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(-19,95 - 45,75) + (4,95 + 5,75)
=[(-19,95)+4,95]+[(-45,75)+5,75]
=(-15)+(-40)
=-55
[(-19,95) + (-45,75)] + [(4,95) +(5,75)]
=[-65,7] + [10,7]
=-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 !
\(\begin{array}{l}a) - 1,2 + ( - 0,8) + 0,25 + 5,75 - 2021\\ = [ - 1,2 + ( - 0,8)] + (0,25 + 5,75) - 2021\\ = ( - 2) + 6 - 2021\\ = 4 - 2021\\ = - 2017\\b) - 0,1 + \frac{{16}}{9} + 11,1 + \frac{{ - 20}}{9}\\ = [( - 0,1) + 11,1] + \left( {\frac{{16}}{9} + \frac{{ - 20}}{9}} \right)\\ = 11 + \frac{{ - 4}}{9}\\ = \frac{{99}}{9} + \frac{{ - 4}}{9}\\ = \frac{{95}}{9}\end{array}\)
\(\begin{array}{l}0,65.78 + 2\dfrac{1}{5}.2020 + 0,35.78 - 2,2.2020\\ = 0,65.78 + 2,2.2020 + 0,35.78 - 2,2.2020\\ = (0,65.78 + 0,35.78) + (2,2.2020 - 2,2.2020)\\ = 78.(0,65 + 0,35) + 0\\ = 78.1\\ = 78\end{array}\)
\(2+3.\left(\frac{-1}{3}\right)^2\)
=\(2+3.\frac{1}{9}\)
= \(2+\frac{1}{3}\)
=\(\frac{7}{3}\)
\(\begin{array}{l}\frac{7}{6}.3\frac{1}{4} + \frac{7}{6}.( - 0,25)\\ = \frac{7}{6}.\frac{{13}}{4} + \frac{7}{6}.\frac{{ - 25}}{{100}}\\ = \frac{7}{6}.\frac{{13}}{4} + \frac{7}{6}.\frac{{ - 1}}{4}\\ = \frac{7}{6}.[\left( {\frac{{13}}{4} + ( - \frac{1}{4})} \right)]\\ = \frac{7}{6}.\frac{{12}}{4}\\ = \frac{7}{6}.3\\ = \frac{7}{2}\end{array}\)
=(-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