câu 1 :chia X^2 -6x+15 cho x-3 được kết quả là:
câu 2 cho P= (x+1)^3+(x+1)(6-x2)-12
a. rút gọn P
b. tính P khi x = -1/2
c. tìm x để P =0
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23 tháng 10 2021
a: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}+1\right)+x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2x-2\sqrt{x}+x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-x-4\sqrt{x}+1}{x-1}\)
HH
14 tháng 10 2018
Câu 1: Đặt tính chia, kết quả là x - 3 dư 6
Câu 2:
a) \(P=\left(x+1\right)^3+\left(x+1\right)\left(6-x^2\right)-12\)
\(\Leftrightarrow P=\left(x+1\right)\left(x^2+2x+1\right)+\left(x+1\right)\left(6-x^2\right)-12\)
\(\Leftrightarrow P=\left(x+1\right)\left(x^2+2x+1+6-x^2\right)-12\)
\(\Leftrightarrow P=\left(x+1\right)\left(7+2x\right)-12\)
b) \(x=-\dfrac{1}{2}\) thì giá trị P là:
\(\Leftrightarrow P=\left(-\dfrac{1}{2}+1\right)\left(7-2.\dfrac{1}{2}\right)-12\)
\(\Leftrightarrow P=3-12\)
\(\Leftrightarrow P=-9\)
c) \(P=0\)
\(\Leftrightarrow\left(x+1\right)\left(7+2x\right)-12=0\)
\(\Leftrightarrow\left(x+1\right)\left(7+2x\right)=12\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
(Sai sót mong bạn thông cảm)
17 tháng 10 2022
Câu 1:
\(\dfrac{x^2-6x+15}{x-3}=\dfrac{x^2-6x+9+6}{x-3}=\left(x-3\right)+\dfrac{6}{x-3}\)
=>Số dư là 6
Câu 2:
a: \(P=x^3+3x^2+3x+1+6x-x^3+6-x^2-12\)
\(=2x^2+9x-5\)
b: Khi x=-1/2 thì \(P=2\cdot\dfrac{1}{4}-\dfrac{9}{2}-5=\dfrac{1}{2}-\dfrac{9}{2}-5=-9\)
c: Để P=0 thì 2x^2+9x-5=0
hay \(x\in\left\{\dfrac{1}{2};-5\right\}\)
27 tháng 7 2021
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
9 tháng 5 2023
a: \(=\dfrac{x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{-5\sqrt{x}-5+x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-3\sqrt{x}-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
b: khi x=6-2căn 5 thì \(P=\dfrac{6-2\sqrt{5}-3\sqrt{5}+3-5}{\left(\sqrt{5}-3\right)\left(\sqrt{5}-4\right)\cdot\sqrt{5}}\)
\(=\dfrac{-5\sqrt{5}+4}{\sqrt{5}\left(\sqrt{5}-3\right)\left(\sqrt{5}-4\right)}\)
21 tháng 8 2021
\(a,P=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}\right)\cdot\left(\dfrac{\sqrt{x}+2}{2}\right)^2\left(x\ge0;x\ne4\right)\\ P=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}\\ P=\dfrac{4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)
\(b,\)Ta có \(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
Thay vào \(P\), ta được:
\(P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+2}{\sqrt{\left(\sqrt{5}-1\right)^2}-2}=\dfrac{\sqrt{5}-1+2}{\sqrt{5}-1-2}=\dfrac{\sqrt{5}+1}{\sqrt{5}-3}\)
\(c,\)Để \(P< 1\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}< 1\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-1< 0\\ \Leftrightarrow\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\sqrt{x}-2}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-2}< 0\\ \Leftrightarrow\sqrt{x}-2< 0\left(4>0\right)\\ \Leftrightarrow\sqrt{x}< 2\\ \Leftrightarrow x< 4\)
Vậy để \(P< 1\) thì \(x< 4\)
Tick nha
21 tháng 8 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}\right)\cdot\left(\dfrac{\sqrt{x}+2}{2}\right)^2\)
\(=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)
b: Thay \(x=6-2\sqrt{5}\) vào P, ta được:
\(P=\dfrac{\sqrt{5}+1+2}{\sqrt{5}+1-2}=\dfrac{3+\sqrt{5}}{\sqrt{5}+1}=\dfrac{1+\sqrt{5}}{2}\)
1 tháng 12 2021
\(a,P=\dfrac{2x^2+2x+2+2x-1+x^2+6x+2}{\left(x-1\right)\left(x^2+x+1\right)}\\ P=\dfrac{3x^2+10x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
1/\(\left(x^2-6x+15\right):\left(x-3\right)\)
Đặt cột dọc ta được x-3 dư 6
2/a/\(p=\left(x+1\right)^3+\left(x+1\right)\left(6-x^2\right)-12\)
\(=x^3+3x^2+3x+1+6x-x^3+6-x^2-12\)
\(=2x^2+9x-11\)
b/thay x = -1/2 ta đc \(2.-\left(\frac{1}{2}\right)^2+9.-\frac{1}{2}-11\)
\(=\frac{1}{2}+\left(-\frac{9}{2}\right)-11\)
\(=\left(-15\right)\)