\(\dfrac{4}{7}-\dfrac{4}{9}\)

\(=\dfrac{36}{63}-\dfrac{28}{63}\)

\(=\dfrac{8}{63}\)

14,36 x 1,78 x 2,45 x ( 9 x 0,9 + 0,9 - 0,9 x 10 )

=14,36 x 1,78 x 2,45 x [0,9x(9+1-10)]

=14,36 x 1,78 x 2,45 x [0,9x0]

=14,36 x 1,78 x 2,45 x 0 

=0 

=14,36x1,78x2,45x0

=0

14,36 x 1,78 x 2,45 x ( 9 x 0,9 + 0,9 - 0,9 x 10 )

=14,36 x 1,78 x 2,45 x [0,9x(9+1-10)]

=14,36 x 1,78 x 2,45 x [0,9x0]

=14,36 x 1,78 x 2,45 x 0 

=0 vì 0 nhân với số nào cũng =0

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17+18 = 171 nhé

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17+18.  = 171

\(1023456^3=1023456.1023456.1023456=\)kết quả lớn lắm

vay co cach nao tinh dc k

1023456³=1072031456922402800

\(3^{15}-9^6=3^{15}-\left(3^2\right)^6=3^{15}-3^{12}=3^{12}\left(3^3-1\right)=3^{12}\cdot26=3^{12}\cdot13\cdot2⋮13\)

1000A = 1451,(51)

=> 990A = 1451,(51) - 14,(51)

=> 990A = 1437

=> A = 1437/990 = 479/330

=479/330

=479/330

=-2+0,(412)=-2+412/999=-2410/999

đề?

\(\sqrt{x^2-8x-9}=\sqrt{\left(x-9\right)\left(x+1\right)}\)

ĐKXĐ: \(\left(x-9\right)\left(x+1\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge9\end{matrix}\right.\)

đề đâu 

( 3x-1) ( x²+ 9) = (3x-1) (7x-10)

⇒( 3x-1) ( x²+ 9) - (3x-1) (7x-10) = 0

⇒( 3x-1) (( x²+ 9)-(7x-10)) = 0

⇒( 3x-1)(x²+9-7x+10)=0

⇒( 3x-1)(x²-7x+19)=0

⇒\(\left[{}\begin{matrix}3x-1=0\\x^2-7x+19=0\end{matrix}\right.\)

3x-1=0

⇒x=\(\dfrac{1}{3}\)

x²-7x+19=0

⇒ \(x^2-\dfrac{7}{2}x-\dfrac{7}{2}x+\left(\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)

⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)

vì \(\left(x-\dfrac{7}{2}\right)^2\ge0\); \(\dfrac{27}{4}>0\)

⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}>0\)

⇒ x vô nghiệm

Vậy x= \(\dfrac{1}{3}\)

\(\left(3x-1\right)\left(x^2+9\right)=\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2+9\right)-\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Ta có: \(\left(3x-1\right)\left(x^2+9\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+9-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+19\right)=0\)

\(\Leftrightarrow3x-1=0\)

hay \(x=\dfrac{1}{3}\)