=(-19,95+4,95) +(-45,75+5,75) 

[(-19,95) + (-45,75)] + [(4,95) + (+5,75)]

=> [-65,7] + [10,7]

=>        -55

nếu mn thấy đúng nhớ cho mk nha

= -15+-40 =-55

(-19,95 - 45,75) + (4,95 + 5,75)

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

=(-15)+(-40)

=-55

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 !

à mình nhầm =5,76 nha

\(^{x^2=5,75}\)

=>\(x=\frac{\sqrt{23}}{2}\)

bạn ơi mình viết sai đề nó phải =5,76 ạ

Bài 1:

11,75+4,95+8,05

= (4,95+8,05)+11,75

= 13+11,75

=24,75

\(19,95\times2000+4001\times39,90-19,95\times2\)

\(=19,95\times2000+4001\times2\times19,95-19,95\times2\)

\(=19,95\times2000+8002\times19,95-19,95\times2\)

\(=19,95\times\left(2000+8002-2\right)\)

\(=19,95\times10000\)

\(=199500\)

= 19,95 x 2 x 1000 + 4001 x 39,90 - 19,95 x 2

= 39,90 x 1000 + 4001 x 39,90 - 39,90

= 39,90 x (1000 + 4001 - 1)

= 39,90 x 5000

= 19,95 x 2 x 5000

= 19,95 x 10000

= 199500

lời giải:

19,95 x 2000 + 4001 x 39,90 -019,95 x2 = 19,95 x 2000 + 4001 x 2 x 19,95 - 19,56 x 2 = 19,95 x ( 2000+ 4001x2-2)

                                                                                                                               = 19,95 x 9982 = 19,95 x 10000 = 199500

đung rồi đấy nhớ cho mình nghe