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\(a.\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=\)\(0\)
\(\Leftrightarrow\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2.\left(x+1\right).\left(x-3\right)}=0\)
\(\Leftrightarrow2x^2-6=0\)
\(\Leftrightarrow2x^2=6\)
\(\Leftrightarrow x^2=3\)
\(\Leftrightarrow x=\sqrt{3}\)
\(b.2x^3-5x^2+3x=0\)
\(\Leftrightarrow x.\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x.\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x.\left[2x.\left(x-1\right)-3.\left(x-1\right)\right]=0\)
\(\Leftrightarrow x.\left(x-1\right).\left(2x-3\right)=0\)
Đến đây tự làm nhé có việc bận
Bài 2:
a) \(x^2-y^2+3x-3y=\left(x^2-y^2\right)+\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
b) \(5x-5y+x^2-2xy+y^2=\left(5x-5y\right)+\left(x^2-2xy+y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)^2=\left(x-y\right)\left(x-y+5\right)\)
c) \(x^2-5x+4=x^2-x-4x+4=\left(x^2-x\right)-\left(4x-4\right)\)
\(=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
Đặt \(a=x^2-3x+5;b=x^2-3x-1\Rightarrow a-b=6.\)
Đặt biểu thức đã cho là A
\(\Rightarrow A=a^2-2ab+b^2=\left(a-b\right)^2=6^2=36\)
=> Biểu thức A không phụ thuộc vào biến x (đpcm)
(x2-3x+5)2-2(x2-3x+5)(x2-3x-1)+(x2-3x-1)2=(x2-3x-5-x2+3x+1)2=(-4)2=42
=> Không phụ thuộc vào x
Chúc bạn học tốt !
\(ab-a+b-a^2=\left(ab-a^2\right)+\left(b-a\right)=a.\left(b-a\right)+\left(b-a\right)=\left(b-a\right).\left(a+1\right)\\ \)
\(8-\left(x-1\right)^3=2^3-\left(x-1\right)^3=\left(2-x+1\right).\left(2^2+2.\left(x-1\right)+\left(x-1\right)^2\right)\)
\(=\left(3-x\right).\left(4+2x-2+x^2-2x+1\right)=\left(3-x\right).\left(3+x^2\right)\)
\(3x^2-8x-16=\left(3x^2+4x\right)-\left(12x+16\right)=x.\left(3x+4\right)-4.\left(3x+4\right)=\left(3x+4\right).\left(x-4\right)\)
\(ab-a+b-a^2\)
\(=\left(ab-a^2\right)+\left(b-a\right)\)
\(=a\left(b-a\right)+\left(b-a\right)\)
\(=\left(a+1\right)\left(b-a\right)\)
\(\left(x^2-x\right)-\left(2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow x=2;x=1\)
\(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy x=1; x=2