trả lời:=2031

đây là ngày sinh của thg Hùng

em họ tôi

hok tốt

2031 ngày sinh của .....!

đừng đăng linh tinh ko bị trừ điểm hỏi đáp đấy

Xl bn nha, mink lớp 7 nên ko giúp bn dk, nếu = tuổi thì mink sẵn sàng giúp đỡ bn!

Thông cảm!

#Girl 2k6#

\(P=\dfrac{x^4+5x^3-20x^2-27x+30}{x^2+4x-21}\left(1\right)\)

Điều kiện xác định khi và chỉ khi

\(x^2+4x-21\ne0\)

\(\Leftrightarrow x^2+7x-3x-21\ne0\)

\(\Leftrightarrow x\left(x+7\right)-3\left(x+7\right)\ne0\)

\(\Leftrightarrow\left(x-3\right)\left(x+7\right)\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-7\end{matrix}\right.\)

Theo đề bài : \(\)

\(x=\sqrt[]{31-12\sqrt[]{3}}=\sqrt[]{27-12\sqrt[]{3}+4}=\sqrt[]{\left(3\sqrt[]{3}-2\right)^2}=\left|3\sqrt[]{3}-2\right|=3\sqrt[]{3}-2\)

\(\left(1\right)\Leftrightarrow P=\dfrac{x^4-3x^3+8x^3-24x^2+4x^2-12x-15x+45-15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{x^3\left(x-3\right)+8x^2\left(x-3\right)+4x\left(x-3\right)-15\left(x-3\right)-15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{\left(x-3\right)\left(x^3+8x^2+4x-15\right)-15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{x^3+8x^2+4x-15}{x+7}-\dfrac{15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{x^3+7x^2+x^2+7x-3x-15}{x+7}-\dfrac{15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{x^2\left(x+7\right)+x\left(x+7\right)-3\left(x+7\right)+6}{x+7}-\dfrac{15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=\dfrac{\left(x^2+x-3\right)\left(x+7\right)+6}{x+7}-\dfrac{15}{\left(x-3\right)\left(x+7\right)}\)

\(\Leftrightarrow P=x^2+x-3+\dfrac{6}{x+7}-\dfrac{15}{\left(x-3\right)\left(x+7\right)}\)

Thay \(x=3\sqrt[]{3}-2\) vào \(P\) ta được

\(\Leftrightarrow P=\left(3\sqrt[]{3}-2\right)^2+3\sqrt[]{3}-2-3+\dfrac{6}{3\sqrt[]{3}-2+7}-\dfrac{15}{\left(3\sqrt[]{3}-2-3\right)\left(3\sqrt[]{3}-2+7\right)}\)

\(\Leftrightarrow P=31-12\sqrt[]{3}+3\sqrt[]{3}-5+\dfrac{6}{3\sqrt[]{3}+5}-\dfrac{15}{\left(3\sqrt[]{3}-5\right)\left(3\sqrt[]{3}+5\right)}\)

\(\Leftrightarrow P=26-9\sqrt[]{3}+\dfrac{6\left(3\sqrt[]{3}-5\right)}{\left(3\sqrt[]{3}+5\right)\left(3\sqrt[]{3}-5\right)}-\dfrac{15}{\left(3\sqrt[]{3}\right)^2-5^2}\)

\(\Leftrightarrow P=26-9\sqrt[]{3}+\dfrac{6\left(3\sqrt[]{3}-5\right)}{2}-\dfrac{15}{2}\)

\(\Leftrightarrow P=\dfrac{37}{2}-9\sqrt[]{3}+3\left(3\sqrt[]{3}-5\right)\)

\(\Leftrightarrow P=\dfrac{37}{2}-9\sqrt[]{3}+9\sqrt[]{3}-15\)

\(\Leftrightarrow P=\dfrac{37}{2}-15=\dfrac{7}{2}\)

P = 7/2

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne3\\x\ge1\end{matrix}\right.\)

Ta có : \(P=\dfrac{x-1-2}{\sqrt{x-1}-\sqrt{2}}=\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\)

\(=\sqrt{x-1}+\sqrt{2}\)

b, Thấy : \(\sqrt{x-1}\ge0\)

\(\Rightarrow P=\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\)

Vậy \(Min=\sqrt{2}\Leftrightarrow x=1\)

Vậy ,...
 

100 + 2x = 144

<=> 2x     = 144 - 100

<=>  2x    = 44

<=>   x     = 44 : 2

<=>   x     = 22

vì 44:2= 22

nên x =22

ĐK:\(x\ge0;x\ne9\)

a) \(P=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+x-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)

b)\(P=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=1+\dfrac{2}{\sqrt{x}+2}\le1+\dfrac{2}{0+2}=2\)

Dấu "=" xảy ra khi \(x=0\)

Vậy \(P_{max}=2\)