Cho \(A=\dfrac{x^2+2x-3}{\left(x-1\right)\left(x+1\right)}\)
a) Vs giá trị nào của x thì a xác định
b) Rút gọn A
c) Tính A tại x=3
d) Tìm x để A=4
e) Tìm x ϵ Z, để A ϵ Z
f) Tìm x để A > 1
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Bổ sung phần c và d luôn:
c, C = \(\dfrac{2}{5}\)
\(\Leftrightarrow\) \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{2}{5}\)
\(\Leftrightarrow\) 5(x2 - 1) = 2(2x2 + 3)
\(\Leftrightarrow\) 5x2 - 5 = 4x2 + 6
\(\Leftrightarrow\) x2 = 11
\(\Leftrightarrow\) x2 - 11 = 0
\(\Leftrightarrow\) (x - \(\sqrt{11}\))(x + \(\sqrt{11}\)) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-\sqrt{11}=0\\x+\sqrt{11}=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\sqrt{11}\left(TM\right)\\x=-\sqrt{11}\left(TM\right)\end{matrix}\right.\)
d, Ta có: \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{x^2+\dfrac{3}{2}-\dfrac{5}{2}}{2\left(x^2+\dfrac{3}{2}\right)}\) = \(\dfrac{1}{2}\) - \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\)
C nguyên \(\Leftrightarrow\) \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\) nguyên \(\Leftrightarrow\) 5 \(⋮\) 4(x2 + \(\dfrac{3}{2}\))
\(\Leftrightarrow\) 4(x2 + \(\dfrac{3}{2}\)) \(\in\) Ư(5)
Xét các TH:
4(x2 + \(\dfrac{3}{2}\)) = 5 \(\Leftrightarrow\) x2 = \(\dfrac{-1}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{1}{4}\) = 0 (Vô nghiệm)
4(x2 + \(\dfrac{3}{2}\)) = -5 \(\Leftrightarrow\) x2 = \(\dfrac{-11}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{11}{4}\) = 0 (Vô nghiệm)
4(x2 + \(\dfrac{3}{2}\)) = 1 \(\Leftrightarrow\) x2 = \(\dfrac{-5}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{5}{4}\) = 0 (Vô nghiệm)
4(x2 + \(\dfrac{3}{2}\)) = -1 \(\Leftrightarrow\) x2 = \(\dfrac{-7}{4}\) \(\Leftrightarrow\) x2 + \(\dfrac{7}{4}\) = 0 (Vô nghiệm)
Vậy không có giá trị nào của x \(\in\) Z thỏa mãn C \(\in\) Z
Chúc bn học tốt! (Ko bt đề sai hay ko nữa :v)
a) Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)^2}\)
a) Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}:\dfrac{x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{x+9}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{x+9}\)
Bài 1:
\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{16}{3}\\ d,P\in Z\Leftrightarrow x+5\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{-6;-4\right\}\)
Bài 2:
\(a,\Leftrightarrow\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}=0\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow\dfrac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=0\Leftrightarrow\dfrac{-x}{x+2}=0\Leftrightarrow x=0\)
`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 = \(\dfrac{-1}{-1+2}=-1\)
Phạm Nguyễn Tất Đạt Nhã Doanh Akai Haruma ngonhuminh kuroba kaito Nguyễn Huy Tú
mấy bn chỉ cần làm câu a , d, e, f nx thôi