tìm x
(x-1)3 =27
giúp mình với! help me
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\(x^3+x^2+\dfrac{1}{3}x+\dfrac{1}{27}\)
\(=x^3+3\cdot\dfrac{1}{3}\cdot x^2+3\cdot\left(\dfrac{1}{3}\right)^2\cdot x+\left(\dfrac{1}{3}\right)^3\)
\(=\left(x+\dfrac{1}{3}\right)^3\)
\(54-9\left(x+2\right)=27\\ \Rightarrow9\left(x+2\right)=27\\ \Rightarrow x+2=3\\ \Rightarrow x=1\)
a) Vì |1/3x+4|≥0
=>-2|1/3x+4|≤0
=>-2|1/3x+4|+5/3≤5/3
Dấu "=" xảy ra <=>1/3x+4=0
<=>x=-12
Vậy Amin=5/3 <=>x=-12
b) Vì 3>0
=>(3x+1)²-2 đạt gtnn
Vì (3x+1)²≥0
=>(3x+1)²-2≥-2
Dấu "=" xảy ra <=>3x+1=0
<=>3x=-1
<=>x=-1/3
a: \(\dfrac{x}{6}=\dfrac{8}{3}\)
=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)
b: \(\dfrac{5}{x}=\dfrac{4}{9}\)
=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)
c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)
=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)
=>x=-1-3=-4
d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)
=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)
=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)
=>\(x=-\dfrac{69}{8}\)
f: ĐKXĐ: x<>1
\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)
=>\(\left(x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)
\(\frac{3x^2+6x+3-2x^2-5x-2}{x^2+2x+1}=3-\frac{2\left(x^2+\frac{2.5}{4}x+\frac{25}{16}+\frac{7}{16}\right)}{\left(x+1\right)^2}=3-\frac{2\left(x+\frac{5}{4}\right)^2+\frac{7}{8}}{\left(x+1\right)^2}\)
lập luận giải nốt nha
(x-1)^3=27
(x-1)^3=3^3
=> x-1=3
x=3+1
x=4
\(\left(x-1\right)^3=27=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)