Tìm x, biết: |4*x + 3| - x = 15
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a
\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)
b
\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)
c
\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)
a: =>(2x+15)(x^2+4)=0
=>2x+15=0
=>2x=-15
=>x=-15/2
b; =>(x-2)(5x-3)=0
=>x=2 hoặc x=3/5
c: =>(x+3)(2-x)=0
=>x=2 hoặc x=-3
a) x + 1/15 = 1/3 - 1/4 = 4/12 - 3/12 = 1/12.
=> x = 1/12 - 1/15 = 5/60 - 4/60 = 1/60.
b) x - 5/12 = 1/3 + 1/4 = 4/12 + 3/12 = 7/12.
=> x = 7/12 + 5/12 = 1
c) x x 2/7 = 3/4 x 4/15 = 1/5
=> x = 1/5 : 2/7 = 1/5 x 7/2 = 7/10
d) x : 3/8 = 4/3 : 3/6 = 4/3 x 2 = 8/3
=> x = 8/3 x 3/8 = 1
\(-4+x=-4+7\)
\(\Rightarrow x=-4+7+4\)
\(\Rightarrow x=7\)
\(\left(-3\right)-\left(-5-x\right)=-15+\left(-6+15\right)\)
\(\Rightarrow x=-15-6+15+3-5\)
\(\Rightarrow x=-8\)
\(-\left(x+2\right)+2=\left(8-15\right)+15\)
\(\Rightarrow-x+2+2=-7+15\)
\(\Rightarrow-x=-7+15-2-2\)
\(\Rightarrow x=-4\)
\(4+\left(x-4\right)=-\left(11-2\right)+\left(11+4\right)\)
\(\Rightarrow4+x-4=-11+2+11+4\)
\(\Rightarrow x=-11+2+11+4-4+4\)
\(\Rightarrow x=6\)
a) -4 + x = -4 + 7 b) (-3) - (-5-x) = -15 + (-6+15)
-4 + x + 4 - 7 = 0 (-3) + 5 + x = -6
x - 7 = 0 2 + 6 = -x
x= 7 8 = -x
Vậy x= 7 x= -8 =) Vậy x= -8
c) -(x+2)+2=(8-15)+15 d) 4+(x-4)= -(11-2) + (11+4 )
-x - 2 + 2 = 8 4+x-4 = 6
-x = 8 x=6
x= -8 Vậy x=6
Vậy x= -8
Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
a ) x ∈ ± 23 28 . b ) x ∈ − 23 6 ; 35 6 . c ) x ∈ − 17 60 ; 97 60 . d ) x ∈ 67 140 ; 143 140 .
|4x + 3| - x = 15
=> |4x + 3| = 15 + x
=> \(\orbr{\begin{cases}4x+3=x+15\\4x+3=-x-15\end{cases}}\)
=> \(\orbr{\begin{cases}3x=12\\5x=-12\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=-\frac{12}{5}\end{cases}}\)
\(x=-3,6\)