Áp dụng TCDTSBN ta có:

\(\dfrac{3x}{2}=\dfrac{2y}{5}=\dfrac{6x-6y}{4-15}=-\dfrac{108}{11}\)

\(\dfrac{3x}{2}=-\dfrac{108}{11}\Rightarrow x=-\dfrac{72}{11}\\ \dfrac{2y}{5}=-\dfrac{108}{11}\Rightarrow y=-\dfrac{270}{11}\)

26

20

26

20

\(18\times\dfrac{13}{9}=26\)

\(24\times\dfrac{5}{6}=20\)

a: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)

Đáp án là : 0

Tk nha!!

9 x 9 x 9 x 9 x 9 x 9 x 9 x 9 x 9 x 9 x 9 x 9 x 9 x 9 x 1 x 1 x 5 x 8 x 5 x 4 x 0 x 2 x 4 x 3 x 6 = 0

Blink chính hãng nhé!

_Học_tốt_

9. .... . 9 .1 . 1 . 5 . 8 . 5 . 4 . 0 . 2 . 4 . 3 . 6 = 0

Kết quả là :

\(9x9x9x9x9x0=0\)

Đáp số : 0

9 x 9 x 9 x 9 x 9  x 0 = 0 k mk mk k lại cho

9 x 9 x 9 x 9 x 9 x 0

= \(9^5.0\)

= \(0\)

9*(19/7-5/7)

9*14/7=9*2=18

Cảm ơn nhen

`!`

\(a,\dfrac{8}{9}\times\dfrac{7}{5}-\dfrac{7}{9}\times\dfrac{2}{5}+\dfrac{2}{9}\\ =\dfrac{8}{9}\times\left(\dfrac{7}{5}-\dfrac{2}{5}\right)+\dfrac{2}{9}\\ =\dfrac{8}{9}\times\dfrac{5}{5}+\dfrac{2}{9}\\ =\dfrac{8}{9}+\dfrac{2}{9}\\ =\dfrac{10}{9}\)

\(b,\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{3}{12}\\ =\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{1}{4}\\ =\dfrac{1}{4}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\\ =\dfrac{1}{4}\times\dfrac{9}{9}\\ =\dfrac{1}{4}\)

Ta có : \(\frac{4}{11}.\frac{-2}{9}+\frac{4}{11}\frac{-8}{9}+\frac{4}{11}.\frac{1}{9}\)

= \(\frac{4}{11}.\left(\frac{-2}{9}+\frac{-8}{9}+\frac{1}{9}\right)\)

= \(\frac{4}{11}.\frac{-9}{9}\)

= \(\frac{4}{11}.\left(-1\right)\)

= \(\frac{-4}{11}\)

\(\frac{4}{11}\left(-\frac{2}{9}+-\frac{8}{9}+\frac{1}{9}\right)=\frac{4}{11}1=\frac{4}{11}\)

xin lỗi =-4/11

(11 / 29 + 2 / 29 + 16 / 29 ) . 9 / 10

1 . 9/10 = 9/10

\(\dfrac{9}{10}\times\left(\dfrac{11}{29}+\dfrac{2}{29}+\dfrac{16}{29}\right)=\dfrac{9}{10}\times\dfrac{29}{29}=\dfrac{9}{10}\times1=\dfrac{9}{10}\)

= 9/10 x ( 11/29 + 2/9 + 16/29 )

= 9/10 x 1

= 9/10

\(\left(x^2-9\right)^2-9\left(x-3\right)^2=0\)

\(< =>\left(x^2-9\right)^2-\left[3\left(x-3\right)\right]^2=0\)

\(< =>\left(x^2-9\right)^2-\left(3x-9\right)^2=0\)

\(< =>\left(x^2-9+3x-9\right)\left(x^2-9-3x+9\right)=0\)

\(< =>\left(x^2+3x-18\right)\left(x^2-3x\right)=0\)

\(=>\left[{}\begin{matrix}x^2+3x-18=0\\x^2-3x=0\end{matrix}\right.< =>\left[{}\begin{matrix}\left(x+6\right)\left(x-3\right)=0\\x\left(x-3\right)=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=-6\\x=3\\x=0\end{matrix}\right.\)