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\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
ĐKXĐ : \(x\ne2,x\ne4\)
Pt \(\Leftrightarrow\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-12\left(\frac{x-2}{x-4}\right)^2=0\) (2)
Đặt \(\frac{x+1}{x-2}=a,\frac{x-2}{x-4}=b\Rightarrow ab=\frac{x+1}{x-4}\)
Khi đó pt (2) trở thành :
\(a^2+ab-12b=0\)
\(\Leftrightarrow a^2-3ab+4ab-12b=0\)
\(\Leftrightarrow a\left(a-3b\right)+4b\left(a-3b\right)=0\)
\(\Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=3b\\a=-4b\end{cases}}\)
Bạn thay vào tính, được nghiệm là \(S=\left\{3,\frac{4}{3}\right\}\)
\(\left(x+5\right)^2-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x+5-3\right)\)
\(=\left(x+5\right)\left(x+2\right)\)
\(2x\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(2x-x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)\)
Ta có: (x^2 + 9y^2 + 4- 6xy -12y+ 4x)+(x^2 -10x+25) =0
(x-3y+2)^2 +(x-5)^2 =0
Vì vế trái luôn luôn lớn hơn hoặc bằng 0 với mọi x,y nên dấu"=" xảy ra khi:
x-3y+2 =0 và x-5=0
5-3y+2 =0 và x=5
y=7/3 và x=5
Vậy x=5 và y=7/3.
Chúc bạn học tốt.
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne3;x\ne-1\right)\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x^2-6x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=0\)
=> 2x=0
<=> x=0
Vậy x=0
+ Ta có: \(\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}=\frac{2x}{\left(x+1\right).\left(x-3\right)}\)\(\left(ĐKXĐ: x\ne-1, x\ne3\right)\)
\(\Leftrightarrow\frac{x.\left(x+1\right)+x.\left(x-3\right)}{2.\left(x-3\right).\left(x+1\right)}=\frac{4x}{2.\left(x-3\right).\left(x+1\right)}\)
\(\Rightarrow x^2+x+x^2-3x=4x\)
\(\Leftrightarrow\left(x^2+x^2\right)+\left(x-3x-4x\right)=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x.\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(TM\right)\\x=6\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{0,6\right\}\)
+ Ta có: \(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)\(\left(ĐKXĐ:x\ne1,x^2+x+1\ne0\right)\)
\(\Leftrightarrow\frac{\left(x^2+x+1\right)+2.\left(x-1\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1+2x-2=3x^2\)
\(\Leftrightarrow\left(x^2-3x^2\right)+\left(x+2x\right)+\left(1-2\right)=0\)
\(\Leftrightarrow-2x^2+3x-1=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right).\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(TM\right)\\x=1\left(L\right)\end{cases}}\)
Vậy \(S=\left\{\frac{1}{2}\right\}\)
\(a.\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=\)\(0\)
\(\Leftrightarrow\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2.\left(x+1\right).\left(x-3\right)}=0\)
\(\Leftrightarrow2x^2-6=0\)
\(\Leftrightarrow2x^2=6\)
\(\Leftrightarrow x^2=3\)
\(\Leftrightarrow x=\sqrt{3}\)
\(b.2x^3-5x^2+3x=0\)
\(\Leftrightarrow x.\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x.\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x.\left[2x.\left(x-1\right)-3.\left(x-1\right)\right]=0\)
\(\Leftrightarrow x.\left(x-1\right).\left(2x-3\right)=0\)
Đến đây tự làm nhé có việc bận
\(\left(x-2\right)^3+\left(5-2x\right)^3=0\)
\(\Leftrightarrow\left(x-2+5-2x\right)\left[\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right]=0\)
\(\Leftrightarrow\left(3-x\right)\left(x^2-4x+4+2x^2-9x+10+4x^2-20x+25\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(7x^2-33x+39\right)=0\)
\(\Leftrightarrow3-x=0\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
Wish you study well !!
\(\left(x-2\right)^3+\left(5-2x\right)^3=0\)
\(\Leftrightarrow\left(x-2+5-2x\right)\left(\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(x^2-4x+4-\left(5x-4x^2-10+4x\right)+25-20x+4x^2\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(x^2-4x+4-5x+4x^2+10-4x+25-20x+4x^2\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(9x^2-33x+39\right)=0\)
Phân tích tiếp nhé
Bạn ơi, mình chỉ làm đc đến đây rồi ko biết làm tiếp ntn đó