\(\frac{xy}{x^2+y^2}=\frac{5}{8}\)

\(\Rightarrow5\left(x^2+y^2\right)=8xy\)

Ta có : \(P=\frac{x^2-2xy+y^2}{x^2+2xy+y^2}=\frac{5\left(x^2+y^2-2xy\right)}{5\left(x^2+y^2+2xy\right)}\)

\(=\frac{5\left(x^2+y^2\right)-10xy}{5\left(x^2+y^2\right)+10xy}=\frac{8xy-10xy}{8xy+10xy}=\frac{-2xy}{18xy}=\frac{-1}{9}\)

Ta có: \(P=\frac{x^2-2xy+y^2}{x^2+2xy+y^2}=\frac{\frac{x^2+y^2-2xy}{x^2+y^2}}{\frac{x^2+y^2+2xy}{x^2+y^2}}=\frac{\frac{x^2+y^2}{x^2+y^2}-\frac{2xy}{x^2+y^2}}{\frac{x^2+y^2}{x^2+y^2}+\frac{2xy}{x^2+y^2}}\)

                                                          \(=\frac{1-\frac{2xy}{x^2+y^2}}{1+\frac{2xy}{x^2+y^2}}=\frac{1-\frac{2.5}{8}}{1+\frac{2.5}{8}}=\frac{-1}{9}\)

Vậy \(P=\frac{-1}{9}\)

a) \(P=\frac{3x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

\(P=\frac{3\left(x-9\right)}{\left(x-3\right)\left(x-2\right)}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

\(P=\frac{3}{x-2}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

\(P=\frac{3\left(3-x\right)-\left(x+3\right)\left(3-x\right)-\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(3-x\right)}\)

\(P=\frac{9-3x-9+x^2-2x^2+4x-x+2}{\left(x-2\right)\left(3-x\right)}\)

\(P=\frac{2-x^2}{\left(x-2\right)\left(3-x\right)}\) (*)

b) Thay \(x=-\frac{1}{2}\) vào (*) ta có:

\(P=\frac{2-\left(-\frac{1}{2}\right)^2}{\left[\left(-\frac{1}{2}\right)-2\right]\left[3-\left(-\frac{1}{2}\right)\right]}=\frac{2-\frac{1}{4}}{-\frac{5}{2}.\frac{7}{2}}=-\frac{\frac{7}{4}}{\frac{5}{2}.\frac{7}{2}}=-\frac{7}{35}=-\frac{1}{5}\)

c) \(\frac{2-x^2}{\left(x-2\right)\left(3-x\right)}< 0\)

\(\Leftrightarrow2-x^2< 0\)

\(\Leftrightarrow-x^2< -2\)

\(\Leftrightarrow x^2>2\)

\(\Leftrightarrow\hept{\begin{cases}x< -\sqrt{2}\\-\sqrt{2}< x< \sqrt{2}\\x>2\end{cases}}\)

Vậy: ...

(x+5)(x²-6x+25)

x³ - x² - 5x + 125

=x³-6x²+25x+5x²-30x+125

=x(x²-6x+25)+5(x²-6x+25)

=(x+5)(x²-6x+25)

Mình thu gọn 2 đa thức trước r mới cộng nhé

\(P\left(x\right)=3x^2+7+2x^4-3x^2-4-5x+2x^3\)

\(P\left(x\right)=\left(3x^2-3x^2\right)+\left(7-4\right)+2x^4-5x+2x^3\)

\(P\left(x\right)=2x^4+2x^3-5x+3\)

\(Q\left(x\right)=-3x^3+2x^2-x^4+x+x^3+4x-2+5x^4\)

\(Q\left(x\right)=\left(-3x^3+x^3\right)+2x^2+\left(-x^4+5x^4\right)+\left(x+4x\right)-2\)

\(Q\left(x\right)=-2x^3+4x^4+2x^2+5x-2\)

\(P\left(x\right)+Q\left(x\right)=2x^4+2x^3-5x+3-2x^3+4x^4+2x^2+5x-2\)

\(P\left(x\right)+Q\left(x\right)=\left(2x^4+4x^4\right)+\left(2x^3-2x^3\right)+\left(-5x+5x\right)+\left(3-2\right)+2x^2\)

\(P\left(x\right)+Q\left(x\right)=6x^4+1+2x^2\)

@Nguyễn Huy Thắng@Mysterious Person@bảo nam trần@Lightning Farron@Thiên Thảo@Sky SơnTùng

a/ đk: \(\left[{}\begin{matrix}x\le\frac{-5-3\sqrt{5}}{10}\\x\ge\frac{-5+3\sqrt{5}}{10}\end{matrix}\right.\)\(\sqrt{x^2+x+1}+\sqrt{3x^2+3x+2}=\sqrt{5x^2+5x-1}\)

\(\Leftrightarrow\sqrt{x^2+x+1}+\sqrt{3\left(x^2+x+1\right)-1}=\sqrt{5\left(x^2+x+1\right)-6}\)

đặt\(x^2+x+1=t\left(t>0\right)\)

\(\sqrt{t}+\sqrt{3t-1}=\sqrt{5t-6}\)

bình phương 2 vế pt trở thành:

\(t+3t-1+2\sqrt{t\left(3t-1\right)}=5t-6\)

\(\Leftrightarrow2\sqrt{3t^2-t}=t-5\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left(2\sqrt{3t^2-t}\right)^2=\left(t-5\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\11t^2+6t-25=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left[{}\begin{matrix}t=\frac{-3+2\sqrt{71}}{11}\\t=\frac{-3-2\sqrt{71}}{11}\end{matrix}\right.\end{matrix}\right.\)=> không có gtri t nào t/m

vậy pt vô nghiệm

a/ ĐKXĐ: ...

Đặt \(x^2+x+1=a>0\)

\(\sqrt{a}+\sqrt{3a-1}=\sqrt{5a-6}\)

\(\Leftrightarrow4a-1+2\sqrt{3a^2-a}=5a-6\)

\(\Leftrightarrow2\sqrt{3a^2-a}=a-5\) (\(a\ge5\))

\(\Leftrightarrow4\left(3a^2-a\right)=a^2-10a+25\)

\(\Leftrightarrow11a^2+6a-25=0\)

Nghiệm xấu quá, chắc bạn nhầm lẫn đâu đó

b/

Đặt \(x^2+x+1=a>0\)

\(\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)

\(\Leftrightarrow2a+3+2\sqrt{a^2+3a}=2a+7\)

\(\Leftrightarrow\sqrt{a^2+3a}=2\)

\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x^2+x+1=1\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{3x^2-5x+7}=3-\sqrt{3x^2-7x+2}\)

\(\Rightarrow3x^2-5x+7=3x^2-7x+11-6\sqrt{3x^2-7x+2}\)

\(\Leftrightarrow3\sqrt{3x^2-7x+2}=2-x\) (\(x\le2\))

\(\Leftrightarrow9\left(3x^2-7x+2\right)=x^2-4x+4\)

\(\Leftrightarrow26x^2-59x+14=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{26}\end{matrix}\right.\)

Do biến đổi ko tương đương nên cần thay lại nghiệm vào pt ban đầu kiểm tra

d/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{x^2+3x+2}+\sqrt{x^2+6x+5}=\sqrt{2x^2+9x+7}\)

\(\Leftrightarrow2x^2+9x+7+2\sqrt{\left(x^2+3x+2\right)\left(x^2+6x+5\right)}=2x^2+9x+7\)

\(\Leftrightarrow\sqrt{\left(x+1\right)^2\left(x+2\right)\left(x+5\right)}=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\left(l\right)\\x=-5\end{matrix}\right.\)

P(x)=2x^4+2x^3-5x-4

Q(x)=4x^4-2x^3+2x^2+5x-2

P(x)+Q(x)

=2x^4+2x^3-5x-4+4x^4-2x^3+2x^2+5x-2

=6x^4+2x^2-6

2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)

\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)

Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)

\(\Rightarrow x=3\)

c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}\right)=0\)

\(\Rightarrow x=1\)

\(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}=0\)

bạn làm nốt pần này nhá