Bài 10:

Đặt \(\dfrac{a}{5}=\dfrac{b}{4}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=4k\end{matrix}\right.\)

Ta có: \(a^2-b^2=36\)

\(\Leftrightarrow25k^2-16k^2=36\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

\(\Leftrightarrow\left\{{}\begin{matrix}a=5k=10\\b=4k=8\end{matrix}\right.\)

Trường hợp 2: k=-2

\(\Leftrightarrow\left\{{}\begin{matrix}a=5k=-10\\b=4k=-8\end{matrix}\right.\)

(2,468+1,057)x0,72 =

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

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

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

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

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

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

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

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

\(---\)

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

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

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

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

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

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

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

\(---\)

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

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

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

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

\(--\)

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

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

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

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

\(---\)

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

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

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

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

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

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

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

\(Toru\)

Hiệu số tuổi của 2 anh em là : 

9 - 4 = 5 (tuổi)

Tuổi em khi đó là : 

5 : (2 - 1) x 1 = 5 (tuổi)

Sau số năm nữa tuổi anh gấp đội tuổi em là : 

5 - 4 = 1 (năm)

Đáp số : 1 năm

vẫn 5 tuổi

gọi só cần tìm là a,ta có

a+9=4+(2xa)

a+9=8xa

9-8=a

1=a

(chú ý:mình giải theo lớp 6,ko hiểu thì thôi)

b= 806

c= 5

Tick nhé

MỜ QUÁ !!!!!!!!

mờ lắm

là 3/5