Help me!Ai giải chi tiết mk tick cho 3 lần
X nhân X nhân X = 64
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(x-1/2)=19/4:5/3
(x-1/2)=19/4x3/5
x-1/2=57/20
x=57/20+1/2
x=67/20
tk cho mk nha
\(\frac{19}{4}:\left(x-\frac{1}{2}\right)=\frac{5}{3}\)
\(x-\frac{1}{2}=\frac{19}{4}:\frac{5}{3}=\frac{57}{20}\)
\(x=\frac{57}{20}+\frac{1}{2}\)
\(x=\frac{67}{20}\)
\(-3^2-5.\left(x-1\right)=4x+7\)
\(\Rightarrow-9-5x+5=4x+7\)
\(\Rightarrow-9+5-7=5x+4x\)
\(\Rightarrow9x=-11\)
\(\Rightarrow x=\frac{-11}{9}\)
Vậy....
\(!x+\frac{1}{x}!\ge2\Rightarrow!a!\ge2\\ \)
Với IaI>=2
ta có: \(\left(x+\frac{1}{x}\right)^2=a^2\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)
Đặt \(x^2+x+1=t\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t^2+t+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(t-3\right)\left(t+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
= \(\left(x^2+x+1\right)\left[\left(x^2+x+1\right)+1\right]-12\)
= \(\left(x^2+x+1\right)^2\left(x^2+x+1\right)-12\)
= \(\left(x^2+x+1\right)\left(x^2+x+1\right)-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-4.3\)
= \(\left(x^2+x+1\right)\left(x^2+x-2\right)+4\left(x^2+x-2\right)\)
= \(\left(x^2+x+5\right)\left(x^2+x-2\right)\)
a) ta có : \(x^3-64=x^3-4^3=\left(x-4\right)\left(x^2+4x+16\right)\)
b) ta có : \(125x^3+y^6=\left(5x\right)^3+\left(y^2\right)^3=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
c) ta có : \(125\left(x+1\right)^3-1=\left(5x+5\right)^3-1\)
\(=\left(5x+5-1\right)\left(\left(5x+5\right)^2+\left(5x+5\right)+1\right)\)
\(=\left(5x+4\right)\left(25x^2+55x+31\right)\)
d) ta có : \(2x\left(x+1\right)+2\left(x+1\right)=\left(2x+2\right)\left(x+1\right)=2\left(x+1\right)\left(x+1\right)\)
X x X x X =64
\(X^3=64\)
Vậy X = 4
X=4nha bạn