?????????????????

a) \(x=2\)

b) \(x=\frac{5}{36}\)

a) \(1-\left(\frac{2x}{3}+2\right)=-1\cdot\frac{1}{3}\)

\(1-\frac{2}{3}x-2=-\frac{1}{3}\)

\(-\frac{2}{3}x-1=-\frac{1}{3}\)

\(-\frac{2}{3}x=\frac{2}{3}\)

\(x=-1\)

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b) \(\frac{2}{5}x-1\cdot\frac{1}{2}x+x=\frac{1}{3}\)

\(\left(\frac{2}{5}-\frac{1}{2}+1\right)x=\frac{1}{3}\)

\(\frac{9}{10}x=\frac{1}{3}\)

\(x=\frac{1}{3}:\frac{9}{10}\)

\(x=\frac{10}{27}\)

Sửa đề: 1+3+...+2n-1\(=n^2\)

Số số hạng của dãy số là:

\(\left(2n-1-1\right):2+1=\frac{2n-2}{2}+1=n-1+1=n\) (số)

Tổng của dãy số là:

\(\left(2n-1+1\right)\cdot\frac{n}{2}=2n\cdot\frac{n}{2}=n^2\)

Ta có: \(11^{99}-\left\lbrace1\frac13-\left\lbrack5\cdot2^3-\left(-7\right)^2+\frac13+11^9\left(27^4-81^3-11^{90}\right)\right\rbrack\right\rbrace\)

\(=11^{99}-\left\lbrace\frac43-\left\lbrack5\cdot8-49+\frac13+11^9\left(3^{12}-3^{12}-11^{90}\right)\right\rbrack\right\rbrace\)

\(=11^{99}-\left\lbrace\frac43-\left\lbrack40-49+\frac13-11^9\cdot11^{90}\right\rbrack\right\rbrace\)

\(=11^{99}-\left\lbrace\frac43-\left\lbrack-9+\frac13-11^{99}\right\rbrack\right\rbrace\)

\(=11^{99}-\frac43+\left\lbrack-9+\frac13-11^{99}\right\rbrack\)

\(=11^{99}-\frac43+\left(-9\right)+\frac13-11^{99}=-1-9=-10\) \(\)

\(x\times2+x\times\dfrac{1}{5}=1.\dfrac{1}{3}\\ \Rightarrow x\times\left(2+\dfrac{1}{5}\right)=\dfrac{1}{3}\\ \Rightarrow x\times\dfrac{11}{5}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{5}{33}\)

Ta có:

 S = 1.1!+2.2!+3.3!+.....+100.100!

 S = (1+1-1).1!+(2+1-1).2!+...+(100+1-1).100!

 S = 2!-1+3!-2!+4!-3!+...+101!-100!

 S = 101!+(100!-100!)+(99!-99!)+...+(2!-2!)-1

 S = 101!-1

k cho mk nha bạn 

                       S=101! - 1

giải giùm e vs ạ

1 is played

2 was given

3 are watered

4 was invited

5 is being prepared

6 be cleaned

7 has been explained

8 was being done

9 be visited

10 be found