Tìm x, biết:
x - 152 : 2 = 46
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\(x-152:x=46\)
\(x^2\) - 152\(\) - 46\(x\) = 0
\(x^2\) - 23\(x\) - 23\(x\) + 529 - 681 = 0
\(x\)(\(x-23\)) - 23(\(x\) - 23) = 681
(\(x-23\))(\(x-23\)) = 681
\(\left[{}\begin{matrix}x-23=-\sqrt{681}\\x+23=\sqrt{681}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=23-\sqrt{681}\\23+\sqrt{681}\end{matrix}\right.\)
\(2x+3=15\\ \Rightarrow2x=15-3=12\\ \Rightarrow x=\dfrac{12}{2}=6\\ ---\\ 28-3x=13\\ \Rightarrow3x=28-13=15\\ \Rightarrow x=\dfrac{15}{3}=5\\ ---\\ \left(x-1954\right).5=50\\ \Rightarrow x-1954=50:5=10\\ \Rightarrow x=10+1954=1964\\ ---\\ x-152:2=46\\ \Rightarrow x-76=46\\ \Rightarrow x=46+76=122\)
a, 2\(x\) + 3 = 15
2\(x\) = 12
\(x\) = 6
b, 28 - 3\(x\) = 13
3\(x\) = 28 - 13
3\(x\) = 15
\(x\) = 5
`x-(5/6 -x) =x-2/3`
`x-5/6 +x -x+2/3 =0`
`x = 5/6-2/3 = 5/6 -4/6 = 1/6`
`x+(x+1)+(x+2)+...+(x+30)=1240`
`=> (x + x + x + ... + x) + (1 + 2 + 3 +... + 30) = 1240`
`=> 31x + 465 = 1240`
`=> 31 x = 1240 - 465`
`⇒ 31x = 775`
`⇒ x = 775 : 31`
`⇒ x = 25`
x³ - x² - x = 1/3
<=> x³ = x² + x + 1/3
<=> 3x³ = 3(x² + x + 1/3)
<=> 3x³ = 3x² + 3x + 1
<=> 3x³ + x³ = x³ + 3x² + 3x + 1
<=> 4x³ = (x + 1)³
<=> ³√(4x³) = ³√(x + 1)³
<=> ³√4.x = x + 1
<=> ³√4.x - x = 1
<=> x(³√4 - 1) = 1
<=> x = 1/(³√4 - 1)
Ta có \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow x^2-x^2+2x=-4+3\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
x - 152 : 2 = 46
x - 76 = 46
x = 46 + 76
x = 122
x - 152 : 2 = 46
x - 76 = 46
x = 76 + 46
x = 122