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Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
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: Ta có: \(P=\left(x-1\right)^2-4x\left(x+1\right)\left(x-1\right)+3\)
\(=x^2-2x+1-4x\left(x^2-1\right)+3\)
\(=x^2-2x+4-4x^3+4x\)
\(=-4x^3+x^2+2x+4\)
b: Thay x=-2 vào P, ta được:
\(P=-4\cdot\left(-8\right)+4-4+4=36\)
\(\left(\dfrac{\dfrac{x}{x+1}}{\dfrac{x^2}{x^2+x+1}}-\dfrac{2x+1}{x^2+x}\right)\dfrac{x^2-1}{x-1}\)ĐK : \(x\ne\pm1\)
\(=\left(\dfrac{x}{x+1}.\dfrac{x^2+x+1}{x^2}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)=\left(\dfrac{x^2+x-1}{x^2+x}-\dfrac{2x+1}{x\left(x+1\right)}\right)\left(x+1\right)\)
\(=\left(\dfrac{x^2+x-1-2x-1}{x\left(x+1\right)}\right)\left(x+1\right)=\dfrac{x^2-3x-2}{x}\)
à xin lỗi mình nhầm dòng cuối
\(=\dfrac{x^2-x-2}{x}=\dfrac{\left(x+1\right)\left(x-2\right)}{x}\)
Để biểu thức trên nhận giá trị dương khi
\(\dfrac{\left(x+1\right)\left(x-2\right)}{x}>0\)bạn tự xét TH cả tử và mẫu nhé, mình đánh trên này bị lỗi
a, ĐKXĐ:\(\left\{{}\begin{matrix}x^2-1\ne0\\x+1\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm1\\x\ne-1\\x\ne1\end{matrix}\right.\Leftrightarrow x\ne\pm1\)
b, \(P=\dfrac{2x^2}{x^2-1}+\dfrac{x}{x+1}-\dfrac{x}{x-1}\)
\(\Rightarrow P=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2-x}{\left(x+1\right)\left(x-1\right)}-\dfrac{x^2+x}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x^2+x^2-x-x^2-x}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x^2-2x}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x}{x+1}\)
c, Thay x=2 vào P ta có:
\(P=\dfrac{2x}{x+1}=\dfrac{2.2}{2+1}=\dfrac{4}{3}\)
Bài `1:`
`a)`
Để `P` có nghĩa thì:
`{(x^2-1\ne0),(x+1\ne0),(x-1\ne0):}`
`<=>x\ne+-1`
`b)`
`P=(2x^2)/(x^2-1)+x/(x+1)-x/(x-1)(x\ne+-1)`
`P=(2x^2)/((x-1)(x+1))+(x.(x-1))/((x+1)(x-1))-(x.(x+1))/((x-1)(x+1))`
`P=(2x^2+x^2-x-x^2-x)/((x-1)(x+1))`
`P=(2x^2-2x)/((x-1)(x+1))`
`P=(2x.(x-1))/((x-1)(x+1))=2x/(x+1)`
`c)`
Với `x=2`
`P=(2.2)/(2+1)=4/3`
\(a,A=\left(x-2\right)\left(x+2\right)-\left(x-1\right)^2+2x\)
\(\Rightarrow A=x^2-4-x^2+2x-1+2x\)
\(\Rightarrow A=4x-5\)
b, thay x=2 vào ta được
\(A=4x-5=4.2-5=8-5=3\)
A = ( x - 2 )( x + 2 ) - ( x - 1)2 + 2x
a) A = x2 - 4 - ( x2 - 2x + 1 )2 + 2x
A = x2 - 4 - 2x2 + 4x - 2 + 2x
A = -x2 + 6x - 6
b) Ta có x = 2
=> -x2 + 6x - 6 = - 4 + 12 - 6
A = 2
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm1\end{matrix}\right.\)
Ta có : \(A=\left(\dfrac{x}{x-1}-\dfrac{1}{x^2-x}\right):\left(\dfrac{1}{x+1}+\dfrac{2}{x^2-1}\right)\)
\(=\left(\dfrac{x}{x-1}-\dfrac{1}{x\left(x-1\right)}\right):\left(\dfrac{1}{x+1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\left(\dfrac{x^2-1}{x\left(x-1\right)}\right):\left(\dfrac{x-1+2}{\left(x-1\right)\left(x+1\right)}\right)\)\(=\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)}:\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1}{x}:\dfrac{1}{x-1}=\dfrac{x+1}{x}.\left(x-1\right)=\dfrac{x^2-1}{x}\)
b, Thay x = 0,5 vào A ta được : A = -3/2
Vậy ...
a) Ta có: \(A=\left(\dfrac{x}{x-1}-\dfrac{1}{x^2-x}\right):\left(\dfrac{1}{x+1}+\dfrac{2}{x^2-1}\right)\)
\(=\left(\dfrac{x^2}{x\left(x-1\right)}-\dfrac{1}{x\left(x-1\right)}\right):\left(\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}+\dfrac{2}{\left(x+1\right)\left(x-1\right)}\right)\)
\(=\dfrac{x^2-1}{x\left(x-1\right)}:\dfrac{x-1+2}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2-1}{x}\)
b) Thay \(x=\dfrac{1}{2}\) vào A, ta được:
\(A=\left(\dfrac{1}{4}-1\right):\dfrac{1}{2}=\dfrac{-3}{4}\cdot2=-\dfrac{3}{2}\)
Vậy: Khi \(x=\dfrac{1}{2}\) thì \(A=-\dfrac{3}{2}\)
-2 nhớ link cho mình nha
A=x(1-x)+(x+1)(x-2)
A=x-x^2+x^2-x-2
A=-2
Nhớ k nha