rút gọn A=(x-2)(x+2)-(x+1)^2
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10 tháng 11 2021
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
19 tháng 1 2022
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 \(x\in\left\{3;1;5;-1\right\}\)
27 tháng 10 2021
Bài 1:
a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{2x}{x-1}\)
10 tháng 11 2021
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
10 tháng 11 2021
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
AN
1
31 tháng 7 2021
a) Ta có: \(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{x^2-4}+\dfrac{1}{x+2}\right)\cdot\left(\dfrac{2}{x}-1\right)\)
\(=\dfrac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)}{x}\)
\(=\dfrac{-4}{x+2}\)
b) Để A=1 thì x+2=-4
hay x=-6(nhận)
MN
3 tháng 3 2020
a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne1\end{cases}}\)
\(A=\frac{2x+1}{x^2-3x+2}+\frac{x+1}{1-x}-\frac{x^2+5}{x^2-3x+2}+\frac{x^2+x}{x-1}\)
\(\Leftrightarrow A=\frac{2x+1}{\left(x-1\right)\left(x-2\right)}-\frac{x+1}{x-1}-\frac{x^2+5}{\left(x-2\right)\left(x-1\right)}+\frac{x^2+x}{x-1}\)
\(\Leftrightarrow A=\frac{2x+1-\left(x+1\right)\left(x-2\right)-x^2-5+\left(x^2+x\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\frac{2x+1-x^2+x+2-x^2-5+x^3-x^2-2x}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\frac{x^3-3x^2+x-2}{\left(x-1\right)\left(x-2\right)}\)
b) Khi \(x^2-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=.0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}\)
\(\Leftrightarrow A=\frac{\left(-1\right)^3-3\left(-1\right)^2-1-2}{\left(-1-2\right)\left(-1-1\right)}=\frac{\left(-1\right)-3-1-2}{\left(-3\right)\left(-2\right)}=\frac{7}{6}\)
c) Để A = 0
\(\Leftrightarrow\frac{x^3-3x^2+x-2}{\left(x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^3-3x^2+x-2=0\)2.89328919
Phần này mik k biết phân tích như thế nào, tính ra :
\(\Leftrightarrow x\approx2,89328919\)
Nhưng nếu đề bắt tìm nghiệm nguyên của x thì \(S=\varnothing\)nhé !
d) Để \(A\inℤ\)
\(\Leftrightarrow x^3-3x^2+x-2⋮\left(x-2\right)\left(x-1\right)\)
\(\Leftrightarrow\hept{\begin{cases}x^3-3x^2+x-2⋮x-2\\x^3-3x+x-2⋮x-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x^2-x-1\right)\left(x-2\right)-4⋮x-2\\\left(x^2-2x-1\right)\left(x-1\right)-3⋮x-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}4⋮x-2\\3⋮x-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\\x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in\left\{1;3;0;4;-2;6\right\}\\x\in\left\{0;2;-2;4\right\}\end{cases}}\)
\(\Leftrightarrow x\in\left\{0;-2;4\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{0;-2;4\right\}\)
14 tháng 10 2021
1) \(A=x^2-6x+9-2x^3+2x=-2x^3+x^2-4x+9\)
2) \(B=x^3-3x+2x^2-6-x^3+1=2x^2-3x-5\)
\(A=\left(x-2\right)\left(x+2\right)-\left(x+1\right)^2\)
\(=x^2-4-\left(x^2+2x+1\right)\)
\(=x^2-4-x^2-2x-1=-2x-5\)
`A = (x - 2)(x+2) - (x+1)^2`
`= (x^2 - 4) - (x^2 + 2x + 1)`
`= x^2 - 4 - x^2 -2x - 1`
`= -2x - 5`
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Áp dụng HDT sau nhé:
`a^2 - b^2 = (a+b)(a-b)`
`(a+b)^2 = a^2 + 2ab + b^2`