\(\dfrac{x^2+3x}{x^2+5x+6}\)
rút gọn phân thức
Rút gọn các phân thức: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}\)
ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)
Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)
Rút gọn phân thức :
\(1)\dfrac{x\left|x-2\right|}{x^2-5x+6}\)
\(2)\dfrac{a^{2x}-b^{2x}}{a^x-b^x}\)
1) Ta có: \(\dfrac{x\left|x-2\right|}{x^2-5x+6}\)
\(=\left[{}\begin{matrix}\dfrac{-x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\left(x< 2\right)\\\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\left(x>2\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}\dfrac{-x}{x-3}\\\dfrac{x}{x-3}\end{matrix}\right.\)
2) Ta có: \(\dfrac{a^{2x}-b^{2x}}{a^x-b^x}\)
\(=\dfrac{\left(a^x\right)^2-\left(b^x\right)^2}{a^x-b^x}\)
\(=\dfrac{\left(a^x-b^x\right)\left(a^x+b^x\right)}{a^x-b^x}=a^x+b^x\)
Rút gọn biểu thức M = \(\dfrac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}:2\sqrt{1+\dfrac{2x}{3-x}}\)
\(M=\dfrac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}:2\sqrt{\dfrac{3-x+2x}{3-x}}\left(-3\le x< 3;x\ne-1\right)\\ M=\dfrac{\sqrt{x+3}\left(x+2+x\sqrt{3-x}\right)}{\sqrt{3-x}\left[x+\left(x+2\right)\sqrt{3+x}\right]}:2\sqrt{\dfrac{x+3}{3-x}}\\ M=\dfrac{\sqrt{x+3}\left(x+2+x\sqrt{3-x}\right)}{\sqrt{3-x}\left[x+\left(x+2\right)\sqrt{3+x}\right]}\cdot\dfrac{3-x}{2\sqrt{\left(3-x\right)}\sqrt{\left(x+3\right)}}\)
\(M=\dfrac{x+2+x\sqrt{3-x}}{x+\left(x+2\right)\sqrt{3-x}}\cdot\dfrac{\sqrt{3-x}}{2\sqrt{3-x}}\\ M=\dfrac{\left(x+2\right)\sqrt{3-x}+x\left(3-x\right)}{2x\sqrt{3-x}+2\left(x+2\right)\sqrt{3-x}}\\ M=\dfrac{\sqrt{3-x}\left(2x+2\right)}{\sqrt{3-x}\left(2x+2x+4\right)}=\dfrac{2\left(x+1\right)}{4\left(x+1\right)}=\dfrac{1}{2}\)
1a. rút gọn biểu thức sau A = \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{4-9x^2}\)
b. biến đổi biểu thức sau thành phân thức đại số B = \(\dfrac{1}{2}+\dfrac{x}{1-\dfrac{x}{x+2}}\)
\(a,A=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\\ b,B=\dfrac{1}{2}+\dfrac{x}{\dfrac{x+2-x}{x+2}}=\dfrac{1}{2}+\dfrac{x}{\dfrac{2}{x+2}}=\dfrac{1}{2}+\dfrac{x\left(x+2\right)}{2}\\ B=\dfrac{1+x^2+2x}{2}=\dfrac{\left(x+1\right)^2}{2}\)
Cho biểu thức B =(\(\dfrac{x^3}{x^3-4x}+\dfrac{6}{^{6-3x}}+\dfrac{1}{2+x}\)): (x+2+\(\dfrac{10-x^2}{x-2}\))
a) Rút gọn B
b) Tìm B biết x2-5x+6=0
c) Tìm x ∈ Z để B ∈ Z
d) Tìm x biết |B|>1
Cho biểu thức :P= \(\dfrac{1}{x^2-x}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}\)
a, Tìm điều kiện của x để biểu thức P có giá trị .
b, Rút gọn biểu thức P.
a, ĐK : \(x\ne1;2;3;4;5\)
b, \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)
\(=\dfrac{1}{x}-\dfrac{1}{x-5}=\dfrac{x-5-x}{x\left(x-5\right)}=\dfrac{-5}{x\left(x-5\right)}\)
Cho biểu thức P= \(\dfrac{1}{x^2-x}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}\)
a) tìm điều kiện của x để biểu thức P có giá trị
b) rút gọn biểu thúc P
a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)
b: \(P=\dfrac{1}{\left(x-1\right)\cdot x}+\dfrac{1}{\left(x-2\right)\left(x-1\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}\)
\(=\dfrac{1}{x-5}-\dfrac{1}{x}=\dfrac{x-x+5}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}\)
Cho biểu thức: K=(\(\dfrac{x^2}{x^2-5x+6}\)+\(\dfrac{x^2}{x^2-3x+2}\)).\(\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
a, Tìm đkxđ rồi rút gọn K
b, Tìm GTLN của K
a: ĐKXĐ: x<>1; x<>2; x<>3
\(K=\left(\dfrac{x^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{x^2}{\left(x-1\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+2x^2+1-x^2}\)
\(=\dfrac{x^3-x^2+x^3-3x^2}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}\)
\(=\dfrac{2x^3-4x^2}{\left(x-2\right)}\cdot\dfrac{1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{2x^2\left(x-2\right)}{\left(x-2\right)\left(x^4+x^2+1\right)}=\dfrac{2x^2}{x^4+x^2+1}\)
b:
rút gọn rồi tính giá trị biểu thức
\(\dfrac{3x^2-12x+12}{x^2-4}\) tại x=\(-\dfrac{1}{4}\)
\(\dfrac{x^2-5x-6}{x^2-9}\) tại x=-1
\(\dfrac{x^2-9y^2}{x^2-6xy+9y^2}\) tại x=1, y=-\(\dfrac{2}{3}\)
a) Ta có: \(\dfrac{3x^2-12x+12}{x^2-4}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)}{x+2}\)
\(=\dfrac{3\cdot\left(\dfrac{-1}{4}-2\right)}{\dfrac{-1}{4}+2}=-\dfrac{27}{7}\)
b) Ta có: \(\dfrac{x^2-5x-6}{x^2-9}\)
\(=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(-1-6\right)\left(-1+1\right)}{\left(-1-3\right)\left(-1+3\right)}\)
=0