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để A xác định
\(\Rightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x^2\ne4\end{cases}}\Rightarrow x\ne\pm2\)
\(A=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}\)
\(A=\frac{4.x-8}{\left(x+2\right).\left(x-2\right)}+\frac{3.x+6}{\left(x-2\right).\left(x+2\right)}-\frac{5x-6}{\left(x-2\right).\left(x+2\right)}\)
\(A=\frac{4x-8+3x+6-5x+6}{\left(x+2\right).\left(x-2\right)}=\frac{2.\left(x+2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{2}{x-2}\)
\(\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{x^2-4}=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4x-8}{\left(x+2\right)\left(x-2\right)}+\frac{3x+4}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}=\frac{4x-8+3x+4-5x+6}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{2x+2}{\left(x+2\right)\left(x-2\right)}=\frac{2x+2}{x^2-4}\)
C, \(x=4\Rightarrow A=\frac{2x+2}{x^2-4}=\frac{-6}{12}=\frac{-1}{2}\)
d, \(A\inℤ\Leftrightarrow2x+2⋮x^2-4\Leftrightarrow2x^2+2x-2x^2+8⋮x^2-4\Leftrightarrow2x+8⋮x^2-4\)
\(\Leftrightarrow2x^2+8x⋮x^2-4\Leftrightarrow16⋮x^2-4\)
\(x^2-4\inℕ\)
\(\Rightarrow x^2\in\left\{0;4;12\right\}\)
Thử lại thì 12 ko là số chính phương vậy x=0 hoặc x=2 thỏa mãn
mk học lớp 6 mong mn thông cảm nếu có sai sót
a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)
b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)
\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}=\dfrac{x+1}{2x}\)
c: Thay x=2017 vào C, ta được:
\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) Thay x=0 vào A, ta được:
\(A=\dfrac{15\cdot\sqrt{0}-11}{0+2\sqrt{0}-3}-\dfrac{3\sqrt{0}-2}{\sqrt{0}-1}-\dfrac{2\sqrt{0}+3}{\sqrt{0}+3}\)
\(=\dfrac{-11}{-3}-\dfrac{-2}{-1}-\dfrac{3}{3}\)
\(=\dfrac{11}{3}-2-1\)
\(=\dfrac{11}{3}-\dfrac{9}{3}=\dfrac{2}{3}\)
a) ĐKXĐ: \(x\notin\left\{0;-\dfrac{1}{2};\dfrac{1}{2}\right\}\)
Ta có: \(A=\left(\dfrac{1}{2x-1}+\dfrac{3}{1-4x^2}-\dfrac{2}{2x+1}\right):\left(\dfrac{x^2}{2x^2+x}\right)\)
\(=\left(\dfrac{2x+1}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{3}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}\right):\left(\dfrac{x^2}{x\left(2x+1\right)}\right)\)
\(=\dfrac{2x+1-3-4x+2}{\left(2x-1\right)\left(2x+1\right)}:\dfrac{x}{2x+1}\)
\(=\dfrac{-2x}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{2x+1}{x}\)
\(=\dfrac{-2}{2x-1}\)
b) Ta có: \(\left|2x-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(nhận\right)\\x=-\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
Thay \(x=\dfrac{3}{2}\) vào biểu thức \(A=\dfrac{-2}{2x-1}\), ta được:
\(A=-2:\left(2\cdot\dfrac{3}{2}-1\right)=-2:\left(3-1\right)=-2:2=-1\)
Vậy: Khi \(\left|2x-1\right|=2\) thì A=-1
c) Để \(A=\dfrac{1}{3}\) thì \(\dfrac{-2}{2x-1}=\dfrac{1}{3}\)
\(\Leftrightarrow2x-1=-6\)
\(\Leftrightarrow2x=-5\)
hay \(x=-\dfrac{5}{2}\)(thỏa ĐK)
Vậy: Để \(A=\dfrac{1}{3}\) thì \(x=-\dfrac{5}{2}\)
a: |x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=-2(nhận) hoặc x=4(loại)
Khi x=-2 thì \(A=\dfrac{4+4}{-2-4}=\dfrac{8}{-6}=\dfrac{-4}{3}\)
b: ĐKXĐ: x<>4; x<>-4
\(B=\dfrac{-\left(x+4\right)}{x-4}+\dfrac{x-4}{x+4}-\dfrac{4x^2}{\left(x-4\right)\left(x+4\right)}\)
\(=\dfrac{-x^2-8x-16+x^2-8x+16-4x^2}{\left(x-4\right)\left(x+4\right)}=\dfrac{-4x^2-16x}{\left(x-4\right)\left(x+4\right)}\)
=-4x/x-4
c: A+B
=-4x/x-4+x^2+4/x-4
=(x-2)^2/(x-4)
A+B>0
=>x-4>0
=>x>4
a) ĐKXĐ: \(x\ne-3,x\ne2\)
b) \(A=\dfrac{\left(x-2\right)\left(x+2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
c) \(A=\dfrac{x-4}{x-2}=\dfrac{3-4}{3-2}=-1\)
a: ĐKXĐ: a<>3; a<>-3; a<>-1
b: \(P=\dfrac{2a^2-3a+3a+9-2a^2-3}{\left(a-3\right)\left(a+3\right)}\cdot\dfrac{a-3}{a+1}\)
\(=\dfrac{6}{\left(a+3\right)\left(a+1\right)}\)
c: |a|=2
=>a=2 hoặc a=-2
Khi a=-2 thì \(P=\dfrac{6}{\left(-2+3\right)\left(-2+1\right)}=-6\)
Khi a=2 thì \(P=\dfrac{6}{\left(2+3\right)\left(2+1\right)}=\dfrac{6}{5\cdot3}=\dfrac{2}{5}\)
a) ĐKXĐ: \(x\ne3;x\ne\pm2\)
\(C=\frac{2a-a^2}{a+3}\cdot\left(\frac{a-2}{a+2}-\frac{a+2}{a-2}+\frac{4a^2}{4-a^2}\right)\)
\(C=\frac{-a^2+2a}{a+3}\cdot\left(-\frac{4a}{a-2}\right)\)
\(C=-\frac{2a-a^2}{a+3}\cdot\frac{4a}{a-2}\)
\(C=-\frac{\left(2a-a^2\right)\cdot4a}{\left(a+3\right)\left(a-2\right)}\)
\(C=\frac{4a^2}{a+3}\)
b) \(C=\frac{4.4^2}{4+3}=\frac{46}{7}\)
c) \(\frac{4a^2}{a+3}=1\)
<=> 4a2 = a + 3
<=> 4a2 - a - 3 = 0
<=> 4a2 - 3a - 4a - 3 = 0
<=> a(4a + 3) - (4a + 3) = 0
<=> (4a + 3)(a - 1) = 0
<=> 4a + 3 = 0 hoặc a - 1 = 0
<=> a = -3/4 hoặc a = 1
sửa đáp án câu b thành \(\frac{64}{7}\) nhé