Cho \(P=\left(\dfrac{x}{x+2}-\dfrac{x^3-8}{x^3+8}.\dfrac{x^2-2x+4}{x^2-4}\right):\dfrac{4}{x+2}\)
a ) Rút gọn P
b ) Tìm x để P<0
c ) Tìm x để \(P=\dfrac{1}{x}+1\)
d ) Tính P khi \(\left|2x-1\right|=3\)
e ) Tính giá trị nhỏ nhất của P
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a. \(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\) \(\left(ĐKXĐ:x\ge0\right)\)
\(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\)
\(\text{}\text{}N=\dfrac{\sqrt{x}+1}{x\sqrt{x}+1}.\dfrac{4\sqrt{x}}{3}\)
\(N=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b.\(N=\dfrac{8}{9}\Leftrightarrow\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)
\(\Leftrightarrow3\sqrt{x}=2x-2\sqrt{x}+2\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=4\end{matrix}\right.\)
c.\(\dfrac{1}{N}>\dfrac{3\sqrt{x}}{4}\Leftrightarrow\dfrac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}>\dfrac{3\sqrt{x}}{4}\)
\(\Leftrightarrow x-\sqrt{x}+1>x\)
\(\Leftrightarrow x< 1\)
a: ĐKXĐ: \(x\ge0\)
Ta có: \(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)
a: ĐKXĐ: x<>2; x<>0
b: \(M=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-2x^2-2x^2+4x}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
c: M>=-3
=>(x+1+6x)/2x>=0
=>(7x+1)/x>=0
=>x>0 hoặc x<=-1/7
Câu 1:
1: Ta có: \(P=\left(\dfrac{x^2}{x^2-3}+\dfrac{2x^2-24}{x^4-9}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\left(\dfrac{x^2\left(x^2+3\right)}{\left(x^2-3\right)\left(x^2+3\right)}+\dfrac{2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+3x^2+2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+5x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+8x^2-3x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^2\left(x^2+8\right)-3\left(x^2+8\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{\left(x^2+8\right)\left(x^2-3\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{7}{x^2+3}\)
Câu 2a đề sai, pt này ko giải được
2b.
\(P\left(x\right)=\left(2x+7\right)\left(x^2-4x+4\right)+\left(a+20\right)x+\left(b-28\right)\)
Do \(\left(2x+7\right)\left(x^2-4x+4\right)⋮\left(x^2-4x+4\right)\)
\(\Rightarrow P\left(x\right)\) chia hết \(Q\left(x\right)\) khi \(\left(a+20\right)x+\left(b-28\right)\) chia hết \(x^2-4x+4\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+20=0\\b-28=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-20\\b=28\end{matrix}\right.\)
3a.
\(VT=\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}=1+\dfrac{1-x^2y^2}{1+x^2+y^2+x^2y^2}\le1+\dfrac{1-x^2y^2}{1+2xy+x^2y^2}\)
\(VT\le1+\dfrac{\left(1-xy\right)\left(1+xy\right)}{\left(xy+1\right)^2}=1+\dfrac{1-xy}{1+xy}=\dfrac{2}{1+xy}\) (đpcm)
3b
Ta có: \(n^3-n=n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên luôn chia hết cho 6
\(\Rightarrow n^3\) luôn đồng dư với n khi chia 6
\(\Rightarrow S\equiv2021^{2022}\left(mod6\right)\)
Mà \(2021\equiv1\left(mod6\right)\Rightarrow2021^{2020}\equiv1\left(mod6\right)\)
\(\Rightarrow2021^{2022}-1⋮6\)
\(\Rightarrow S-1⋮6\)
`đk:x ne +-3,x ne -2`
`B=(21/(x^2-9)-(x-4)/(3-x)-(x-1)/(3+x)):(1-1/(x+3))`
`=(21/(x^2-9)+(x-4)/(x-3)-(x-1)/(x+3)):((x+3-1)/(x+3))`
`=((21+x^2-x-12-x^2+4x-3)/((x-3)(x+3))):(x+2)/(x+3)`
`=(3x+6)/((x-3)(x+3))*(x+3)/(x+2)`
`=(3x+6)/((x-3)(x+2))`
`=3/(x-3)`
`b)|2x+1|=5`
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2(tm)\\x=-3(l)\end{array} \right.\)
`=>B=3/(2-3)=-3`
`c)B=-3/5`
`<=>3/(x-3)=3/(-5)`
`<=>x-3=-5`
`<=>x=-2(l)`
`d)B<0`
`<=>3/(x-3)<0`
Mà `3>0`
`=>x-3<0<=>x<3`
a) đk: \(x\ne\pm3\)
\(B=\left[\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-4}{x-3}-\dfrac{x-1}{x+3}\right]:\left(\dfrac{x+3-1}{x+3}\right)\)
= \(\left[\dfrac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\dfrac{x+2}{x+3}\)
= \(\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}\)
= \(\dfrac{3x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}=\dfrac{3}{x-3}\)
b) Để \(\left|2x+1\right|=5\)
<=> \(\left[{}\begin{matrix}2x+1=5< =>x=2\left(c\right)\\2x+1=-5< =>x=-3\left(l\right)\end{matrix}\right.\)
Thay x = 2, ta có;
B = \(\dfrac{3}{2-3}=-3\)
c) Để B = \(\dfrac{-3}{5}\)
<=> \(\dfrac{3}{x-3}=\dfrac{-3}{5}\)
<=> x - 3 = -5
<=> x = -2
d) Để B < 0
<=> \(\dfrac{3}{x-3}< 0\)
<=> x - 3 < 0
<=> x < 3
`a)P=(x/(x+2)-(x^3-8)/(x^3+8)*(x^2-2x+4)/(x^2-4)):4/(x+2)`
`đk:x ne 0,x ne -2`
`P=(x/(x+2)-((x-2)(x^2+2x+4))/((x+2)(x^2-2x+4))*(x^2-2x+4)/((x-2)(x+2)))*(x+2)/4`
`=(x/(x+2)-(x^2+2x+4)/(x+2)^2)*(x+2)/4`
`=(x^2+2x-x^2-2x-4)/(x+2)^2*(x+2)/4`
`=-4/(x+2)^2*(x+2)/4`
`=-1/(x+2)`
`b)P<0`
`<=>-1/(x+2)<0`
Vì `-1<0`
`<=>x+2>0`
`<=>x> -2`
`c)P=1/x+1(x ne 0)`
`<=>-1/(x+2)=1/x+1`
`<=>1/x+1+1/(x+2)=0``
`<=>x+2+x(x+2)+x=0`
`<=>x^2+4x+2=0`
`<=>` \(\left[ \begin{array}{l}x=\sqrt2-2\\x=-\sqrt2-2\end{array} \right.\)
`d)|2x-1|=3`
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2(l)\\x=-1(tm)\end{array} \right.\)
`x=-1=>P=-1/(-1+2)=-1`
`e)P=-1/(x+2)` thì nhỏ nhất cái gì nhỉ?
a) đk: \(x\ne-2;2\)
\(P=\left[\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{4}{x+2}\)
= \(\left[\dfrac{x}{x+2}-\dfrac{x^2+2x+4}{\left(x+2\right)^2}\right].\dfrac{x+2}{4}\)
= \(\dfrac{x^2+2x-x^2-2x-4}{\left(x+2\right)^2}.\dfrac{x+2}{4}\) = \(\dfrac{-4}{4\left(x+2\right)}=\dfrac{-1}{x+2}\)
b) Để P < 0
<=> \(\dfrac{-1}{x+2}< 0\)
<=> x +2 > 0
<=> x > -2 ( x khác 2)
c) Để P= \(\dfrac{1}{x}+1\)
<=> \(\dfrac{-1}{x+2}=\dfrac{1}{x}+1\)
<=> \(\dfrac{1}{x}+\dfrac{1}{x+2}+1=0\)
<=> \(\dfrac{x+2+x+x\left(x+2\right)}{x\left(x+2\right)}=0\)
<=> x2 + 4x + 2 = 0
<=> (x+2)2 = 2
<=> \(\left[{}\begin{matrix}x=\sqrt{2}-2\left(c\right)\\x=-\sqrt{2}-2\left(c\right)\end{matrix}\right.\)
d) Để \(\left|2x-1\right|=3\)
<=> \(\left[{}\begin{matrix}2x-1=3< =>x=2\left(l\right)\\2x-1=-3< =>x=-1\left(c\right)\end{matrix}\right.\)
Thay x = -1, ta có:
P = \(\dfrac{-1}{-1+2}=-1\)