x^3+27=9-x^2
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a: \(27^{2-x}< =9\)
=>\(\left(3^3\right)^{2-x}< =3^2\)
=>\(3^{6-3x}< =3^2\)
=>6-3x<=2
=>-3x<=-4
=>\(x>=\dfrac{4}{3}\)
b: \(7^{3-x}< 49\)
=>\(7^{3-x}< 7^2\)
=>3-x<2
=>-x<2-3=-1
=>x>1
c: \(27^{3-x}>9\)
=>\(\left(3^3\right)^{3-x}>3^2\)
=>\(3^{9-3x}>3^2\)
=>9-3x>2
=>-3x>-7
=>\(x< \dfrac{7}{3}\)
d: \(2^{3-x}< 2^3\)
=>3-x<3
=>-x<0
=>x>0
e: \(27^{3-x^2}< 27^{x+1}\)
=>\(3-x^2< x+1\)
=>\(-x^2-x+2< 0\)
=>\(x^2+x-2>0\)
=>(x+2)(x-1)>0
=>\(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)
1/
\(3\left(-1-4x^2+5x\right)+4\left(3x^2+7x-6\right)=-27\)
\(\Leftrightarrow-3-12x^2+15x+12x^2+28x-24=-27\)
\(\Leftrightarrow43x=0\Rightarrow x=0\)
2/
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-1\right)=27\)
\(\Leftrightarrow x^3+27-x^3+x=27\)
\(\Leftrightarrow x=0\)
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9-x.x^2+x.9=27\\ x^3-3x^2+9x-x^3+9x=27\\ 3x^2+18x=27\\ 21x^2=27\\ x^2=\dfrac{9}{7}\\ \Rightarrow x=\sqrt{\dfrac{9}{7}}\)
\(\frac{2}{x-3}\)-\(\frac{27}{x^3-27}\)=\(\frac{3}{x^2+3x+9}\)
\(\frac{2}{x-3}\)-\(\frac{27}{\left(x-3\right)\left(x^2+3x+9\right)}\)=\(\frac{3}{x^2+3x+9}\)
\(\frac{2\left(x^2+3x+9\right)}{\left(x-3\right)\left(x^2+3x+9\right)}\)-\(\frac{27}{\left(x-3\right)\left(x^2+3x+9\right)}\)=\(\frac{3\left(x-3\right)}{\left(x-3\right)\left(x^2+3x+9\right)}\)
2x2+6x+18-27=3x-9 2x2+6x-3x=27-18-9 2x2+3x=0 x(2x+3)=0 x=0 hoặc 2x+3=0 x=0 hoặc x=\(\frac{-3}{2}\)Hướng dẫn giải:
a) 5 x (10 + 9)
Cách 1:
5 x (10 + 9) = 5 x 19 = 95
Cách 2:
5 x (10 + 9) = 5 x 10 + 5 x 9 = 50 + 45 = 95
b) 10 x (30 + 5)
Cách 1:
10 x (30 + 5) = 10 x 35 = 350
Cách 2:
10 x (30 + 5) = 10 x 30 + 10 x 5 = 300 + 50 = 350
Hướng dẫn giải nè :
a) 5 x (10 + 9)
Cách 1:
5 x (10 + 9) = 5 x 19 = 95
Cách 2:
5 x (10 + 9) = 5 x 10 + 5 x 9 = 50 + 45 = 95
b) 10 x (30 + 5)
Cách 1:
10 x (30 + 5) = 10 x 35 = 350
Cách 2:
10 x (30 + 5) = 10 x 30 + 10 x 5 = 300 + 50 = 350
2/3 x 4/5 + 4/5 x 8/3 = 4/5 x (2/3 + 8/3) = 4/5 x 10/3 = 8/3
27/32 x 16/9 -27/32 x7/9 + 27/32
= 27/32 x (16/9 - 7/9 + 1 )
=27/32 x 2
=27/16
\(\frac{2}{3}\) x \(\frac{4}{5}\) + \(\frac{4}{5}\) x \(\frac{8}{3}\)
=\(\frac{4}{5}\) x ( \(\frac{2}{3}\) + \(\frac{8}{3}\) )
= \(\frac{4}{5}\) x \(\frac{10}{3}\)
= \(\frac{40}{15}\) = \(\frac{8}{5}\)
x^3 +3^3=3^2 -x^2
\(X^3+27=9-x^2\)
\((X+3)(x^2-3x+9)- (9-x^2)=0\)
\((X+3)(x^2-3x+9)-(x+3)(3-x)=0\)
\((X+3)(x^2-3x+9-3+x)=0\)
\((X+3)(x^2-2x+6)=0\)
th1 x+3=0
X=-3
Th2 \(X^2-2x+6=0\)
\((X^2-2x+4)+2=0\)
\((X-2)^2+2=0\)
ta có \((X-2)^2> hoặc = 0\)
Mà 2>0. Suy ra \((X-2)^2+2>0 ( vô nghiệm)\)
suy ra x=-3