Tìm x biết:
a) x + (x + 1) + (x + 2) + .......+ (x + 50) = 2040
b) 1 + 2 + 3 + .... + x = 300
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a: =>x-3=9
=>x=12
b: =>10-x=-26
=>x=36
c: =>x:4-1=2
=>x:4=3
=>x=12
d: =>x^2=4
=>x=2 hoặc x=-2
e: =>(x-2)^2=100
=>x-2=10 hoặc x-2=-10
=>x=12 hoặc x=-8
a: \(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a, \(\sqrt{\left(2x-3\right)^2}=7\\ \Rightarrow\left|2x-3\right|=7\\ \Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c, \(\sqrt{x^2-9}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\sqrt{x+3}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
Giải :
a) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 50 ) = 2040
=> 51x + ( 1 + 2 + 3 + ... + 50 ) = 2040
=> 51x + ( 50 + 1 ) . 50 : 2 = 2040
=> 51x + 1275 = 2040
=> 51x = 765
=> x = 15
b) 1 + 2 + 3 + ... + x = 300
=> ( x + 1 ) . x : 2 = 300
=> ( x + 1 ) . x = 600
=> ( x + 1 ) . x = 25 . 24
=> x = 24
a) x3-1-(x2+2x)(x-2)=5
⇔ x3-1-x3+4x=5
⇔ 4x=6
⇔ \(x=\dfrac{3}{2}\)
(2-x)^3+(2+x)^3-12x(x+1)=0
=>\(8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x\left(x+1\right)=0\)
=>\(12x^2+16-12x^2-12x=0\)
=>16-12x=0
=>4-3x=0
=>x=4/3
a: \(x\cdot\dfrac{2}{5}+\dfrac{1}{2}\cdot x=9\)
=>\(x\left(\dfrac{2}{5}+\dfrac{1}{2}\right)=9\)
=>\(x\cdot\dfrac{9}{10}=9\)
=>\(x=9:\dfrac{9}{10}=10\)
b: \(\dfrac{1}{9}:x+\dfrac{3}{9}:x=\dfrac{5}{7}\)
=>\(\left(\dfrac{1}{9}+\dfrac{3}{9}\right):x=\dfrac{5}{7}\)
=>\(\dfrac{4}{9}:x=\dfrac{5}{7}\)
=>\(x=\dfrac{4}{9}:\dfrac{5}{7}=\dfrac{4}{9}\cdot\dfrac{7}{5}=\dfrac{28}{45}\)
Giải :
a) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 50 ) = 2040
=> 51x + ( 1 + 2 + 3 + ... + 50 ) = 2040
=> 51x + ( 50 + 1 ) . 50 : 2 = 2040
=> 51x + 1275 = 2040
=> 51x = 765
=> x = 15
b) 1 + 2 + 3 + ... + x = 300
=> ( x + 1 ) . x : 2 = 300
=> ( x + 1 ) . x = 600
=> ( x + 1 ) . x = 25 . 24
=> x = 24