftrightarrow 2^{3|2x+6|}=2^{12}\)
\(\Leftrightarrow 3|2x+6|=12\Leftrightarrow |2x+6|=4\)
\(\Rightarrow\left[{}\begin{matrix}2x+6=4\\2x+6=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
Bài 2: (Đề1) Tính
\(4\dfrac{2}{3}+3\dfrac{2}{7}\)=..........................
\(8\dfrac{5}{9}:5\dfrac{1}{2}\) =..........................
\(6\dfrac{5}{7}:2\dfrac{1}{6}\) =..........................
\(1\dfrac{3}{4}x2\dfrac{5}{6}\) =..........................
\(5\dfrac{3}{4}-2\) =..........................
\(4\dfrac{2}{3}+3\dfrac{2}{7}=\dfrac{14}{3}+\dfrac{23}{7}=\dfrac{14x7+23x3}{21}=\dfrac{167}{21}\)
\(8\dfrac{5}{9}:5\dfrac{1}{2}=\dfrac{77}{9}:\dfrac{11}{2}=\dfrac{77}{9}x\dfrac{2}{11}=\dfrac{14}{9}\)
\(6\dfrac{5}{7}:2\dfrac{1}{6}=\dfrac{47}{7}:\dfrac{13}{6}=\dfrac{47}{7}x\dfrac{6}{13}=\dfrac{282}{91}\)
\(1\dfrac{3}{4}x2\dfrac{5}{6}=\dfrac{7}{4}x\dfrac{17}{6}=\dfrac{119}{24}\)
\(5\dfrac{3}{4}-2=\dfrac{23}{4}-2=\dfrac{23}{4}-\dfrac{8}{4}=\dfrac{15}{4}\)
\(\left(\dfrac{2}{3}+\dfrac{2}{13}+\dfrac{2}{11}+\dfrac{2}{6}\right)\) \(\left(\dfrac{5}{4}+\dfrac{5}{7}+\dfrac{5}{6}+\dfrac{5}{11}\right)\)
___________________ X ___________________
\(\left(\dfrac{4}{3}+\dfrac{4}{13}+\dfrac{4}{11}+\dfrac{4}{6}\right)\) \(\left(\dfrac{9}{4}+\dfrac{9}{7}+\dfrac{9}{6}\dfrac{9}{11}\right)\)
ta có
\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)
_______________________ X ________________________
\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)
= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)
1.tính
\(3\dfrac{1}{4}-2\dfrac{1}{3}=\)
\(6\dfrac{5}{9}+2\dfrac{5}{6}=\)
\(6\dfrac{5}{9}-2\dfrac{5}{6}=\)
Lời giải:
1.
$3\frac{1}{4}-2\frac{1}{3}=3+\frac{1}{4}-(2+\frac{1}{3})$
$=3-2+\frac{1}{4}-\frac{1}{3}$
$=1+\frac{1}{4}-\frac{1}{3}=\frac{5}{4}-\frac{1}{3}=\frac{11}{12}$
2.
$6\frac{5}{9}+2\frac{5}{6}=6+\frac{5}{9}+2+\frac{5}{6}=8+\frac{5}{6}+\frac{5}{9}=8+\frac{25}{18}=8+1+\frac{7}{18}=9+\frac{7}{18}=9\frac{7}{18}$
3.
$6\frac{5}{9}-2\frac{5}{6}=6+\frac{5}{9}-(2+\frac{5}{6})$
$=6+\frac{5}{9}-2-\frac{5}{6}$
$=(6-2)+\frac{5}{9}-\frac{5}{6}$
$=4+\frac{5}{9}-\frac{5}{6}=\frac{41}{9}-\frac{5}{6}=\frac{67}{8}$
1, \(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6+\sqrt{6}}{\sqrt{6}}\)
2, \(\dfrac{6-6\sqrt{3}}{1-\sqrt{3}}+\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\)
3, \(\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
4, \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\dfrac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
5, \(\left(\dfrac{3\sqrt{125}}{15}-\dfrac{10-4\sqrt{5}}{\sqrt{5}-2}\right)\cdot\dfrac{1}{\sqrt{5}}\)
1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)
2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
3: \(=\sqrt{3}+1-\sqrt{3}=1\)