Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c, x^3 - y^3 = xy + 8
1) Nếu x-y <= -1
(x -y)(x^2 + xy + y^2) = xy +8
=> (x -y)(x^2 + xy + y^2) <= -(x^2 + xy +y^2)
=> xy +8 <= -(x^2 + xy +y^2)
=> (x+y)^2 + 8 <=0 => Vô nghiệm
2) Nếu x-y =0 => x=y , Vô nghiệm
3) x- y>=1
=> (x -y)(x^2 + xy + y^2) >= x^2 + xy + y^2
=> xy + 8 >= x^2 + xy + y^2
=> x^2 + y^2 <=8
=> x^2 <=8
=> x=0 => y= -2
=> x= 1 => y + y^3 + 7 =0 (loại)
a,\(x^2+2y^2+z^2-2xy-2y+2z+2=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2+2x+1\right)=0\)\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z+1\right)^1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-y=0\\y-1=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=1\\z=-1\end{matrix}\right.\)
\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}=\dfrac{2}{x+6}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{2}{x+6}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+5}=\dfrac{2}{x+6}\)
\(\Leftrightarrow\dfrac{4}{\left(x+1\right)\left(x+5\right)}=\dfrac{2}{x+6}\)
\(\Leftrightarrow2\left(x+6\right)=\left(x+1\right)\left(x+5\right)\)
\(\Leftrightarrow2x+12=x^2+6x+5\)
\(\Leftrightarrow x^2+4x-7=0\)
\(\Delta'=b'^2-ac\)
\(\Delta'=11\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b'+\sqrt{\Delta'}}{a}=-2+\sqrt{11}\\x_2=\dfrac{-b'-\sqrt{\Delta'}}{a}=-2-\sqrt{11}\end{matrix}\right.\)
\(1+\dfrac{1}{6}+\dfrac{120-x}{x}=\dfrac{120}{x}\)
\(1+\dfrac{1}{6}+\dfrac{126-\left(x+6\right)}{x+6}=\dfrac{120}{x}\)
\(1+\dfrac{1}{6}-1+\dfrac{126}{x+6}=\dfrac{120}{x}\)
\(\dfrac{1}{6}+\dfrac{126}{x+6}=\dfrac{120}{x}\)
\(\dfrac{126}{x+6}=\dfrac{120}{x}-\dfrac{1}{6}=\dfrac{120.6}{6x}-\dfrac{x}{6x}\)
\(\dfrac{126}{x+6}=\dfrac{126.6-x}{6x}\)
\(126.6.x=\left(126.6.-x\right)\left(x+6\right)\)ok
đk: x khác -6 ,làm toán là khôn khéo, bn tim msc vế trái =6(x+6)
có: (6(x+6) + (x+6) + 6(120-x)) /6(x+6) = 120/x
bây gio bn rut gon r cho tich trung tỷ = ngoai ty la tim dc x
\(\dfrac{x+1}{3}>\dfrac{2x-1}{6}-2\)
\(\Leftrightarrow2\left(x+1\right)>2x-1-12\)
\(\Leftrightarrow2x+2>2x-13\) \(\Leftrightarrow2x-2x>-13-2\)
\(\Leftrightarrow0x>-15\) ( luôn đúng)
Vậy bpt trên có vô số nghiệm
\(\Rightarrow\) k cần phải biểu diễn trên trục số
1)\(ĐKXĐ:x\ne0\)
Đặt \(\left(x+\dfrac{1}{x}\right)^2=a\)
\(\Rightarrow x^2+\dfrac{1}{x^2}=a-2\)
\(\Rightarrow VT=2a+\left(a-2\right)^2-\left(a-2\right)a\)
\(=2a+a^2-4a+4-a^2+2a=4\)
\(\Rightarrow\left(x+2\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loai\right)\\x=-4\end{matrix}\right.\)
b) \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+6\right)+7\left(x+6\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{x+7}{\left(x+4\right)\left(x+7\right)}-\dfrac{x+4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x+28-54=0\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow x\left(x-2\right)+13\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\) x - 2 = 0 hoặc x + 13 = 0
\(\Leftrightarrow\) x = 2 hoặc x = -13
Vậy x = 2 hoặc x = -13.
ĐKXĐ: \(xy\ne0\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{6xy}=\dfrac{1}{6}\)
\(\Rightarrow6x+6y+1=xy\)
\(\Leftrightarrow xy-6x-6y+36=37\)
\(\Leftrightarrow x\left(y-6\right)-6\left(y-6\right)=37\)
\(\Leftrightarrow\left(x-6\right)\left(y-6\right)=37\)
\(\Rightarrow\left(x-6;y-6\right)=\left(-37;-1\right);\left(-1;-37\right);\left(1;37\right);\left(37;1\right)\)
\(\Rightarrow\left(x;y\right)=\left(-31;5\right);\left(5;-31\right);\left(7;43\right);\left(43;7\right)\)