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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: \(=-15,5\left(20,8+9,2\right)+3,5\left(9,2+20,8\right)\)
\(=30\cdot\left(-12\right)=-360\)
b: \(=\dfrac{21}{20}:\dfrac{27}{10}+2+\left(\dfrac{2}{5}:\dfrac{5}{2}\right)\cdot\left(\dfrac{21}{5}-\dfrac{13}{10}\right)\)
\(=\dfrac{21}{10}\cdot\dfrac{10}{27}+2+\dfrac{4}{25}\cdot\dfrac{29}{10}\)
\(=\dfrac{7}{9}+2+\dfrac{58}{125}=\dfrac{3647}{1125}\)
a) Ta có : (1/16)10 = [(1/2)4]10 = (1/2)40
Vì (1/2)40 < (1/2)50 nên (1/16)10 < (1/2)50
b) Ta có : 430 = ( 2 . 2)30 = 230 . 230 = (22)15 . (23)10 > 315 . 810 > 3 . 310 .810 = 3 . (3 . 8)10 = 3 .2410
Vậy nên 230 + 330 + 430 > 2410 . 3
Mình chỉ giải thế thôi, còn đâu bn tự làm tiếp
`a)2^{300}=(2^3)^100=8^100`
`3^200=(3^2)^100=9^100`
Vì `9^100>8^100`
`=>2^300<3^200`
`b)3xx24^10`
`=3.(3.8)^10`
`=3^{11}.8^10`
`=3^{11}.2^30`
`2^300=2^{30}.2^{270}`
`=2^{30}.8^{90}`
Vì `3^11<8^90`
`=>3^{11}.2^30<8^{90}.2^30=2^300`
`=>3xx24^{10}<2^300+3^20+4^30`
(2x-5)-(\(\frac{3}{2}\) . 6x + \(\frac{3}{2}\))=4
2x -5 - 9x -\(\frac{3}{2}\) =4
2x - 9x = 4+ 5+ \(\frac{3}{2}\)
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 !