\(-15,5.20,8+3,5.9,2-15,5.9,2+3,5.20,8\)

\(=-15,5\left(20,8+9,2\right)+3,5\left(9,2+20,8\right)\)

\(=\left(-15,5\right).30+3,5.30\)

\(=30.\left(-12,5\right)\)

\(=-375\)

= -15,5 .( 20,8 - 9,2) + 3,5 . (9,2+20,8)

=-15,5 . 11,6 + 3,5 .30

=-179,8+105

=-74,8

\(-15,5.20,8+3,5.9,2-15,5.9,2+3,5.20,8\)

\(=-15,5\left(20,8+9,2\right)+3,5\left(9,2+20,8\right)\)

\(=\left(9,2+20,8\right)\left(-15,5+3,5\right)\)

\(=30.\left(-12\right)=-360\)

A) -15,5 x 20,8 +2,5 x 9,2 -15,5 x 9,2 + 3,5 x 20,8

=-15,5 x 20,8 + 2,5 x 9,2 + -15,5 x 9,2 +3,5 x 20,8

=[20,8x(-15,5+3,5)]+[9,2x(2,5+-15,5)]

=[20,8x(-12)]+[9,2x(-13)]

=-249,6+-119,6

=-369,2

a) -369,2

b) -7

a: \(=-15,5\left(20,8+9,2\right)+3,5\left(9,2+20,8\right)\)

\(=30\cdot\left(-12\right)=-360\)

b: \(=\dfrac{21}{20}:\dfrac{27}{10}+2+\left(\dfrac{2}{5}:\dfrac{5}{2}\right)\cdot\left(\dfrac{21}{5}-\dfrac{13}{10}\right)\)

\(=\dfrac{21}{10}\cdot\dfrac{10}{27}+2+\dfrac{4}{25}\cdot\dfrac{29}{10}\)

\(=\dfrac{7}{9}+2+\dfrac{58}{125}=\dfrac{3647}{1125}\)

2)

a) \(2\left|2x-3\right|=1\)

=> \(\left|2x-3\right|=1:2\)

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

=> \(\left[{}\begin{matrix}2x-3=\frac{1}{2}\\2x-3=-\frac{1}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}2x=\frac{1}{2}+3=\frac{7}{2}\\2x=\left(-\frac{1}{2}\right)+3=\frac{5}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\frac{7}{2}:2\\x=\frac{5}{2}:2\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{5}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{7}{4};\frac{5}{4}\right\}.\)

b) \(7,5-3\left|5-2x\right|=-4,5\)

=> \(4,5\left|5x-2\right|=-4,5\)

=> \(\left|5x-2\right|=\left(-4,5\right):4,5\)

=> \(\left|5x-2\right|=-1\)

Ta luôn có: \(\left|x\right|>0\forall x\)

=> \(\left|5x-2\right|>-1\)

=> \(\left|5x-2\right|\ne-1\)

Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.

c) \(\left|3x-4\right|+\left|3y+5\right|=0\)

Ta có: \(\left|3x-4\right|>\) hoặc \(=0\forall x\)

\(\left|3y+5\right|>\) hoặc \(=0\forall y.\)

=> \(\left|3x-4\right|+\left|3y+5\right|=0\)

=> \(\left[{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}3x=0+4=4\\3y=0-5=-5\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=4:3\\y=\left(-5\right):3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{4}{3}\\y=-\frac{5}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{4}{3}\right\};y\in\left\{-\frac{5}{3}\right\}.\)

Chúc bạn học tốt!

Bài 1:

a) \(-15,5.20,8+3,5.9,2-15,5.9,2+3,5.20,8\)

\(=20,8.\left(-15,5+3,5\right)+9,2.\left(-15,5+3,5\right)\)

\(=\left(-15,5+3,5\right).\left(20,8+9,2\right)\)

\(=\left(-12\right).30=-360\)

b) \(\left[\left(-19,95\right)+\left(-45,75\right)\right]+\left[4,95+5,75\right]\)

\(=\left[\left(-19,95\right)+4,95\right]+\left[\left(-45,75\right)+5,75\right]\)

\(=-15+\left(-40\right)=-55\)

Bài 2 :

\(a,2.\left|2x-3\right|=1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=\frac{1}{2}\\2x-3=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{1}{2}+3\\2x=-\frac{1}{2}+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{7}{2}\\2x=\frac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{5}{4}\end{matrix}\right.\)

Vậy : \(x\in\left\{\frac{7}{4},\frac{5}{4}\right\}\)

\(b,7.5-3\left|5-2x\right|=-4.5\)

\(\Leftrightarrow3.\left|5-2x\right|=7.5-\left(-4.5\right)=12\)

\(\Leftrightarrow\left|5-2x\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{9}{2}\end{matrix}\right.\)

Vậy : \(x\in\left\{\frac{1}{2},\frac{9}{2}\right\}\)

\(c,\left|3x-4\right|+\left|3y+5\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=-\frac{5}{3}\end{matrix}\right.\)

Vậy : \(\left(x,y\right)=\left(\frac{4}{3},-\frac{5}{3}\right)\)

Bài 3 :

a) \(2^{300}\) và \(3^{200}\)

Ta có : \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

mà : \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)

Vậy : \(3^{200}>2^{300}\)

b) \(2^{30}+3^{30}+4^{30}\) và \(3.2.4^{10}\)

Ta có : \(3.2.4^{10}=6.\left(2^2\right)^{10}=6.2^{20}=3.2^{21}\)

Ta thấy : \(2^{30}>3.2^{21}\Rightarrow2^{30}+3^{30}+4^{30}>3.2^{21}\)

hay : \(2^{30}+3^{30}+4^{30}>3.2.4^{10}\)

Vậy : \(2^{30}+3^{30}+4^{30}>3.2.4^{10}\)

Chúc bạn học tốt !