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a) = \(\frac{2x}{\left(x-2\right)\left(x-3\right)}\)-\(\frac{1}{\left(x-2\right)\left(x-3\right)}\)
các bài sau tt
Bài 1 yêu cầu gì em?
Bài 2:
\(a,x\left(x-1\right)+5\left(x-1\right)=\left(x+5\right)\left(x-1\right)\\ b,3x\left(x+1\right)+3\left(x+1\right)=\left(3x+3\right)\left(x+1\right)=3\left(x+1\right)\left(x+1\right)=3\left(x+1\right)^2\\ c,x\left(x-3\right)+xy\left(x-3\right)=\left(x+xy\right)\left(x-3\right)=x\left(y+1\right)\left(x-3\right)\\ d,2x\left(x-2\right)-6\left(x-2\right)=\left(2x-6\right)\left(x-2\right)=2\left(x-3\right)\left(x-2\right)\)
Bài 1:
a) \(3xy+6y\)
\(=3y\left(x+2\right)\)
b) \(3x^2+9x\)
\(=3x\left(x+3\right)\)
c) \(6x-9y^2\)
\(=3\left(2x-3y^2\right)\)
d) \(10xy^2-6x^2y\)
\(=2xy\left(5y-3x\right)\)
Bài 2:
a) \(x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x+5\right)\)
b) \(3x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(3x+3\right)\)
\(=3\left(x+1\right)\left(x+1\right)\)
\(=3\left(x+1\right)^2\)
c) \(x\left(x-3\right)+xy\left(x-3\right)\)
\(=\left(x+xy\right)\left(x-3\right)\)
\(=x\left(1+y\right)\left(x-3\right)\)
d) \(2x\left(x-2\right)-6\left(x-2\right)\)
\(=\left(2x-6\right)\left(x-2\right)\)
\(=2\left(x-3\right)\left(x-2\right)\)
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
a)
\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}=\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}\)
\(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\)
MTC: \(2\left(x-1\right)\left(x+1\right)\left(x-5\right)\)
\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}\\ =\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}=\dfrac{2\left(x+1\right)\left(3x-6\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)
\(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x-5\right)\left(5x-5\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)
Câu 5: B
Câu 6:
a: ĐKXĐ: \(x-2\ne0\)
=>\(x\ne2\)
b: ĐKXĐ: \(x+1\ne0\)
=>\(x\ne-1\)
8:
\(A=\dfrac{x^2+4}{3x^2-6x}+\dfrac{5x+2}{3x}-\dfrac{4x}{3x^2-6x}\)
\(=\dfrac{x^2+4-4x}{3x\left(x-2\right)}+\dfrac{5x+2}{3x}\)
\(=\dfrac{\left(x-2\right)^2}{3x\left(x-2\right)}+\dfrac{5x+2}{3x}\)
\(=\dfrac{x-2+5x+2}{3x}=\dfrac{6x}{3x}=2\)
7:
\(\dfrac{8x^3yz}{24xy^2}\)
\(=\dfrac{8xy\cdot x^2z}{8xy\cdot3y}\)
\(=\dfrac{x^2z}{3y}\)
Chọn D