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\(D=\left(\frac{3}{7}\right)^{21}:\left(\left(\frac{3}{7}\right)^2\right)^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{2.6}=\left(\frac{3}{7}\right)^9\)
\(E=\left(-\frac{1}{3}\right)^{7+9}:\left(-\frac{1}{3}\right)^{5.3}+\left(-2\right)^{12+3}:\left(-2\right)^{15}=\left(-\frac{1}{3}\right)^{16}:\left(-\frac{1}{3}\right)^{15}+\left(-2\right)^{15}:\left(-2\right)^{15}=-\frac{1}{3}+1=\frac{2}{3}\)
\(\left|x+1\right|,\left|x-2\right|,\left|x+3\right|\ge0\)
\(6\ge0\Rightarrow x\ge0\)
\(\left|x+1\right|+\left|x-2\right|+\left|x+3\right|=6\)
\(\Rightarrow\left(x+1\right)+\left(x-2\right)+\left(x+3\right)=6\)
\(\Rightarrow\left(x+x+x\right)+\left(1-2+3\right)=6\)
\(\Rightarrow3x+2=6\)
\(\Rightarrow3x=6-2\)
\(\Rightarrow3x=4\)
\(\Rightarrow x=\frac{4}{3}\)
Ta có: \(S=\left(5-\frac{2}{3}+\frac{3}{2}\right)-\left(7-\frac{5}{4}-\frac{1}{2}\right)-\left(1-\frac{4}{3}+\frac{2}{5}\right).\)
\(\Rightarrow S=\left(\frac{13}{3}+\frac{3}{2}\right)-\left(\frac{23}{4}-\frac{1}{2}\right)-\left(\frac{-1}{3}+\frac{2}{5}\right)\)
\(\Rightarrow S=\frac{35}{6}-\frac{21}{4}-\frac{1}{15}\)
\(\Rightarrow S=\frac{7}{12}-\frac{1}{15}=\frac{31}{60}\)
Vậy \(S=\frac{31}{60}\)
a) 0,16 + 1,3
= \(\frac{4}{25}+\frac{13}{10}\)
= \(\frac{40}{250}+\frac{325}{250}\)
= \(\frac{73}{50}\)
b) 1,3 + 0,12 . \(2\frac{8}{11}\)
= \(\frac{13}{10}+\frac{3}{25}\times\frac{30}{11}\)
= \(\frac{13}{10}+\frac{18}{55}\)
= \(\frac{143}{110}+\frac{36}{110}\)
= \(\frac{179}{110}\)
c) 0,6 + 1,6
= \(\frac{3}{5}+\frac{8}{5}\)
= \(\frac{11}{5}\)
d) 3,6 + 1,36 \(\times\)\(2\frac{1}{5}\)
= \(\frac{18}{5}+\frac{34}{25}\times\frac{11}{5}\)
= \(\frac{18}{5}+\frac{374}{125}\)
= \(\frac{450}{125}+\frac{374}{125}\)
= \(\frac{824}{125}\)
Áp dụng công thức \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\) ta được:
\(\frac{x+2}{x+6}=\frac{3}{x+1}\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)=3\left(x+6\right)\)
\(\Leftrightarrow x^2+3x+2=3x+18\)
\(\Leftrightarrow x^2=16\)
Vậy \(x\in\left\{4;-4\right\}\)
(x+2)/(x+6)=3/(x+1)
<=> (x+2)(x+1)/(x+6)(x+1)=3(x+6)/(x+6)(x+1)
=>(x+2)(x+1)=3(x+6)
<=> x^2+x+2x+2=3x+18
<=> x^2=16
<=>x^2=4^2 hoặc (-4)^2
<=> x=4 hoặc x=-4
Vậy.........
cách 2:
a=\(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
a=(6-5-3)-(2/3+5/3-7/3)+(1/2+3/2-5/2)
a=-2-1/2
a=-5/2
ta xét thấy 3^n+1 = 27
nếu n bằng 0 thì tức là 3^0+1 =3^1 =3
nếu n bằng 1 thì tức là 3^1+1=3^2=9
neu n bang 2 thi tuc la 3^2+1=3^3=27
..........................
vậy n =2
\(\text{Đổi:}0,1\left(6\right)=0+\frac{16-1}{90}=\frac{15}{90}=\frac{1}{6}\)
=> 3-0,1(6)+1/2
= 3-1/6+1/2
= \(\frac{18-1+3}{6}\)
=20/6
=10/3
3-0,1(6)+1/2
=3-1/6+1/2
=17/6+1/2
=10/3