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\(a,P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\left(x\ne\pm3;x\ne-2\right)\\ P=\dfrac{2x^2-7x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\\ P=\dfrac{3x^2+6x}{\left(x-3\right)\left(x+2\right)}=\dfrac{3x\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}=\dfrac{3x}{x-3}\\ b,x^2-7x+12=0\\ \Leftrightarrow\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow x=4\left(x\ne3\right)\\ \Leftrightarrow A=\dfrac{3\cdot4}{4-3}=12\\ c,P=\dfrac{3\left(x-3\right)+9}{x-3}=3+\dfrac{9}{x-3}\in Z\\ \Leftrightarrow x-3\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;4;6;12\right\}\)
a) \(\frac{x^2+2x+4}{4x^3-32}=\frac{x^2+2x+4}{4\left(x^3-8\right)}=\frac{x^2+2x+4}{4\left(x-2\right)\left(x^2+2x+4\right)}=\frac{1}{4\left(x-2\right)}.\)
b) \(\frac{10x-15}{4x^2-9}=\frac{5\left(2x-3\right)}{\left(2x\right)^2-3^2}=\frac{5\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}=\frac{5}{2x+3}.\)
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HAND!!!!
\(\frac{x^2+2x+4}{4x^3-32}=\frac{\left(x+2\right)^2}{4\left(x^3-8\right)}=\frac{\left(x+2\right)^2}{4\left(x-2\right)\left(x^2+2x+4\right)}=\frac{x+2}{4\left(x^2+2x+4\right)}.\)
\(\frac{10x-15}{4x^2-9}=\frac{5\left(2x-3\right)}{\left(2x\right)^2-3^2}=\frac{5\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}=\frac{5}{2x+3}\)
a. \(x\ne5\) là ĐKXĐ của biểu thức P
b. P =\(\dfrac{\left(x-5\right)^2}{x-5}\)=\(x-5\)
c. P = -1 <=> x-5 =-1 <=> x=4
a) \(\frac{2x^2-4x+8}{x^3+8}\Rightarrow\) ĐKXĐ: \(x^3+8\ne0 \Leftrightarrow x^3\ne-8 \Leftrightarrow x\ne-2 \)
b) \(\frac{2x^2-4x+8}{x^3+8}=\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{2}{x+2}\)
c) \(\frac{2}{x+2}\Rightarrow f\left(2\right)=\frac{2}{2+2}=\frac{1}{2}\)
d) \(\frac{2}{x+2}=2\)
\(\Leftrightarrow x+2=1\)
\(\Leftrightarrow x=-1\)
uum, mik nghĩ phần C chỗ x+2=1 thì phải gt tại sao x+2=1 thì đúng hơn
\(P=\dfrac{3x^2+6x+3}{x+1}\)
\(a,\) Điều kiện xác định: \(x+1\ne0\Leftrightarrow x\ne-1\)
\(b,P=\dfrac{3x^2+6x+3}{x+1}=\dfrac{3\left(x^2+2x+1\right)}{x+1}=\dfrac{3\left(x+1\right)^2}{x+1}=3\left(x+1\right)=3x+3\)
\(c,x=1\Rightarrow P=3.1+3=6\)
Bài 1 :
\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)
\(=x^2-4x+4-x+9=x^2-5x+13\)
Bài 2 :
a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)
\(=\frac{-\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)^2}=\frac{-\left(2x+1\right)}{2x-1}=\frac{-2x-1}{2x-1}\)
b, Thay x = -4 ta được :
\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)
a:
ĐKXĐ: \(x\notin\left\{5;-5;-1;0\right\}\)
\(P=\left(\dfrac{15-x}{x^2-25}+\dfrac{2}{x+5}\right):\dfrac{x+1}{2x^2-10x}\)
\(=\left(\dfrac{15-x}{\left(x-5\right)\left(x+5\right)}+\dfrac{2}{x+5}\right)\cdot\dfrac{2x\left(x-5\right)}{x+1}\)
\(=\dfrac{15-x+2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\cdot\dfrac{2x\left(x-5\right)}{x+1}\)
\(=\dfrac{x+5}{\left(x+5\right)}\cdot\dfrac{2x}{x+1}=\dfrac{2x}{x+1}\)
b: Thay x=1 vào P, ta được:
\(P=\dfrac{2\cdot1}{1+1}=\dfrac{2}{2}=1\)
ah giúp em bài toán lớp 6 em đăng trên trang của em đc ko ạ?