7) x + ( x + 5 ) x 3 = 75
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NN
1
NH
28 tháng 3 2023
\(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{11}{75}\left(x\ne0;x\ne-2\right)\)
\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{22}{75}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{22}{75}\)
\(\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{22}{75}\)
\(\dfrac{1}{x+2}=\dfrac{1}{3}-\dfrac{22}{75}\)
\(\dfrac{1}{x+2}=\dfrac{1}{25}\)
suy ra
\(x+2=25\\ x=23\left(tm\right)\)
28 tháng 3 2023
=>2/3*5+2/5*7+...+2/x(x+2)=22/75
=>1/3-1/5+1/5-1/7+...+1/x-1/x+2=22/75
=>1/3-1/x+2=22/75
=>1/x+2=25/75-22/75=3/75=1/25
=>x=23
8 tháng 7 2021
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
NT
1
NA
Nguyễn An Ninh
VIP
24 tháng 4 2023
\((\)125-120\()\)x45
5x45=225
\(\dfrac{30x25x7x8}{75x8x12x14}\)=\(\dfrac{10\cdot1\cdot7\cdot1}{1\cdot1\cdot1\cdot12\cdot2}\)=\(\dfrac{70}{24}\)
TT
2
4 tháng 8 2021
a) Ta có: \(3\left(x+5\right)-3=2\left(x+1\right)+7\)
\(\Leftrightarrow3x+15-3=2x+2+7\)
\(\Leftrightarrow3x+12=2x+9\)
hay x=-3
b) Ta có: \(15-\left(x+2\right)^2=-1\)
\(\Leftrightarrow\left(x+2\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
c) Ta có: \(75-\left(2x+1\right)^3=11\)
\(\Leftrightarrow\left(2x+1\right)^3=64\)
\(\Leftrightarrow2x+1=4\)
hay \(x=\dfrac{3}{2}\)
LN
3
TA
4 tháng 8 2021
`3(x+5)-3=2(x+1)+7`
`3x+15-3=2x+2+7`
`x=-3`
.
`15-(x+2)^2=-1`
`(x+2)^2=16`
`(x+2)^2=4^2=(-4)^2`
`[(x+2=4),(x+2=-4):}`
`[(x=2),(x=-6):}`
.
`75-(2x+1)^3=11`
`(2x+1)^3=64`
`(2x+1)^2=4^3`
`2x+1=4`
`x=3/2`
R
3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 tháng 8 2023
(\(x\) - 5)2 = 16
(\(x-5\))2 = 42
\(\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy \(x\) \(\in\) { 1; 9 }
B. 4\(x\) + 33 = 72
4\(^x\) + 33 = 49
4\(^x\) = 49 - 33
4\(^x\) = 16
4\(^x\) = 16
4\(^x\) = 42
\(x\) = 2
c. 2\(^x\).4 = 128
2\(^x\) = 128 : 4
2\(^x\) = 32
2\(^x\) = 25
\(x\) = 5
D, 5\(^x\) . 3 - 75 = 0
5\(^x\).3 = 75
5\(^x\) = 75: 3
5\(^x\) = 25
5\(^x\) = 52
\(x\) = 2
9 tháng 8 2023
A) \(\left(x-5\right)^2=16=4^2\Rightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
B) \(4^x+33=7^2\Rightarrow4^x=49-33\Rightarrow4^x=16=4^2\Rightarrow x=2\)
C) \(2^x.4=128\Rightarrow2^x=32=2^5\Rightarrow x=5\)
D) \(5^x.3-75=0\Rightarrow5^x.3=75\Rightarrow5^x=25=5^2\Rightarrow x=2\)
x + ( x + 5 ) x 3 = 75
x + 3x + 15 = 75
4x = 60
x = 15
\(x+\left(x+5\right)\cdot3=75\)
\(x+3x+15=75\)
\(x+3x=75-15\)
\(4x=60\)
\(x=60:4\)
\(x=15\)