Lời giải:

Ta có:

$\sqrt[3]{3x-2}+\sqrt{x+1}=3$

$\iff (\sqrt[3]{3x-2}-1)+(\sqrt{x+1}-2)=0$

$\iff \frac{x-3}{\sqrt[3]{3(x-2{)}^2}+\sqrt[3]{3x-2}+1}+\frac{x-3}{\sqrt{x+1}+2}=0$

$\iff (x-3)\left[\frac{1}{\sqrt[3]{3(x-2{)}^2}+\sqrt[3]{3x-2}+1}+\frac{1}{\sqrt{x+1}+2}\right]=0$

$\mathrm{Dễ\; thấy\; biểu\; thức\; trong\; ngoặc\; vuông\; luôn\; lớn\; hơn\; \$0\$\; với\; mọi x\in ĐKXĐx\in ĐKXĐ nên\; PT\; có\; nghiệm\; duy\; nhất x=3}$

rfyfyrx3ryer8uei3yr4

góc \(_{O_3}\)=\(87^{_{^{^{ }}}^0}\)

góc \(_{O_4}\)=\(^{93^0}\)

Vì \(O^4\)và \(O^3\)là 2 góc kề bù mà  \(O^4\)-\(O^3\)=\(6^0\)

=>\(o^3\)=\(87^0\)    vì (\(180^0\)-\(6^0\)):2= \(87^0\)

=>\(O^4\)= \(93^0\)  vì   \(87^0\)+\(6^0\)=\(93^0\)

(  mk trình bày đại á)

\(\frac{-1}{10}\)-\(\frac{2}{5}\)x+\(\frac{7}{20}\)=\(\frac{1}{10}\)

              \(\frac{2}{5}\)x+\(\frac{7}{20}\)=\(\frac{-1}{10}\)-\(\frac{1}{10}\)

              \(\frac{2}{5}\)x+\(\frac{7}{20}\)=\(\frac{-2}{10}\)

              \(\frac{2}{5}\)x              =\(\frac{-4}{20}\)-\(\frac{7}{20}\)

              \(\frac{2}{5}\)x              =\(\frac{-11}{20}\)

                       x               =\(\frac{-11}{20}\).\(\frac{5}{2}\)

                      x                =-1\(\frac{3}{8}\)

vậy...

-1/10-2/5x+7/20=1/10

2/5x+7/10=-1/10-1/10

2/5x+7/10=0

2/5x=0-7/10

2/5x=-3/10

x=-3/10:2/5

x=-3/4

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{2}\)

\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\right)\)

\(=\frac{8}{9}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{9-8}{8.9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)=\frac{8}{9}-\frac{8}{9}=0\)

3B=3/1.4+3/4.7+3/7.10+...+3/100.103

3B=(4-1)/1.4+(7-4)/4.7+(10-7)/7.10+...+(103-100)/100.103

3B=1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103=1-1/103=102/103

B=102/(3.103)=34/103

HT

\(\frac{1}{1.4}+\frac{1}{4.7}+.....+\frac{1}{100.103}\)

Đặt :

\(A=\frac{1}{1.4}+\frac{1}{4.7}+....+\frac{1}{100.103}\)

\(3A=\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{100.103}\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{100}-\frac{1}{103}\)

\(3A=1-\frac{1}{103}\)

\(3A=\frac{102}{103}\)

\(A=\frac{34}{103}\)

giúp mình với ToT

trả lời nhớ k nhé!

=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1999\cdot2000}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...-\frac{1}{999}+\frac{1}{999}-\frac{1}{2000}\)

=\(\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{999}+\frac{1}{999}\right)-\frac{1}{2000}\)

=\(\frac{1}{1}+0+0+...+0-\frac{1}{2000}\)

=\(\frac{1}{1}-\frac{1}{2000}\)

=\(\frac{2000}{2000}-\frac{1}{2000}\)

=\(\frac{1999}{2000}\)

Ta có :

b - a = 57

\(\frac{a}{-9}=\frac{b}{10}\)

Áp dụng tính chất dãy tỉ số bănhg nhau , ta có :

\(\frac{a}{-9}=\frac{b}{10}=\frac{b-a}{10-\left(-9\right)}=\frac{57}{19}=3\)

\(\Rightarrow\hept{\begin{cases}a=3.\left(-9\right)=-27\\b=3.10=30\end{cases}}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
a/-9=b/10=b-a/10+9=57/19=3
=>a/-9=3=>a=-27
=>b/10=3=>b=30

   *CHÚC BẠN HỌC TỐT*

minh hoc lop 12

thì trả lời đi chị

[4/5] 5 x = [-4/5]^7

\(\left(\frac{4}{5}\right)^5.x=\left(\frac{-4}{5}\right)^7\)

\(\Rightarrow x=\left(\frac{-1.4}{5}\right)^7:\left(\frac{4}{5}\right)^5\)

\(\Rightarrow x=\left(-1\right)^7.\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5\)

\(\Rightarrow x=\left(-1\right).\left(\frac{4}{5}\right)^2\)

\(\Rightarrow x=\left(-1\right).\frac{16}{25}\)

\(\Rightarrow x=\frac{-16}{25}\)