tim x
6x +4x =2010
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\(J=\frac{2010}{4x+20\sqrt{x}+30}\)
\(=\frac{2010}{\left(2\sqrt{x}\right)^2+2.2\sqrt{x}.5+25+5}\)
\(=\frac{2010}{\left(2\sqrt{x}+5\right)^2+5}\)
\(A_{max}\Leftrightarrow\frac{2010}{\left(2\sqrt{x}+5\right)^2+5}\)lớn nhất
\(\Rightarrow\left(2\sqrt{x}+5\right)^2+5\)nhỏ nhất
\(\Rightarrow\left(2\sqrt{x}+5\right)^2\)nhỏ nhất
Mà \(2\sqrt{x}+5\ge5\Rightarrow2\sqrt{x}+5=5\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\)
Với x = 0 \(\Rightarrow J_{max}=\frac{2010}{4.0+20\sqrt{0}+30}=\frac{2010}{30}=67\)
Giải:
a) \(6x+4x=2010\)
\(\Leftrightarrow10x=2010\)
\(\Leftrightarrow x=\dfrac{2010}{10}\)
\(\Leftrightarrow x=201\)
b) \(1+2+...+x=45\)
\(\Leftrightarrow\dfrac{\left(x+1\right).x}{2}=45\)
\(\Leftrightarrow\left(x+1\right).x=90\)
Mà \(90=9.10\)
\(\Leftrightarrow x=9\)
c) \(1+3+5+...+x=36\)
\(\Leftrightarrow\dfrac{\left[\left(x-1\right):2+1\right].\left(x+1\right)}{2}=36\)
\(\Leftrightarrow\left[\left(x-1\right):2+1\right].\left(x+1\right)=72\)
\(\Leftrightarrow\left[\left(x-1\right).\dfrac{1}{2}+1\right].\left(x+1\right)=72\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-\dfrac{1}{2}+1\right).\left(x+1\right)=72\)
Mà \(72=6.12\)
\(\Leftrightarrow x=11\)
Chúc bạn học tốt!
\(6x(x-2)=x-2\)
\(< =>6x\left(x-2\right)-x+2=0\\ < =>6x\left(x-2\right)-\left(x-2\right)=0\\ < =>\left(x-2\right)\left(6x-1\right)=0\)
<=> x-2=0 hoặc 6x-1=0
Nếu x-2=0 thì x=2
Nếu 6x-1=0 thì 6x=1 \(=>x=\dfrac{1}{6}\)
Vậy \(x=2,x=\dfrac{1}{6}\)
Học tốt!
\(p=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+2010\)\(=\left(x^3+1\right)+x-\left(x^3-1\right)+2010=x^3+1+x-x^3+1+2010=x+2012\)Với \(x=-2010\Rightarrow p=-2010+2012=2\)
\(q=16x\left(4x^2-5\right)-\left(4x+1\right)\left(16x^2-4x+1\right)=64x^3-80x-64x^3-1=-80x-1\)Với \(x=\dfrac{1}{5}\Rightarrow q=-80.\dfrac{1}{5}-1=-17\)
=>10x=2010
=>x=2010:10
x=201
\(6x+4x=2010\)
\(10x=2010\)
\(x=2010:10=201\)