a: \(-\frac{10}{3}+\frac{13}{10}-\frac16+\frac{7}{10}\)

\(=\left(-\frac{10}{3}-\frac16\right)+\left(\frac{13}{10}+\frac{7}{10}\right)\)

\(=-\frac{21}{6}+\frac{20}{10}=-\frac72+2=-\frac32\)

b: \(1-\frac{20}{6}+\frac{39}{30}-\frac{4}{24}+\frac{34}{50}\)

\(=1-\frac{20}{6}+\frac16+1,3+0,68\)

\(=1-\frac{19}{6}+1,98=2,98-\frac{19}{6}=\frac{298}{100}-\frac{19}{6}=\frac{-56}{300}=\frac{-14}{75}\)

c: \(\frac{15}{24}-\frac{6}{26}+\frac{28}{48}-\frac{30}{39}+\frac{10}{24}\)

\(=\left(\frac{15}{24}+\frac{10}{24}+\frac{28}{48}\right)-\left(\frac{3}{13}+\frac{10}{13}\right)\)

\(=\left(\frac{25}{24}+\frac{14}{24}\right)-1=\frac{39}{24}-1=\frac{15}{24}=\frac58\)

d: \(-\frac{5}{10}+\frac{12}{28}-\frac{3}{27}-\frac{7}{18}+\frac{20}{35}\)

\(=-\frac12+\frac37-\frac19-\frac{7}{18}+\frac47\)

\(=\left(-\frac12-\frac19-\frac{7}{18}\right)+\left(\frac37+\frac47\right)=-1+1=0\)

a)\(2\left(x-\frac{1}{2}\right)^3-\frac{1}{4}=0\)

\(\left(x-\frac{1}{2}\right)^3=\frac{1}{8}\)

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

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

       \(\Rightarrow x=1\)

b)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

c)\(\left(2n+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

     \(\left(2n+\frac{3}{5}\right)^2=\frac{9}{25}\)

      \(\left(2n+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2=\left(-\frac{3}{5}\right)^2\)

\(\Rightarrow\hept{\begin{cases}2n+\frac{3}{5}=\frac{3}{5}\\2n+\frac{3}{5}=-\frac{3}{5}\end{cases}}\Rightarrow\hept{\begin{cases}n=0\\n=-\frac{3}{5}\end{cases}}\)

                       Vậy n=0;-3/5

d)\(3\left(3n-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

   \(\left(3n-\frac{1}{2}\right)^3=-\frac{1}{27}\)

     \(\left(3n-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)

     \(3n-\frac{1}{2}=-\frac{1}{3}\)

     \(\Rightarrow n=\frac{1}{18}\)

Lời giải chi tiết

1²  1                                     1³ 1² – 0²        (0 + 1)²  0²  +1²

2² 1 + 3                                2³ 3² – 1²         (1 + 2)² 1² + 2²

3² 1 + 3 + 5                           3³ 6² – 3²      (2 + 3)²  2² +  3²

                                                     4³ 10² – 6²

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1) (x-1)(x+5)(-3x+8)=0

\(\hept{\begin{cases}\\\\\end{cases}}\)

1) (x-1)(x+5)(-3+8)=0

=  (x-1)(x+5).5       =0

\(\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0+1=1\\x=0-5=-5\end{cases}}\)

\(\Rightarrow x\in\left\{1;-5\right\}\)

2) (x-1)(x-2)(x-3)=0

\(\hept{\begin{cases}x-1=0\\x-2=0\\x-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0+1=1\\x=0+2=2\\x=0+3=3\end{cases}}\)

\(\Rightarrow x\in\left\{1;2;3\right\}\)

3)(5x+3)(x²+4)(x-1)=0

\(\hept{\begin{cases}5x+3=0\\x^2+4=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}5x=0-3=-3\\x^2=0-4=-4\\x=0+1=1\end{cases}}\Rightarrow\hept{\begin{cases}x=-3:5\Rightarrow x\in\varnothing\\x\in\varnothing\\x=1\end{cases}}\)

\(\Rightarrow x=1\)

4)x(x²-1)=0

\(\orbr{\begin{cases}x=0\\x^2-1=0\Rightarrow x^2=0+1=1\Rightarrow x^2=1^2;(-1)^2\Rightarrow x\in\left\{1;-1\right\}\end{cases}}\)

\(\Rightarrow x\in\left\{-1;0;1\right\}\)

Xin lỗi về phần bên trên nha! tại tui ấn nhầm nút.Sorry.

\(a,\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\frac{11}{15}x=\frac{2}{5}\)

\(x=\frac{6}{11}\)

b,\(\left(2x-3\right).\left(6-2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

Vậy

Bước 1: Xét từng nhân tử

Phương trình tích bằng 0 khi ít nhất một trong hai nhân tử bằng 0:

Trường hợp 1:

\(x + \frac{1}{2} = 0 \Rightarrow x = - \frac{1}{2}\)

Trường hợp 2:

\(\frac{2}{3} - 2 x = 0 \Rightarrow 2 x = \frac{2}{3} \Rightarrow x = \frac{1}{3}\)