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\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
a)Ta có : \(\dfrac{x+1}{1-x}\)( giữ nguyên )
\(\dfrac{x^2-2}{1-x}\)( giữ nguyên )
\(\dfrac{2x^2-x}{x-1}=\dfrac{x-2x^2}{1-x}\)
b)Ta có : \(\dfrac{1}{x-1}=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x+1}{x^3-1}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x^2-2x}{x^3-1}\)
\(\dfrac{2x-3x^2}{x^3-1}\)(giữ nguyên )
c) MTC = ( x+ 2)2(x - 2)2
Do đó , ta có : \(\dfrac{1}{x^2+4x+4}=\dfrac{1}{\left(x+2\right)^2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}\)
\(\dfrac{x}{x^2-4}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x^2-2^2\right)}{\left(x+2\right)^2\left(x-2\right)^2}=\dfrac{x^3-4x}{\left(x+2\right)^2\left(x-2\right)^2}\)
d) MTC = xyz( x - y)( y - z)( x - z)
Do đó , ta có : \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}=\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{y\left(y-x\right)\left(y-z\right)}=\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
Cộng các phân thức lại ta có :
\(\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
= \(\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Rightarrow2x^2+3\left(x^2-1\right)=5x^2+5x\)
\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Rightarrow5x^2-3=5x^2+5x\)
\(\Rightarrow-3=5x\)
\(\Rightarrow5x=-3\)
\(\Rightarrow x=-\dfrac{3}{5}\)
Vậy ....
P/s : Làm bừa !
a,(5x-2y)(x2-xy+1)=5x3-5x2+5x-2yx2+2xy2-2y
=5x3-7x2y+2xy2+5x-2y
b,(x-2)(x+2)(\(\dfrac{1}{2}\) x-5)=x2-4.\(\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-2x+20\)
c,\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-1x^2+10x+\dfrac{3}{2}x-15\)
=\(\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
d,\(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)
=\(x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)
=\(-5x+4x-15\)
=\(-x-15\)
Chúc bạn học tốt(mỏi tay quá)
\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
a, Vì x2 ≥ 0 , 2y2 ≥ 0 với mọi x,y
=>x2+2y2+ 1 ≥ 1
=>Phân thức trên luôn có nghĩa
\(k\left(x\right)=\dfrac{5x^2-22x+25}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5x^2-20x+20-x+2-x+2+1}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{\left(5x^2-20x+20\right)-\left(x-2\right)-\left(x-2\right)+1}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x^2-4x+4\right)-\left(x-2\right)-\left(x-2\right)+1}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x-2\right)^2-\left(x-2\right)-\left(x-2\right)+1}{\left(x-2\right)^2}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x-2\right)^2}{\left(x-2\right)^2}-\dfrac{x-2}{\left(x-2\right)^2}-\dfrac{x-2}{\left(x-2\right)^2}+\dfrac{1}{\left(x-2\right)^2}\)
\(\Leftrightarrow k\left(x\right)=5-\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{\left(x-2\right)^2}\)
Đặt \(y=\dfrac{1}{x-2}\)
\(\Rightarrow k\left(x\right)=5-y-y+y^2=y^2-2y+1+4=\left(y-1\right)^2+4\ge4\)
Vậy GTNN của \(k\left(x\right)=4\) khi \(y=1\Rightarrow\dfrac{1}{x-2}=1\Leftrightarrow x=3\)
\(h\left(x\right)=\dfrac{x^2-x+1}{\left(x-1\right)^2}\)
\(\Leftrightarrow h\left(x\right)=\dfrac{x^2-2x+1+x-1+1}{\left(x-1\right)^2}\)
\(\Leftrightarrow h\left(x\right)=\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2}+\dfrac{x-1}{\left(x-1\right)^2}+\dfrac{1}{\left(x-1\right)^2}\)
\(\Leftrightarrow h\left(x\right)=1+\dfrac{1}{x-1}+\dfrac{1}{\left(x-1\right)^2}\)
Đặt \(y=\dfrac{1}{x-1}\)
\(\Rightarrow h\left(x\right)=1+y+y^2\)
\(\Rightarrow h\left(x\right)=y^2+y+1\)
\(\Rightarrow h\left(x\right)=y^2+2.y.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(\Rightarrow h\left(x\right)=\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
=> GTNN của \(h\left(x\right)=\dfrac{3}{4}\) khi \(y+\dfrac{1}{2}=0\Leftrightarrow y=\dfrac{-1}{2}\)
\(\Leftrightarrow\dfrac{1}{x-1}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=-1\)