Xác định công thức hàm số:
a/ f(x) + 3f\(\left(\frac{1}{3}\right)\)=\(x^2\)
b/f(x) +2f \(\left(\frac{1}{x}\right)\)= 2x+\(\frac{1}{x}\)
c/ f(x) +3f (-x) =x+1
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a) \(xy+3x-7y-21\)
\(\Leftrightarrow\left(xy+3x\right)-\left(7y+21\right)\)
\(\Leftrightarrow x\left(y+3\right)-7\left(y+3\right)\)
\(\Leftrightarrow\left(x-7\right)\left(y+3\right)\)
b) \(2xy-15-6x+5y\)
\(\Leftrightarrow\left(2xy-6x\right)-\left(15-5y\right)\)
\(\Leftrightarrow x\left(2y-6\right)-5\left(3-y\right)\)
\(\Leftrightarrow2x\left(y-3\right)+5\left(y-3\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(y-3\right)\)
\(\left(3x-2\right)^2-\left(4-3x\right)^2=20\)
\(\Leftrightarrow\left(3x-2-4+3x\right)\left(3x-2+4-3x\right)=20\)
\(\Leftrightarrow\left(6x-6\right)\cdot2=20\)
\(\Leftrightarrow6x-6=10\)
\(\Leftrightarrow6\left(x-1\right)=10\)
\(\Leftrightarrow x-1=\frac{10}{6}\Leftrightarrow x=\frac{8}{3}\)
\(\left(3x-2\right)^2-\left(4-3x\right)^2=20\)
\(\Leftrightarrow\left(3x-2-4+3x\right)\left(3x-2+4-3x\right)=20\)
\(\Leftrightarrow\left(6x-6\right).2=20\)
\(\Leftrightarrow6x-6=10\)
\(\Leftrightarrow6\left(x-1\right)=10\)
\(\Leftrightarrow x-1=\frac{10}{6}\)
\(\Leftrightarrow x=\frac{8}{3}\)
\(x^2+4y^2+z^2-2x-6z+8y+15=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\)
thấy: \(\left(x-1\right)^2\ge0;\left(2y+2\right)^2\ge0;\left(z-3\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1>0\) (đpcm)
\(B=a^2\left(11-8a\right)+\left(2a-1\right)^3=11a^2-8a^3+\left(2a-1\right)^3=\left[\left(2a-1\right)^3-8a^3\right]+11a^2\)
\(=-12a^2+6a-1+11a^2=-a^2+6a-1=-\left(a^2-6a+9\right)+8=-\left(a-3\right)^2+8\)
Vậy giá trị lớn nhất của B là 8 tại a = 3
b )=x4-2x3-2x3+4x2+4x2-8x-8x+16
=x3(x-2)-2x2(x-2)+4x(x-2)-8(x-2)
=(x-2)(x3-2x2+4x-8)
=(x-2)[x2(x-2)+4(x-2)]
=(x-2)2(x2+4)
a) đề thiếu ko bn?
b) \(x^4-4x^3+8x^2-16x+16=\left(x^4-4x^2\right)-\left(4x^3-12x^2+8x\right)-\left(8x-16\right)\)
\(=x^2\left(x-2\right)\left(x+2\right)-4x\left(x^2-3x+2\right)-8\left(x-2\right)\)
\(=x^2\left(x-2\right)\left(x+2\right)-4x\left(x-2\right)\left(x-1\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left[x^2\left(x+2\right)-4x\left(x-1\right)-8\right]=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)
\(=\left(x-2\right)\left[\left(x^3-8\right)-\left(2x^2-4x\right)\right]=\left(x-2\right)\left[\left(x-2\right)\left(x^2+2x+4\right)-2x\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(x-2\right)\left(x^2+2x+4-2x\right)=\left(x-2\right)^2\left(x^2+4\right)\)