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\(\left(x-1\right)^2+\left|x^2-1\right|=0.\)
Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left|x^2-1\right|\ge0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-1=0\\x^2-1=0\end{cases}\Rightarrow x=1}\)
\(-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-3\right)^2\)
\(=-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-\left(4x^2-12x+9\right)\)
\(=-3x^3-12x^2-12x+x^3-x+3x^2-3-4x^2+12x-9\)
\(=-2x^3-13x^2-x-12\)
a, y=a/x
=>15=a/4
=>a=15.4
=>a=60
b,khi =-5
y=a/x
=>y=60/-5
=>y=-12
lik e nhe nhug bai duoi nua
a,y=a/x
=> 15=a/4
=>a=60
b,khi x=-5
=>y=60/-5
=>y=-12
lik e nhe
\(3-\frac{x}{5}-x=\frac{x}{x-1}\)
\(\Rightarrow\frac{15\left(x-1\right)}{5\left(x-1\right)}-\frac{x\left(x-1\right)}{5\left(x-1\right)}-\frac{5x\left(x-1\right)}{5\left(x-1\right)}=\frac{5x}{5\left(x-1\right)}\)
\(\Rightarrow15\left(x+1\right)-x\left(x-1\right)-5x\left(x-1\right)=5x\)
\(\Rightarrow15x+15-x^2+x-5x^2+5x=5x\)
Bạn tự làm tiếp theo ha
\(\frac{3-x}{5-x}=\frac{x}{x+1}\)
\(\left(3-x\right)\left(x+1\right)=\left(5-x\right)x\)
\(3\left(x+1\right)-x\left(x+1\right)=5x-x^2\)
\(3x+3-x^2-x=5x-x^2\)
\(2x+3-x^2=5x-x^2\)
\(2x+3=5x\)
\(3=5x-2x\)
\(3x=3\)
\(x=1\)
Vậy x = 1
/ x - 2 / >= x - 2
/ x - 5 / = / 5 - x / >= 5 - x
=> / x - 5 / + / x - 2 / >= 5 - x + x - 2
/ x - 5 / + / x - 2 / >= 3
Dấu = xảy ra khi :
x - 2 >= 0 => x >= 2
5 - x >= 0 => x <= 3
Vậy x = 2 ; 3