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\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
B3;a,ĐKXĐ:\(x\ne\pm4\)
A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)
Câu 1:
\(\text{a) }\dfrac{x^2-xy}{3xy-3y^2}=\dfrac{x\left(x-y\right)}{3y\left(x-y\right)}=\dfrac{x}{3y}\)
\(\text{b) }\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\\ =\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\\ =\dfrac{2a\left(x-1\right)^2}{5b\left(1-x\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)^2}{5b\left(x-1\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\\ =-\dfrac{2ax-2a}{5bx+5b}\)
\(\text{c) }\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
\(\text{d) }\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)
\(\text{e) }\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x+y\right)^3}\\ =\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\\ =\dfrac{x^3+y^3}{x^4-xy^3}\)
Câu 3:
\(\text{ a) }\dfrac{\left(a+b\right)^2-c^2}{a+b+c}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
\(\text{b) }\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\\ =\dfrac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}\\ =\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\\ =\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}\\ =\dfrac{a+b-c}{a-b+c}\)
\(\text{c) }\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\\ =\dfrac{2x^3-x^2-6x^2+3x-15x+45}{3x^3-10x^2-9x^2+3x+30x-9}\\ =\dfrac{\left(2x^3-x^2-15x\right)-\left(6x^2-3x-45\right)}{\left(3x^3-10x^2+3x\right)-\left(9x^2-30x+9\right)}\\ =\dfrac{x\left(2x^2-x-15\right)-3\left(2x^2-x-15\right)}{x\left(3x^2-10x+3\right)-3\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-6x+5x-15\right)}{\left(x-3\right)\left(3x^2-9x-x+3\right)}\\ =\dfrac{\left(x-3\right)\left[\left(2x^2-6x\right)+\left(5x-15\right)\right]}{\left(x-3\right)\left[\left(3x^2-9x\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left[x\left(x-3\right)+5\left(x-3\right)\right]}{\left(x-3\right)\left[3x\left(x-3\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left(x-3\right)\left(x+5\right)}{\left(x-3\right)\left(x-3\right)\left(3x-1\right)}\\ =\dfrac{x+5}{3x-1}\)
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
1, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1), (2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\) \(\xrightarrow[]{}\) đpcm
5. a, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)
\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)
\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1),(2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)
mà x+y+z=3
=>\(x^2+y^2+z^2+3\ge2.3=6\)
<=> \(x^2+y^2+z^2\ge6-3=3\)
<=> \(A\ge3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTNN của A=x2+y2+z2 là 3 khi x=y=z=1
b, Ta có: x+y+z=3
=> \(\left(x+y+z\right)^2=9\)
<=> \(x^2+y^2+z^2+2xy+2yz+2xz=9\)
<=> \(x^2+y^2+z^2=9-2xy-2yz-2xz\)
mà \(x^2+y^2+z^2\ge3\) (theo a)
=> \(9-2xy-2yz-2xz\ge3\)
<=> \(-2\left(xy+yz+xz\right)\ge3-9=-6\)
<=> \(xy+yz+xz\le\dfrac{-6}{-2}=3\)
<=> \(B\le3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTLN của B=xy+yz+xz là 3 khi x=y=z=1
Bài 1 rút gọn bc tự làm :
\(B=\dfrac{3y^3-7y^2+5y-1}{2y^3-y^2-4y+3}\)
\(B=\dfrac{3x^3-3y^2-4y^2+4y+y-1}{2y^3-2y^2+y^2-y+3y-3}\)
\(B=\dfrac{3y^2\left(y-1\right)-4y\left(y-1\right)+\left(y-1\right)}{2y^2\left(y-1\right)+y\left(y-1\right)-3\left(y-1\right)}\)
\(B=\dfrac{\left(3y^2-4y+1\right)\left(y-1\right)}{\left(2y^2+y-3\right)\left(y-1\right)}\)
\(B=\dfrac{3y^2-3y-y+1}{2y^2-2y+3y-3}=\dfrac{3y\left(y-1\right)-\left(y-1\right)}{2y\left(y-1\right)+3\left(y-1\right)}\)
\(B=\dfrac{\left(3y-1\right)\left(y-1\right)}{\left(3y+2\right)\left(y-1\right)}=\dfrac{3y-1}{3y+2}\)
Bài 2 )
a ) \(x+\dfrac{1}{x}=3\)
\(\Leftrightarrow x^2+2x\dfrac{1}{x}+\dfrac{1}{x^2}=9\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}=1\)
b ) \(\left(x+\dfrac{1}{x}\right)^3=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+\dfrac{3}{x}+3x=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3\left(\dfrac{1}{x}+x\right)=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}=18\)