a 

b: 

c: =6/7(8/13+1-3/13)

=6/7*18/13

=108/91

d: =9/25*-53/3-9/25*22/3

=9/25(-53/3-22/3)

=9/25*(-25)=-9

e: =2/5(-10/9+1/9)

=-2/5

vì bạn ấy đã đặt các tích riêng thẳng cột vs nhau nên thực chất bạn ấy đã nhân số đó vs :

2+3=5

thừa số thứ nhất là :

460:5=92

tích đúng là :

92x23=2116

tick cho mik bạn nhé

mik cảm ơn bạn khuê rất là nhiều ạ

\(\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{230.233}\)

\(=\frac{2}{5}-\frac{2}{8}+\frac{2}{8}-\frac{2}{11}+...+\frac{2}{230}-\frac{2}{233}\)

\(=\frac{2}{5}-\frac{2}{233}\)

\(=\frac{466}{1165}-\frac{10}{1165}=\frac{456}{1165}\)

\(\dfrac{5}{8}+\dfrac{3}{8}x=1\dfrac{1}{8}\)

\(\Rightarrow\dfrac{5}{8}+\dfrac{3}{8}x=\dfrac{9}{8}\)

\(\Rightarrow\dfrac{3}{8}x=\dfrac{9}{8}-\dfrac{5}{8}\)

\(\Rightarrow\dfrac{3}{8}x=\dfrac{4}{8}\)

\(\Rightarrow x=\dfrac{4}{8}:\dfrac{3}{8}\)

\(\Rightarrow x=\dfrac{4}{8}\cdot\dfrac{8}{3}=\dfrac{4}{3}\)

\(---\)

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

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

\(\Rightarrow\dfrac{1}{3}x-2=-\dfrac{1}{6}\)

\(\Rightarrow\dfrac{1}{3}x=-\dfrac{1}{6}+2\)

\(\Rightarrow\dfrac{1}{3}x=\dfrac{11}{6}\)

\(\Rightarrow x=\dfrac{11}{6}:\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{11}{6}\cdot3=\dfrac{11}{2}\)

\(---\)

\(\left(3x+4\right)\left(2x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x+4=0\\2x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=-4\\2x=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(--\)

\(-4+4x=9x-14\)

\(\Rightarrow4x-9x=-14+4\)

\(\Rightarrow-5x=-10\)

\(\Rightarrow x=\left(-10\right):\left(-5\right)=2\)

\(---\)

\(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)

\(=\left[\left(-\dfrac{14}{25}\right)^2\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}-\dfrac{1}{6}\)

\(=\left(\dfrac{196}{625}\cdot\dfrac{125}{49}\right)\cdot\dfrac{5}{6}-\dfrac{1}{6}\)

\(=\dfrac{4}{5}\cdot\dfrac{5}{6}-\dfrac{1}{6}\)

\(=\dfrac{4}{6}-\dfrac{1}{6}\)

\(=\dfrac{3}{6}\)

\(=\dfrac{1}{2}\)

\(Toru\)

12. \(\dfrac{4\sqrt{3}}{3}\pi\)

13. \(12\pi\)

14. \(\sqrt{6}\pi a^2\)

a: 

\(\dfrac{5}{8}+\dfrac{3}{8}x=1\dfrac{1}{8}\)

\(\Leftrightarrow x\cdot\dfrac{3}{8}=\dfrac{9}{8}-\dfrac{5}{8}=\dfrac{4}{8}\)

=>\(x\cdot3=4\)

=>x=4/3

b: \(\dfrac{1}{2}-\left(\dfrac{1}{3}x-2\right)=\dfrac{2}{3}\)

=>\(\dfrac{1}{3}x-2=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3-4}{6}=-\dfrac{1}{6}\)

=>\(x\cdot\dfrac{1}{3}=2-\dfrac{1}{6}=\dfrac{11}{6}\)

=>\(x=\dfrac{11}{2}\)

c: \(-4+4x=9x-14\)

=>\(-5x=-10\)

=>x=2

d: \(-\dfrac{3}{2}-2x+\dfrac{3}{4}=-2\)

=>\(-2x-\dfrac{3}{4}=-2\)

=>\(2x+\dfrac{3}{4}=2\)

=>\(2x=\dfrac{5}{4}\)

=>\(x=\dfrac{5}{8}\)

e: (3x+4)(2x+3)=0

=>\(\left[{}\begin{matrix}3x+4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Nếu em thay $x=9,10,...$ không ra kết quả thì có nghĩa bài toán không có nghiệm $x=9,10,...$ thôi. 

Em xét 3 TH:

$x\geq 7$

$3\leq x< 7$

$x< 3$

Để phá trị tuyệt đối

Còn không có chuyện phải thay $x\leq 7$

Có ai ko giúp em với