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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 :
\(\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
= \(\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
a) 3x+2(x-5)=-x+2
<=> 3x+2x+x=2+10
<=>6x=12
<=>x=2
b) 3x2-2x=0
<=>x(3x-2)=0
<=>\(\left[{}\begin{matrix}x=0\\3x-2=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
c) \(\dfrac{2x}{3}\)+\(\dfrac{x-4}{6}\)=2-\(\dfrac{x}{2}\)
<=>\(\dfrac{8x+2x-8}{12}\)=\(\dfrac{24-6x}{12}\)
<=> 8x+2x-8=24-6x
<=>8x+2x+6x=24+8
<=>16x=32
<=>x=2
d) \(\dfrac{x-2}{x+2}\)-\(\dfrac{3}{x-2}\)= -\(\dfrac{2\left(x-11\right)}{4-x^2}\) ( ĐKXĐ: x\(\ne\)\(\pm\)2)
<=> \(\dfrac{\left(x-2\right)^2-3\left(x+2\right)}{x^2-4}\)=\(\dfrac{2\left(x-11\right)}{x^2-4}\)
=> (x-2)2-3(x+2)=2(x-11)
<=> x2-4x+4-3x-6=2x-22
<=> x2-4x-3x-2x=-22-4+6
<=> x-9x+20=0
<=> (x-4)(x-5)=0
<=>\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\) ( thỏa mãn diều kiện )
d) (x2+1)(x2-4x+4)=0
=> x2-4x+4=0 (x2+1\(\ge\)1 với mọi x)
=>(x-2)2 =0
=>x=2
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
\(A=\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+9\right)}+\dfrac{1}{\left(x+9\right)\left(x+11\right)}\)\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}+\dfrac{1}{x+9}-\dfrac{1}{x+11}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+11}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{x+11}{\left(x+1\right)\left(x+11\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+11\right)}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{x+11-x-1}{\left(x+1\right)\left(x+11\right)}\right)=\dfrac{1}{2}.\dfrac{10}{\left(x+1\right)\left(x+11\right)}=\dfrac{10}{2\left(x+1\right)\left(x+11\right)}\)
\(k\left(x\right)=\dfrac{5x^2-22x+25}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5x^2-20x+20-x+2-x+2+1}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{\left(5x^2-20x+20\right)-\left(x-2\right)-\left(x-2\right)+1}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x^2-4x+4\right)-\left(x-2\right)-\left(x-2\right)+1}{x^2-4x+4}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x-2\right)^2-\left(x-2\right)-\left(x-2\right)+1}{\left(x-2\right)^2}\)
\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x-2\right)^2}{\left(x-2\right)^2}-\dfrac{x-2}{\left(x-2\right)^2}-\dfrac{x-2}{\left(x-2\right)^2}+\dfrac{1}{\left(x-2\right)^2}\)
\(\Leftrightarrow k\left(x\right)=5-\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{\left(x-2\right)^2}\)
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 \(y+\dfrac{1}{2}=0\Leftrightarrow y=\dfrac{-1}{2}\)
\(\Leftrightarrow\dfrac{1}{x-1}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=-1\)
a: \(A=2x-3-5x+2-3x+1=-6x=-6\cdot\dfrac{-2}{3}=4\)
b: \(B=x^{2n-2n+3}=x^3=\left(-3\right)^3=-27\)
c)(x2+x)2-2(x2+x)-15
đặt x2+x=a ta có
a2-2a-15
=a2+3a-5a-15
=(a2+3a)-(5a+15)
=a(a+3)-5(a+3)
=(a+3)(a-5)
thay a=x2+x
(x2+x+3)(x2+x-5)