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a.\(17-\left(43-\left|x\right|\right)=45\)
\(\Leftrightarrow43-\left|x\right|=17-45\)
\(\Leftrightarrow43-\left|x\right|=-28\)
\(\Leftrightarrow\left|x\right|=43+28\)
\(\Leftrightarrow\left|x\right|=71\)
\(\Leftrightarrow x\in\left\{-71;71\right\}\)
b. \(\left(x-2\right)\left(x+15\right)=0\Leftrightarrow x\in\left\{2;-15\right\}\)
c. \(2x^2-3=29\)
\(\Leftrightarrow2x^2=29+3\)
\(\Leftrightarrow2x^2=32\)
\(\Leftrightarrow x^2=32\div2\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x^2=4^2=\left(-4\right)^2\)
Vậy x = 4 hoặc x = 4
2/5 ST1 = 1/6 ST2 hay 2/5 ST1 = 2/12 ST2
suy ra ST1 x 5 = ST2 x 12
ST1 = ST2 x 5/12
Coi .........
( bạn tự vẽ sơ đồ nhé. ST1 5 phần, ST2 12 phần )
ST1 là : 51 : ( 5 + 12 ) x 5 = 15
ST2 là : 51 - 15 = 36
Đ/s : ....................
Tỉ số giữa số thứ nhất và số thứ hai là:
\(\frac{1}{6}\div\frac{2}{5}=\frac{5}{12}\)( số thứ hai)
Tổng số phần bằng nhau là:
5 + 12 = 17 (phần)
Số thứ nhất là:
51 : 17 . 5 = 15
Số thứ hai là:
51 - 15 = 36
\(\frac{-3}{4}+\frac{3}{7}+\frac{-1}{4}+\frac{4}{9}+\frac{4}{7}\)
\(=\left(\frac{-3}{4}+\frac{-1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)
\(=-1+1+\frac{4}{9}\)
\(=\frac{4}{9}\)
\(\frac{1}{2}x+150\%x=2014\)
\(\frac{1}{2}x+\frac{3}{2}x=2014\)
\(\left(\frac{1}{2}+\frac{3}{2}\right).x=2014\)
\(2.x=2014\)
\(\Rightarrow x=1007\)
\(\frac{1}{2}x+150\%x=2014\)
=>\(\frac{1}{2}x+\frac{3}{2}x=2014\)
=>x(\(\frac{1}{2}+\frac{3}{2}\))=2014
=>x.2=2014
=>x=1007
vậy x=1007
\(2x+\left(x-3\right)=\frac{1}{2}\)
\(2x+x-3=\frac{1}{2}\)
\(3x-3=\frac{1}{2}\)
\(3x=\frac{1}{2}+3\)
\(3x=\frac{7}{2}\)
\(x=\frac{7}{2}:3\)
\(\Rightarrow x=\frac{7}{6}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(2A=2+\frac{2}{2}+\frac{2}{2^2}+...+\frac{2}{2^9}\)
\(2A=2+1+\frac{1}{2}+...+\frac{2}{2^8}\)
=>\(A=2+\frac{-1}{29}\)
=>\(A=\frac{57}{29}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> \(2A=2+1+\frac{1}{2}+...+\frac{1}{2^8}\)
=> \(A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
=> \(A=2+1+\frac{1}{2}+...+\frac{1}{2^8}-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\)
=> \(A=2-\frac{1}{2^9}\)
\(A=\frac{3n+2}{n-2}=\frac{3\left(n-2\right)+8}{n-2}=3+\frac{8}{n-2}\)
A nguyên khi \(\frac{8}{n-2}\) nguyên
=> \(8⋮n-2\) => \(n-2\inƯ\left(8\right)\)
=> \(n-2\in\left\{1;2;4;8;-1;-2;-4;-8\right\}\)
=> \(n\in\left\{3;4;6;10;1;0;-2;-4;-6\right\}\)
\(a,\frac{7}{15}-\frac{22}{60}=\frac{7\cdot4}{60}-\frac{22}{60}=\frac{28}{60}-\frac{22}{60}=\frac{6}{60}=\frac{1}{10}\)
\(b,\frac{37}{40}-\frac{64}{100}=\frac{37}{40}-\frac{16}{25}=\frac{185}{200}-\frac{128}{200}=\frac{57}{200}\)
\(c,130\frac{25}{8}-120\frac{17}{8}\)
\(=130+\frac{25}{8}-120+\frac{17}{8}\)
\(=130-120+\left[\frac{25}{8}+\frac{17}{8}\right]\)
\(=130-120+\frac{21}{4}\)
\(=10+\frac{21}{4}=10+5\frac{1}{4}=(10+5)+\frac{1}{4}=15\frac{1}{4}\)
Ba câu kia tương tự
a)\(\frac{7}{15}-\frac{22}{60}=\frac{7}{15}-\frac{11}{30}=\frac{14}{30}-\frac{11}{30}=\frac{14-11}{30}=\frac{3}{30}=\frac{1}{10}\)
b)\(\frac{37}{40}-\frac{64}{100}=\frac{37}{40}-\frac{16}{25}=\frac{185}{200}-\frac{128}{200}=\frac{185-128}{200}=\frac{57}{200}\)
c)\(130\frac{25}{8}-120\frac{7}{8}=\left(130+\frac{25}{8}\right)-\left(120+\frac{7}{8}\right)=130+\frac{25}{8}-120-\frac{7}{8}\)\(=\left(130-120\right)+\left(\frac{25}{8}-\frac{7}{8}\right)=10+\frac{25-7}{8}=10+\frac{18}{8}=10+\frac{9}{4}\)\(=10+2\frac{1}{4}=10+2+\frac{1}{4}=12+\frac{1}{4}=12\frac{1}{4}\)
d)\(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{6}=1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)=1-\left(\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\right)=1-\frac{6}{6}=1-1=0\)
e)\(\frac{17}{8}-\frac{11}{6}+\frac{2}{9}=\frac{153}{72}-\frac{132}{72}+\frac{16}{72}=\frac{153-132+16}{72}=\frac{37}{72}\)
f)\(\frac{21}{21}+\frac{18}{69}+\frac{3}{5}=1+\frac{6}{23}+\frac{3}{5}=\frac{115}{115}+\frac{30}{115}+\frac{69}{115}=\frac{115+30+69}{115}=\frac{214}{115}\)