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a) \(\left(2x^3y-0,5x^2\right)^3\)
\(=\left(2x^3y\right)^3-3\left(2x^3y\right)^20,5x^2+3.2x^3y\left(0,5x^2\right)^2-\left(0,5x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+1,5x^7y-0,125x^6\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
c) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)
\(=x^3-3^3\)
\(=x^3-27.\)
a,\(\left(2x^3y-0,5x^2\right)^3=\left(2x^3y\right)^3-3.\left(2x^3y\right)^2.\left(0,5x^2\right)+3.\left(0,5x^2\right)^2.\left(2x^3y\right)-\left(0,5x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+\frac{3}{2}x^7y-\frac{1}{8}x^6\)
b,\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3=x^3-27y^3\)
\(\left(x^2-3\right)\left(x^4+3x^2+9\right)=\left(x^2-3\right)\left[\left(x^2\right)^2+3.x^2+3^2\right]\)
\(=\left(x^2\right)^3-3^3=x^6-27\)
A= 4x2 - 3x + 1
= (2x) 2 - 2.2x.4/3 + (4/3) 2 - (4/3) 2 + 1
= (2x - 4/3) 2 - 7/9
Nhận xét: (2x - 4/3) 2 \(\ge\)0 với mọi x
=> (2x - 4/3) 2 - 7/9 \(\le\) 7/9
=> Min A là 9
Dấu "=" xảy ra <=> 2x - 4/3 = 0 <=> 2x = 4/3 <=> x = 2/3
Vậy..
a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2-2x\right)\left(3x-2+2x\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-2\right)\left(5x-2\right)=0\end{matrix}\right.\)
hay x=2
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{10}{3}\\\left(-3,5x-1,5x-5\right)\left(-3,5x+1,5x+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{10}{3}\\\left(-5x-5\right)\left(-2x+5\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-1;\dfrac{5}{2}\right\}\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(3x-1-x-15\right)\left(3x-1+x+15\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(2x-16\right)\left(4x+14\right)=0\end{matrix}\right.\Leftrightarrow x=8\)
d: \(\Leftrightarrow\left|x-2\right|=0,5x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=8\\\left(0,5x-4-x+2\right)\left(0,5x-4+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=8\\\left(-0,5x-2\right)\left(1,5x-6\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Đặt \(a=x^2+x-2\), ta có:
\(G=\left(a-4\right)\left(a+4\right)\)
\(=a^2-16\ge-16\)
Dấu = xảy ra khi a=0
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Rightarrow x=-2;1\)
Vậy...
B = (0,5x^2)^2 - 3.|0,5x^2+x| + 2,25 - 2,25
= ( |0,5x^2+x| - 1,5 ) ^2 - 2,25 >= -2,25
Dấu "=" xảy ra <=> |0,25x^2+x| = 1,5
<=> 0,5x^2+x = 1,5 hoặc 0,5x^2+x = -1,5
Đến đó bạn giải 2 pt đó để tìm x nha
Vậy Min của B = -2,25 <=> x=......