A=[3+\(\dfrac{1}{6}\)]-2/5/[5-1/6]+7/10
B] [\(\dfrac{161616}{212121}\)+\(\dfrac{2022}{3033}\)] nhan 7
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27 tháng 3 2016
(161616\212121+2022\3033)*7
=(\(\frac{161616}{212121}+\frac{2022}{3033}\))*7
=(\(\frac{16}{21}+\frac{2}{3}\))*7
=\(\frac{30}{21}\)*7
=\(\frac{210}{21}\)
=10
12 tháng 3 2023
Sửa đề : \(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\)
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{7}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2}{7}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}\right)}\right):\dfrac{2021}{2020}\\ =\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2021}{2022}=0\)
NV
3
PT
2
S
30 tháng 11 2018
\(\left(\frac{161616}{212121}+\frac{2022}{3033}\right)\times7=\left(\frac{16.10101}{21.10101}+\frac{1011.2}{1011.3}\right)\times7=\left(\frac{16}{21}+\frac{2}{3}\right)\times7\)
\(=\frac{10}{7}\times7=10\)
30 tháng 11 2018
Bài làm
\(\left(\frac{161616}{212121}+\frac{2022}{3033}\right)x7\)
= \(\left(\frac{16}{21}+\frac{2}{3}\right)x7\)
= \(\left(\frac{16}{21}+\frac{14}{21}\right)x7\)
= \(\frac{30}{21}x7\)
= \(\frac{30}{3}\)
= \(10\)
# Chúc bạn học tốt #
23 tháng 1 2022
5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)
\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)
\(\Leftrightarrow x^2-x-72=0\)
=>(x-9)(x+8)=0
=>x=9 hoặc x=-8
6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)
\(\Leftrightarrow2x^2-10x=0\)
=>2x(x-5)=0
=>x=0 hoặc x=5
23 tháng 1 2022
5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x
<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8
6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9
<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5
7, <=> (x-1)^2 = (3x+3)^2
<=> (x-1-3x-3)(x-1+3x+3) = 0
<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2
8, = (x^2-10x-15)(x^2-10x+25)
S
24 tháng 2 2023
a/\(-\dfrac{1}{6}+\left(-\dfrac{5}{8}+\dfrac{7}{6}\right)\)
\(=\left(-\dfrac{1}{6}+\dfrac{7}{6}\right)-\dfrac{5}{8}\)
\(=1-\dfrac{5}{8}\)
\(=\dfrac{8}{8}-\dfrac{5}{8}=\dfrac{3}{8}\)
b/\(\left(\dfrac{2}{5}\right)^2+5\dfrac{1}{2}\cdot\left(4,5-2\right)+\dfrac{2^3}{\left(-4\right)}\)
\(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+-\dfrac{2^3}{2^2}\)
\(=\dfrac{4}{25}+\dfrac{55}{4}+\left(-2\right)\)
\(=\dfrac{16}{100}+\dfrac{1375}{100}-\dfrac{200}{100}\)
\(=\dfrac{1191}{100}=11,91\)
TC
8 tháng 5 2022
bài 1
a)\(=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)
b)\(=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
c)\(=\dfrac{30}{9}=\dfrac{10}{3}\)
d)\(=\dfrac{8}{5}\times\dfrac{7}{4}=\dfrac{56}{20}=\dfrac{14}{5}\)
