2 x 5 + 10 = ?
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`687 + (2 × 5) + 10`
`= 687 + 10 + 10`
`= 697 + 10`
`= 707`
`2 1/2 xx 7/5 + (-9/10) xx 2/3`
`= 5/2 xx 7/10 + (-9/10) xx 2/3`
`= 35/20 + (-18/30)`
`= 7/4 + (-3/5)`
`= 23/20`
__
`x + 45% =2 1/2 - 5/3`
`=> x+ 45/100 = 5/2 - 5/3`
`=>x+ 9/20= 15/6-10/6`
`=> x+9/20 = 5/6`
`=>x= 5/6 - 9/20`
`=>x=23/60`
__
`80% +x=-5/2 +3/4`
`=> 80/100 + x= -10/4 +3/4`
`=> 4/5 + x= -7/4`
`=>x= -7/4-4/5`
`=>x=-51/20`
__
`4/25 -x=-5/2 +(-3/10)`
`=> 4/25 -x= -25/10 +(-3/10)`
`=> 4/25 -x= -28/10`
`=>x= 4/25 -(14/5)`
`=>x= 4/25 + 14/5`
`=>x=74/25`
a) 5.2² + (x + 3) = 5²
5.4 + x + 3 = 25
20 + x + 3 = 25
x + 23 = 25
x = 25 - 23
x = 2
b) 2³ + (x - 3²) = 5³ - 4³
8 + (x - 9) = 125 - 64
8 + x - 9 = 61
x - 1 = 61
x = 61 + 1
x = 62
c) 4.(x - 5) - 2³ = 2⁴.3
4x - 20 - 8 = 16.3
4x - 28 = 48
4x = 48 + 28
4x = 76
x = 76 : 4
x = 19
d) 5.(x + 7) - 10 = 2³.5
5x + 35 - 10 = 8.5
5x + 25 = 40
5x = 40 - 25
5x = 15
x = 15 : 5
x = 3
e) 7² - 7.(13 - x) = 14
49 - 91 + 7x = 14
7x - 42 = 14
7x = 14 + 42
7x = 56
x = 56 : 7
x = 8
a) \(5\cdot2^2+\left(x+3\right)=5^2\)
\(\Rightarrow x+3=5^2-5\cdot2^2\)
\(\Rightarrow x+3=25-5\cdot4\)
\(\Rightarrow x+3=5\)
\(\Rightarrow x=5-3\)
\(\Rightarrow x=2\)
b) \(2^3+\left(x-3^2\right)=5^3-4^3\)
\(\Rightarrow8+\left(x-9\right)=125-64\)
\(\Rightarrow8+x-9=61\)
\(\Rightarrow x-1=61\)
\(\Rightarrow x=61+1\)
\(\Rightarrow x=62\)
c) \(4\left(x-5\right)-2^3=2^4\cdot3\)
\(\Rightarrow4\left(x-5\right)=2^4\cdot3+2^3\)
\(\Rightarrow4\cdot\left(x-5\right)=16\cdot3+8\)
\(\Rightarrow4\cdot\left(x-5\right)=56\)
\(\Rightarrow x-5=56:4\)
\(\Rightarrow x-5=14\)
\(\Rightarrow x=19\)
d) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)=8\cdot5+10\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow5\left(x+7\right)=50\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
e) \(7^2-7\left(13-x\right)=14\)
\(\Rightarrow7\left(13-x\right)=7^2-14\)
\(\Rightarrow7\left(13-x\right)=49-14\)
\(\Rightarrow7\left(13-x\right)=35\)
\(\Rightarrow13-x=5\)
\(\Rightarrow x=13-5\)
\(\Rightarrow x=8\)
f) \(5x-5^2=10\)
\(\Rightarrow5x=10+5^2\)
\(\Rightarrow5x=10+25\)
\(\Rightarrow5x=35\)
\(\Rightarrow x=\dfrac{35}{5}\)
\(\Rightarrow x=7\)
g) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x=3^4+2\cdot3^2\)
\(\Rightarrow9x=81+2\cdot9\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
h) \(10x+2^2\cdot5=10^2\)
\(\Rightarrow10x=10^2-2^2\cdot5\)
\(\Rightarrow10x=100-4\cdot5\)
\(\Rightarrow10x=80\)
\(\Rightarrow x=\dfrac{80}{10}\)
\(\Rightarrow x=8\)
i) \(125-5\left(4+x\right)=15\)
\(\Rightarrow5\left(4+x\right)=125-5\)
\(\Rightarrow5\left(4+x\right)=120\)
\(\Rightarrow4+x=\dfrac{120}{5}\)
\(\Rightarrow4+x=24\)
\(\Rightarrow x=24-4\)
\(\Rightarrow x=20\)
j) \(2^6+\left(5+x\right)=3^4\)
\(\Rightarrow5+x=3^4-2^6\)
\(\Rightarrow5+x=81-64\)
\(\Rightarrow5+x=17\)
\(\Rightarrow x=17-5\)
\(\Rightarrow x=12\)
a) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{5}{7}\times1\)
\(=\dfrac{5}{7}\)
b) \(\dfrac{1}{10}+\dfrac{5}{9}+\dfrac{4}{9}+\dfrac{9}{10}-1\)
\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{1}{10}+\dfrac{9}{10}-1\right)\)
\(=1+0\)
\(=1\)
c) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}+\dfrac{2}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{2}{7}\)
\(=\dfrac{5}{7}+\dfrac{2}{7}\)
\(=1\)
d) \(\dfrac{2}{7}+\dfrac{2}{8}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{4}{7}\)
\(=\left(\dfrac{2}{8}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{4}{7}\right)\)
\(=\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+1\)
\(=\dfrac{1}{2}+1\)
\(=\dfrac{3}{2}\)
e) \(\dfrac{4}{5}+\dfrac{3}{10}+\dfrac{2}{10}+0,7\)
\(=\dfrac{4}{5}+\dfrac{5}{10}+\dfrac{7}{10}\)
\(=\dfrac{4}{5}+\dfrac{12}{10}\)
\(=\dfrac{4}{5}+\dfrac{6}{5}\)
\(=\dfrac{10}{5}\)
\(=2\)
g) \(362\times728+326\times272\)
\(=326\times\left(728+272\right)\)
\(=326\times1000\)
\(=326000\)
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a: \(\dfrac{x-1}{x^2-x+1}-\dfrac{x+1}{x^2+x+1}=\dfrac{10}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)-x\left(x+1\right)\left(x^2-x+1\right)=10\)
\(\Leftrightarrow x\left(x^3-1\right)-x\left(x^3+1\right)=10\)
=>-2x=10
hay x=-5
d: \(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+7\right)\left(x+8\right)}=\dfrac{1}{14}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+8}=\dfrac{1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(x+8\right)=14\left(x+8\right)-14\left(x+1\right)\)
\(\Leftrightarrow x^2+9x+8=14x+112-14x-14=98\)
\(\Leftrightarrow x^2+9x-90=0\)
\(\Leftrightarrow x\in\left\{6;-15\right\}\)
1; 5.22 + (\(x\) + 3) = 52
5.4 + (\(x\) + 3) = 25
20 + (\(x\) + 3) = 25
\(x\) + 3 = 25 - 20
\(x+3\) = 5
\(x\) = 5 - 3
\(x\) = 2
Vậy \(x=2\)
2; 23 + (\(x\) - 32) = 53 - 43
8 + (\(x\) - 9) = 125 - 64
8 + (\(x\) - 9) = 61
\(x\) - 9 = 61 - 8
\(x\) - 9 = 53
\(x\) = 53 + 9
\(x\) = 62
Vậy \(x\) = 62
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
<=> x = 15
bằng 20
20 ok okkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk