\(M=\dfrac{13x^2-x^4-36}{x^3-5x^2+6x}\)
a) Rút gọn M
b) Tìm x để M > 0; M < 0; có nghĩa; vô nghĩa
c) Tìm x thuộc Z để M rút gọn thuộc Z
K
Khách
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Những câu hỏi liên quan
5 tháng 10 2021
a: Ta có: \(x^2=3-2\sqrt{2}\)
nên \(x=\sqrt{2}-1\)
Thay \(x=\sqrt{2}-1\) vào A, ta được:
\(A=\dfrac{\left(\sqrt{2}+1\right)^2}{\sqrt{2}-1}=\dfrac{3+2\sqrt{2}}{\sqrt{2}-1}=7+5\sqrt{2}\)
HH
11 tháng 6 2018
1/
a/ \(D=2x\left(10x^2-5x-2\right)-5x\left(4x^2-2x-1\right)\)
\(D=2x\left[10\left(x^2-\frac{1}{2}x-\frac{1}{5}\right)\right]-5x\left[4\left(x^2-\frac{1}{2}x-\frac{1}{4}\right)\right]\)
\(D=20x\left(x^2-\frac{1}{2}x-\frac{1}{5}\right)-20x\left(x^2-\frac{1}{2}x-\frac{1}{4}\right)\)
\(D=20x^3-10x^2-4x-20x^3+10x^2+5x\)
\(D=x\)
b/ Mình xin sửa lại đề:
Tính giá trị biểu thức \(E\left(x\right)=x^5-13x^4+13x^3-13x^2+13x+2012\)
Tại x = 12
\(E\left(x\right)=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x-1\right)x+2012\)
\(E\left(x\right)=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2-x+2012\)
\(E\left(x\right)=2012-x\)
\(E\left(x\right)=2000\)
2/
a/ \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
<=> \(2x^2-10x-3x-2x^2=26\)
<=> \(-13x=26\)
<=> \(x=-2\)
b/ Bạn vui lòng coi lại đề.
3a/ Ta có \(D=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
\(D=5x^2-3x-x^3+x^2+x^3-6x^2-10+3x\)
\(D=-10\)
Vậy giá trị của D không phụ thuộc vào x (đpcm)
Y
6 tháng 4 2023
\(M=\dfrac{4}{x+2}+\dfrac{3}{x-2}-\dfrac{5x+2}{x^2-4}\left(dkxd:x\ne\pm2\right)\)
\(=\dfrac{4}{x+2}+\dfrac{3}{x-2}-\dfrac{5x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)+3\left(x+2\right)-\left(5x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x-8+3x+6-5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2}{x+2}\)
Để \(M=\dfrac{2}{5}\) thì \(\dfrac{2}{x+2}=\dfrac{2}{5}\)
Suy ra :
\(2.5=2\left(x+2\right)\)
\(\Leftrightarrow2x+4=10\)
\(\Leftrightarrow x=3\)
Vậy \(M=\dfrac{2}{5}\) thì x = 3
11 tháng 3 2020
ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)
\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)
Đề sai à ??
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 12 2020
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne2\end{matrix}\right.\)
\(M=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(M=\dfrac{-8\sqrt{x}}{x-4}\)
\(M< 0\Leftrightarrow-\dfrac{8\sqrt{x}}{x-4}< 0\Leftrightarrow x-4>0\Leftrightarrow x>4\)
25 tháng 10 2023
a: \(M=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b: Khi \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\) thì
\(M=\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}-2}{\sqrt{\left(\sqrt{2}+1\right)^2}}=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}\)
\(=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\left(\sqrt{2}-1\right)^2=3-2\sqrt{2}\)
c: M>0
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\)
mà \(\sqrt{x}>0\)
nên \(\sqrt{x}-2>0\)
=>\(\sqrt{x}>2\)
=>x>4
\(M=\dfrac{13x^2-x^4-36}{x^3-5x^2+6x}\)
\(=\dfrac{-x^4+13x^2-36}{x\left(x^2-5x+6\right)}\)
\(=\dfrac{-x^4+9x^2+4x^2-36}{x\left(x^2-2x-3x+6\right)}\)
\(=\dfrac{-x^2\left(x^2-9\right)+4\left(x^2-9\right)}{x\cdot\left[x\left(x-2\right)-3\left(x-2\right)\right]}\)
\(=\dfrac{\left(-x^2+4\right)\left(x^2-9\right)}{x\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{\left(4-x^2\right)\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{\left(2-x\right)\left(2+x\right)\left(x+3\right)}{x\left(x-2\right)}\)
\(=\dfrac{-\left(x-2\right)\left(2+x\right)\left(x+3\right)}{x\left(x-2\right)}\)
\(=\dfrac{-\left(2+x\right)\left(x+3\right)}{x}\)
\(=\dfrac{-\left(2x+6+x^2+3x\right)}{x}\)
\(=\dfrac{-\left(5x+6+x^2\right)}{x}\)
\(=-\dfrac{5x+6+x^2}{x}\)
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