Rút gọn biểu thức P=3√9^6-2x^3 với x<0 được kết quả là :
A,P =11 x^3
B,P=7x^3
C,P=-25x^3
D,P=-11x^3
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\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)
\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)
\(A=\dfrac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}\left(x>3\right)\\ A=\dfrac{\left(x+3\right)+2\sqrt{\left(x-3\right)\left(x+3\right)}}{2\left(x-3\right)+\sqrt{\left(x-3\right)\left(x+3\right)}}\\ A=\dfrac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2+\sqrt{x+3}\right)}\)
Tới đây chịu rùi, hình như đề sai đk?
1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
a: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
a: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b)
\(P=A-B=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2-9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9-x^2+9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(2-x\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x}{x-3}\)
c)
Để \(P\le1\) thì:
\(-\dfrac{x}{x-3}\le1\)
\(\Leftrightarrow\dfrac{x}{x-3}\ge1\\ \Leftrightarrow x-3-x\ge1\\ \Leftrightarrow-3\ge1\left(vô.lý\right)\)
Vậy không tồn tại giá trị x để \(P\le1\)
`HaNa♬D`
Làm lại nha cái này đúng, kia sai nha=)
b)
Với \(\left\{{}\begin{matrix}x\ne3\\x\ne2\end{matrix}\right.\)
\(P=A-B=(\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)})+\dfrac{2x-1}{x-3}\\ =\left(\dfrac{2x-9-x^2-9}{\left(x-3\right)\left(x-2\right)}\right)+\dfrac{\left(2x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}+\dfrac{2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2+2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-3x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-2x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(x-2\right)-\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\dfrac{x-1}{x-3}\)
c)
Để P\(\ge1\) thì:
\(\dfrac{x-1}{x-3}\ge1\\ \Leftrightarrow x-3-x+1-1\ge0\\ \Leftrightarrow-3\ge0\left(vô.lý\right)\)
Vậy không tồn tại giá trị x để \(P\ge1\)
`HaNa☘D`
a) \(A=\frac{2x}{x+3}+\frac{2}{x-3}+\frac{x^2-x+6}{9-x^2}\left(x\ne\pm3\right)\)
\(\Leftrightarrow A=\frac{2x}{x+3}+\frac{2}{x-3}-\frac{x^2-x+6}{x^2-9}\)
\(\Leftrightarrow A=\frac{2x}{x+3}+\frac{2}{x-3}-\frac{x^2-x+6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x^2-x+6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x^2-6x+2x+6-x^2+x-6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{x}{x+3}\)
Vậy \(A=\frac{x}{x+3}\left(x\ne\pm3\right)\)
b) Ta có \(A=\frac{x}{x+3}\left(x\ne\pm3\right)\)
Để A nhạn giá trị nguyên thì \(\frac{x}{x+3}\)nhận gái trị nguyên
Ta có \(\frac{x}{x+3}=\frac{x+3-3}{x+3}=1-\frac{3}{x+3}\)
=> \(\frac{3}{x+3}\)nguyên thì \(1-\frac{3}{x+3}\)nguyên
=> 3 chia hết cho x+2.
x nguyên => x+3 nguyên => x+3\(\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
Ta có bảng
x+3 | -3 | -1 | 1 | 3 |
x | -6 | -4 | -2 | 0 |
Đối chiếu điều kiện x\(\ne\pm3;x\inℤ\)
=> x={-6;-4;-2;0}
Vậy x={-6;-4;-2;0} thì A nhận giá trị nguyên