Tính giá trị bt:
- A) 50% × 1/1/3 × 10 × 7/35 × 0.75
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\(50\%.1\frac{1}{3}.10.\frac{7}{35}.0,75\)
\(=\frac{1}{2}.\frac{4}{3}.10.\frac{1}{5}.\frac{3}{4}\)
\(=\left(\frac{1}{2}.10.\frac{1}{5}\right).\left(\frac{4}{3}.\frac{3}{4}\right)\)
\(=1.1=1\)
\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)
\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)
Bài 2:
a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)
\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)
\(=-\dfrac{3}{5}\)
b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(\Leftrightarrow8x-1=5\)
\(\Leftrightarrow8x=6\)
hay \(x=\dfrac{3}{4}\)
\(50\%.\frac{4}{3}.10.\frac{7}{35}.0,75\)
\(=\frac{1}{2}.\frac{4}{3}.10.\frac{7}{35}.\frac{3}{4}\)
\(=\left(\frac{1}{2}.10\right)\times\left(\frac{4}{3}.\frac{3}{4}\right).\frac{7}{35}\)
\(=5.1.\frac{7}{35}\)
\(=\frac{35}{35}=1\)
~ Hok tốt ~
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{40}-\frac{1}{40}\right)-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{42}{43}\)
~ Hok tốt ~
a)3 . 103 + 2 . 102 + 5 . 10
= 3 . 102 . 10 + 2 . 10 . 10 + 5 . 10
= 10 . ( 3 . 102 + 2 . 10 + 5 )
= 10 . ( 3 . 100 + 20 + 5 )
= 10 . ( 300 + 20 + 5 )
= 10 . 325
= 3250
a) 3 . 103 + 2 . 102 + 5 . 10
= 3 . 1000 + 2 . 100 + 5 . 10
= 3000 + 200 + 50
= 3250
\(a,2.\left|x+1\right|-3=5\)
\(\Rightarrow2.\left|x+1\right|=5+3\)
\(\Rightarrow2.\left|x+1\right|=8\)
\(\Rightarrow\left|x+1\right|=8:2\)
\(\Rightarrow\left|x+1\right|=4\)
\(\Rightarrow\orbr{\begin{cases}x+1=4\\x+1=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Vậy : x = 3 hoặc x = -5
b) Để A có giá trị nguyên thì n + 1 \(⋮\)n - 2
Ta có : n + 1 = ( n - 2 ) + 3
=> n + 1 \(⋮\)n - 2
khi ( n - 2 ) + 3 \(⋮\) n - 2
=> 3 \(⋮\)n - 2
=> n - 2 \(\in\)Ư ( 3 ) = { 1 ; -1 ; 3 ; -3 }
Với n - 2 = 1 => n = 3
Với n - 2 = -1 => n = 1
Với n - 2 = 3 => n = 5
Với n - 2 = -3 => n = -1
Vậy : n \(\in\){ 3 ; 1 ; 5 ; -1 }