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`a,x^3-8 ne 0`
`=>x^3 ne 8`
`=>x ne 2`
`b,2x^2+5x+3 ne 0`
`=>2x^2+2x+3x+3 ne 0`
`=>2x(x+1)+3(x+1) ne 0`
`=>(x+1)(2x+3) ne 0`
`=>x ne -1,-3/2`
`c,x^2-4 ne 0`
`=>x^2 ne 4`
`=>x ne 2,-2`
a) ĐK:
\(x^3-8\ne0\\ \Leftrightarrow x\ne2\)
b) ĐK:
\(2x^2+5x+3\ne0\\ \Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne-\dfrac{3}{2}\end{matrix}\right.\)
c) ĐK:
\(x^2-4\ne0\\ \Leftrightarrow x\ne\pm2\)
a: \(A=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
b: \(B=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)
c: \(C=\dfrac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\dfrac{3x+4}{x}\)
d: \(D=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}=\dfrac{x+2}{2}\)
e: \(E=\dfrac{-x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{-x}{x+2}\)
f: \(F=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
6:
a: ĐKXĐ: x<>0
\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)
\(=\dfrac{\left(x+1\right)^3}{x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x}\)
b: ĐKXĐ: x<>1
\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)
\(=\dfrac{\left(x-1\right)^3}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{2}\)
c: ĐKXĐ: x<>-2
\(\dfrac{x^2+4x+4}{2x+4}\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)
\(=\dfrac{x+2}{2}\)
d: ĐKXĐ: x<>-2
\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)
\(=\dfrac{\left(-x+1\right)\left(x+2\right)}{x+2}=-x+1\)
e: ĐKXĐ: x<>-y
\(\dfrac{x^2-y^2}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{x+y}=x-y\)
g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)
\(=\dfrac{3x\left(x+2\right)}{\left(x+2\right)\cdot\left(x-2\right)}=\dfrac{3x}{x-2}\)
7:
a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)
\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)
b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)
\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)
c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)
d:
\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)
\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1}{x+3}=\dfrac{x-3}{\left(x+3\right)\left(x-3\right)}\)
a, Để A xác định
\(\Rightarrow\hept{\begin{cases}3x^2\ne0\\4x\ne0\end{cases}}\Rightarrow x\left(3x-4\right)\ne0\)
\(\Rightarrow x\ne0\Rightarrow3x-4\ne0\Rightarrow x\ne0\)
b Để \(B=\frac{x^2+4x+4}{2x+4}=\frac{\left(x+2\right)^2}{2x+4}\)xác định
\(\Rightarrow\hept{\begin{cases}2x\ne0\\4\ne0\end{cases}}\Rightarrow2\left(x+2\right)\ne0\)
\(\Rightarrow x+2\ne0\Rightarrow x\ne-2\)
c,, Để \(C=\frac{2x-x^2}{x^2-4}=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
d, Để D xác định :
\(\Rightarrow\hept{\begin{cases}x^3\ne0\\8\ne0\end{cases}}\Rightarrow x^3-8\ne0\)
\(\Rightarrow x\ne2\)
\(A=\dfrac{2x+6}{\left(x+3\right)\left(x-2\right)}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x+3\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
A=0 <=> 2x+6=0 <=> x=-3(ko tm đkxđ) => ko có x để A=0
\(A=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
\(B=\dfrac{x^2-9}{x^2-6x+9}=\dfrac{x^2-9}{\left(x-3\right)^2}\)
ĐKXĐ: (x-3)2 \(\ne\)0 <=> \(x\ne3\)
B=0 <=> x2-9=0 <=> x=3(ko tm đkxđ) or x=-3(tm đkxđ)=>x=-3
\(B=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)
A.
Để phân thức A xác định <=> x - 2 # 0
<=> x #2
B. để phân thức B xác định <=> x^2 - 6x #0
<=> x(x-6) #0
<=> | x #0 <=> |x #0
| x - 6 #0 | x # 6