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a/ đkxđ: x \(\ne\pm\)2; x≠3
\(P=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}-\dfrac{4x^2}{x^2-4}\right):\dfrac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)
\(=\left(\dfrac{\left(2+x\right)^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}+\dfrac{4x^2}{x^2-4}\right):\dfrac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\)
\(=\dfrac{x^2+4x+4-x^2+4x-4+4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\dfrac{2-x}{x-3}\)
\(=\dfrac{8x+4x^2}{2+x}\cdot\dfrac{1}{x-3}=\dfrac{4x\left(2+x\right)}{2+x}\cdot\dfrac{1}{x-3}=\dfrac{4x}{x-3}\)
b/ x = \(\dfrac{1}{3}\Leftrightarrow P=\dfrac{4\cdot\dfrac{1}{3}}{\dfrac{1}{3}-3}=\dfrac{4}{3}:\left(-\dfrac{8}{3}\right)=\dfrac{4}{3}\cdot\left(-\dfrac{3}{8}\right)=-\dfrac{4}{8}=-\dfrac{1}{2}\)
c/ \(P\in Z\Rightarrow\dfrac{4x}{x-3}\in Z\)
Ta có: \(\dfrac{4x}{x-3}=\dfrac{4x-12+12}{x-3}=\dfrac{4\left(x-3\right)}{x-3}+\dfrac{12}{x-3}=4+\dfrac{12}{x-3}\)
=> \(x-3\inƯ\left(12\right)\) thì P ∈ Z
=> \(x-3=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
\(\Leftrightarrow x=\left\{-9;-3;-1;0;1;2;4;5;6;7;9;15\right\}\)
mà x>4
=> x = {5;6;7;9;15}
a, Ta có:
\(P=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}-\dfrac{4x^2}{x^2-4}\right):\dfrac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)
\(=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}+\dfrac{4x^2}{4-x^2}\right):\left[\dfrac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\right]\)
\(=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}+\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}\right):\dfrac{x-3}{2-x}\)
\(=\dfrac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4+4x+x^2-\left(4-4x+x^2\right)+4x^2}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4+4x+x^2-4+4x-x^2+4x^2}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)
\(=\dfrac{4x}{x-3}\)
B3;a,ĐKXĐ:\(x\ne\pm4\)
A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)
Bài 1:
a) \(x\ne2\)
Bài 2:
a) \(x\ne0;x\ne5\)
b) \(\dfrac{x^2-10x+25}{x^2-5x}=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
c) Để phân thức có giá trị nguyên thì \(\dfrac{x-5}{x}\) phải có giá trị nguyên.
=> \(x=-5\)
Bài 3:
a) \(\left(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\right)\cdot\left(\dfrac{4x^2-4}{5}\right)\)
\(=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{2\left(2x^2-2\right)}{5}\)
\(=\dfrac{\left(x+1\right)^2+6-\left(x-1\right)\left(x+3\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\cdot2\left(x^2-1\right)}{5}\)
\(=\dfrac{\left(x+1\right)^2+6-\left(x^2+3x-x-3\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left[\left(x+1\right)^2+6-\left(x^2+2x-3\right)\right]\cdot\dfrac{2}{5}\)
\(=\left[\left(x+1\right)^2+6-x^2-2x+3\right]\cdot\dfrac{2}{5}\)
\(=\left[\left(x+1\right)^2+9-x^2-2x\right]\cdot\dfrac{2}{5}\)
\(=\dfrac{2\left(x+1\right)^2}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2\left(x^2+2x+1\right)}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+2}{5}+\dfrac{18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+2+18}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
\(=\dfrac{2x^2+4x+20}{5}-\dfrac{2}{5}x^2-\dfrac{4}{5}x\)
c) tự làm, đkxđ: \(x\ne1;x\ne-1\)
ô hô ngộ quá nhìu người bt toán lớp 8 trong khi lớp 7 với lại óc nguyow trở lại r kaka
a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{x^2-x-2}{x^2}\right)\)
\(=\dfrac{x\left(x-2\right)^2+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x^2-x-2\right)}{x^2}\)
\(=\dfrac{x\left[x^2-4x+4+4x\right]}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x+1}{2x}\)
b) Thay \(x=\dfrac{1}{2}\) vào P, ta được:
\(P=\dfrac{1}{2}+1=\dfrac{3}{2}\)
em c.ơn trước
a: \(B=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x-2}{2x-1}\)
\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{1}{2x-1}=\dfrac{-4x\left(x+2\right)}{\left(x+2\right)\left(2x-1\right)}=\dfrac{-4x}{2x-1}\)
b: |x|=3
=>x=3 hoặc x=-3
Khi x=3 thì \(B=\dfrac{-4\cdot3}{2\cdot3-1}=\dfrac{-12}{5}\)
Khi x=-3 thì \(B=\dfrac{-4\cdot\left(-3\right)}{2\cdot\left(-3\right)-1}=\dfrac{12}{-7}=\dfrac{-12}{7}\)
`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`
`b)` Với `x ne -1;x ne -5` có:
`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`
`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`
`A=[x^2-3x-4]/[(x+1)(x+5)]`
`A=[(x+1)(x-4)]/[(x+1)(x+5)]`
`A=[x-4]/[x+5]`
`c)` Với `x ne -5; x ne -1; x ne 4` có:
`P=A.B=[x-4]/[x+5].[-10]/[x-4]`
`=[-10]/[x+5]`
Để `P` nguyên `<=>[-10]/[x+5] in ZZ`
`=>x+5 in Ư_{-10}`
Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`
`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)
đkxđ: x\(\ne\pm3\)
a/ \(P=\left(\dfrac{x}{x+3}-\dfrac{x^2+5}{x^2-9}+\dfrac{7}{x-3}\right)\cdot\dfrac{x+3}{4}=\left(\dfrac{x\left(x-3\right)-x^2-5+7\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\cdot\dfrac{x+3}{4}=\dfrac{x^2-3x-x^2-5+7x+21}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{4}=\dfrac{4x+16}{x-3}\cdot\dfrac{1}{4}=\dfrac{4\left(x+4\right)}{4\left(x-3\right)}=\dfrac{x+4}{x-3}\)
b/ tại x = 5 thì:
\(P=\dfrac{5+4}{5-3}=\dfrac{9}{2}\)
c/ Ta có: \(\dfrac{x+4}{x-3}=\dfrac{x-3+7}{x-3}=\dfrac{x-3}{x-3}+\dfrac{7}{x-3}=1+\dfrac{7}{x-3}\)
để P ∈ Z thì \(\dfrac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)\)
=> x - 3 = {-7;-1;1;7}
=> x = {-4;2;4;10}
Vậy.............
1,
\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)
2.
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
3.
Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)
4.
\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)
\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)
5.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
a: \(P=\dfrac{8+5x-2x-8}{x\left(x+4\right)}=\dfrac{3x}{x\left(x+4\right)}=\dfrac{3}{x+4}\)
b: Khi x=1/2 thì P=3/(1/2+4)=3:9/2=3*2/9=6/9=2/3