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Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
a) \(4x^2\left(5x^3-2x+3\right)\)
\(=20x^5-8x^3+12x^2\)
b) \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)\)
\(=12y^5+2y^4-y^2\)
c) \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-14x^2+8x\)
d) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+22x-55-6x^2-23x-21\)
\(=-x-76\)
1, \(4x^2\left(5x^3-2x+3\right)=20x^5-8x^3+12x^2\)
2, \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)=12y^5+2y^4-y^2\)
3, \(\left(5x^2-4x\right)\left(x-2\right)=5x^3-10x^2-4x^2+8x=5x^3-14x^2+8x\)
4, \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)=6x^2+33x-10x-55-\left(6x^2+14x+9x+21\right)\)
\(=6x^2+23x-55-6x^2-23x-21=-76\)
A = 5(x + 3)(x - 3) + (2x + 3)3 + (x - 6)2
A = 5(x + 3)(x - 3) + 4x2 + 12x + 9 + x2 - 12x + 36
A = 5x2 - 45x + 4x2 + 12x + 9 + x2 - 12x + 36
A = 10x2 (1)
Thay x = -1/5 vào (1), ta có:
A = 10x2 = 10.(-1/5)2 = 2/5
A = 2/5
Vậy:...
a)
\(P=\left(\frac{2x}{2x^2-5x+3}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right)\)
\(P=\left(\frac{2x}{\left(2x-3\right)\left(x-1\right)}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right)\)
\(P=\left(\frac{2x}{\left(2x-3\right)\left(x-1\right)}-\frac{5(x-1)}{\left(2x-3\right)\left(x-1\right)}\right):\left(3-\frac{2}{x-1}\right)\)
\(P=\frac{2x-5(x-1)}{(2x-3)(x-1)}:\frac{3\left(x-1\right)-2}{x-1}\)
\(P=\frac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\frac{3x-3-2}{x-1}\)
\(P=\frac{-3x+5}{\left(2x-3\right)\left(x-1\right)}:\frac{3x-5}{x-1}\)
\(P=\frac{-3x+5}{\left(2x-3\right)\left(x-1\right)}\cdot\frac{x-1}{3x-5}\)
\(P=\frac{-\left(3x-5\right)}{\left(2x-3\right)\left(x-1\right)}\cdot\frac{x-1}{3x-5}\)
\(P=\frac{-1}{2x-3}\)
Vậy \(P=\frac{-1}{2x-3}\)
b)
\(\vert3x-2\vert+1=5\)
\(\Rightarrow\vert3x-2\vert=4\)
\(\Rightarrow\begin{cases}3x-2=4\\ 3x-2=-4\end{cases}\)
\(\Rightarrow\begin{cases}3x=6\\ 3x=-2\end{cases}\)
\(\Rightarrow\begin{cases}x=2\\ x=-\frac23\end{cases}\)
+) TH1: Với \(x=2\) thì \(P=\frac{-1}{2\cdot2-3}=\frac{-1}{4-3}=\frac{-1}{1}=-1\)
+) TH2: Với \(x=-\frac23\) thì \(P=\frac{-1}{2\cdot\left(-\frac23\right)-3}=\frac{-1}{-\frac43-3}=\frac{-1}{-\frac{13}{3}}=\frac{3}{13}\)
Vậy \(P=-1\) khi \(x=2\) , \(P=\frac{3}{13}\) khi \(x=-\frac23\)
a) Sau khi rút gọn, ta có \(P = \frac{- 1}{2 x - 3}\).
b) Ở câu này có 2 trường hợp nha bạn:
+ Trường hợp 1: \(3 x - 2 = 4\) \(3 x = 6\) \(x = 2\) + Trường hợp 2: \(3 x - 2 = - 4\) \(3 x = - 2\) \(x = - \frac{2}{3}\)Kết quả: khi \(x = 2\) thì \(P = - 1\) và khi \(x = - \frac{2}{3}\) thì \(P = \frac{3}{13}\).