Tìm x, biết rằng:
a) 3.(10.x)=111
b)3+(10.x)=111
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a, \(3.\left(10.x\right)=111\Leftrightarrow\left(10.x\right)=111:3\)
\(\Leftrightarrow10.x=37\Leftrightarrow x=37:10\Leftrightarrow x=3,7\)
b,\(3.\left(10+x\right)=111\Leftrightarrow\left(10+x\right)=111:3\)
\(\Leftrightarrow10+x=37\Leftrightarrow x=37-10\Leftrightarrow x=27\)
c,\(3+\left(10.x\right)=111\Leftrightarrow\left(10.x\right)=111-3\)
\(\Leftrightarrow10.x=108\Leftrightarrow x=108:10\Leftrightarrow x=10,8\)
d,\(3+\left(10+x\right)=111\Leftrightarrow\left(10+x\right)=111-3\)
\(\Leftrightarrow10+x=108\Leftrightarrow x=108-10\Leftrightarrow x=98\)
3 + (10.x) = 111
10.x = 111- 3
10.x = 108
x = 108:10
x = 10,8
3 + (10 + x ) = 111
10 + x = 111- 3
10 + x = 108
x = 108 – 10
x = 98
3.(10 + x) = 111
10+ x = 111: 3
10 + x = 37
x = 37 – 10
x = 27
a) \(3.\left(10.x\right)=111\)
\(10.x=37\)
\(x=\dfrac{37}{10}\)
b) \(3.\left(10+x\right)=111\)
\(10+x=37\)
\(x=27\)
c) \(3+\left(10.x\right)=111\)
\(10.x=108\)
\(x=\dfrac{54}{5}\)
d) \(3+\left(10+x\right)=111\)
\(x=111-3-10\)
\(x=98\)
a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)
\(\Leftrightarrow6x=-3\)
hay \(x=-\dfrac{1}{2}\)
b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)
\(\Leftrightarrow2x^3+6x=2x^3+24x\)
\(\Leftrightarrow x=0\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)
\(\Leftrightarrow12x=-11\)
hay \(x=-\dfrac{11}{12}\)
a)\(3.\left(10:x\right)=111\)
\(\Rightarrow10:x=37\)
\(x=10:37\)
\(x=\frac{10}{37}\)
b)\(3.\left(10+x\right)=111\)
\(\Rightarrow10+x=37\)
\(x=37-10\)
\(x=27\)
c)\(3+\left(10.x\right)=111\)
\(\Rightarrow10x=108\)
\(x=108:10\)
\(x=\frac{54}{5}\)
d)\(3+\left(10+x\right)=111\)
\(\Rightarrow10+x=108\)
\(x=108-10\)
\(x=98\)
a)
3 . ( 10 . x ) = 111
3 . 10 . x = 111
30 . x = 111
x = \(\frac{37}{10}\)
b) 3 + ( 10 . x ) = 111
10 . x = 111 - 3
10 . x = 108
x = 108 : 10
x = 10.8
3. ( 10.x ) = 111
10.x = 111 : 3
10.x = 37
x = 37 : 10
x = 3,7
______________________
3 + ( 10.x ) = 111
10.x = 111 - 3
10.x = 108
x = 108 : 10
x = 10,8