HOC24
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\(2x^2-72=0\)
\(2x^2=72\)
\(\Rightarrow x^2=72:2=36\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
Sửa đề x + 113
Đặt x + 56 = a2
x + 113 = b2 ( a;b thuộc N ) -
=> b2 - a2 = 113 - 56 = 57
=> ( b - a ).( b + a ) = 57 = 57 . 1 = 1 . 57 = 17 . 3 = 3.17
rồi bạn lắp vào x, y và giải ra
\(P=x^2+y^2\)
Có \(x^2\ge0;y^2\ge0\)
\(\Rightarrow P\ge0+0=0\)
Dấu "=" xảy ra khi x = 0 ; y = 0
Vậy Min P = 0 <=> x = 0 ; y = 0
\(\left(2x-3\right)^2-\left(2x+1\right)\left(x-1\right)+3\left(2x-3\right)\)
\(=\left(2x-3\right)\left(2x-3+3\right)+\left(2x+1\right)\left(x-1\right)\)
\(=\left(2x-3\right).2x+2x^2-x-1\)
\(=4x^2-6x+2x^2-x-1\)
\(=6x^2-7x-1\)
1) \(n^3+11n=n^3-n+12n=n\left(n^2-1\right)+12n=\left(n-1\right)n\left(n+1\right)+12n\)
Có \(\left(n-1\right)n\left(n+1\right)⋮6;12n⋮6\)
\(\Rightarrow n^3+11n⋮6\)
2)\(n^3-19n=n^3-n-18n=\left(n-1\right)n\left(n+1\right)-18n\)
\(Có\left(n-1\right)n\left(n+1\right)⋮6;18n⋮6\)
\(\Rightarrow n^3-19n⋮6\)
\(x^{16}+x^8+1\)
\(=\left(x^8\right)^2+2.x^8.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(=\left(x^8+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(A=\dfrac{x^2+3x+5}{x-1}=\dfrac{x^2-x+4x-4+9}{x-1}\)
Để \(A\in Z\)
\(\Rightarrow\dfrac{9}{x-1}\in Z\Rightarrow x-1\in\left(1;-1;3;-3;9;-9\right)\)
\(\Rightarrow x\in\left(2;0;4;-2;10;-8\right)\)
\(ĐKXĐ:1-3x\ne0\Leftrightarrow x\ne\dfrac{1}{3};3x+1\ne0\Leftrightarrow x\ne-\dfrac{1}{3}\)
\(A=\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x}\)
\(A=\left(\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(3x+1\right)}\right):\left(\dfrac{6x^2+10x}{1+3x}\right)\)
\(A=\dfrac{3x^2+5x}{3x+1-9x^2-3x}.\dfrac{1+3x}{6x^2+10}\)
\(A=\dfrac{\left(3x^2+5x\right).\left(1+3x\right)}{\left(1-9x^2\right).2.\left(3x^2+5\right)}\)
\(A=\dfrac{1+3x}{\left(1+3x\right)\left(1-3x\right)}=\dfrac{1}{\left(1-3x\right).2}=\dfrac{1}{2-6x}\)
\(a,\left(-64\right)+\left(x-3\right)^3\)
\(=\left(-4\right)^3+\left(x-3\right)^3\)
\(=\left(-4+x-3\right)\left[\left(-4\right)^2-\left(-4\right)\left(x-3\right)+\left(x-3\right)^2\right]\)
\(a,\left(-3x+2\right)^3=\left(2-3x\right)^3=2^3-3.2^2.3x+3.2.9x^2-27x^3=8-36x+54x^2-27x^3\)