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\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
a) 4x -8 ≥ 3(3x-1)-2x +1
⇒4x -8 ≥7x -2
⇒4x -7x ≥ -2 +8
⇒-3x ≥ 6
⇒x≤-2
Vậy bpt có nghiệm là:{x|x≤-2}
b) (x-3)(x+2)+(x+4)2≤ 2x (x+5)+4
⇔ x2+2x - 3x - 6 +x2 + 8x +16≤ 2x2 + 10x +4
⇔ x2 +2x - 3x + x2 + 8x - 2x2- 10x ≤ 4+6-16
⇔ -3x ≤ -6
⇔ x≥ 2
Vậy bpt có tập nghiệm là: {x|x≥2}
câu nào cũng ghi lại đề nha
a) \(x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b)\(x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\dfrac{1}{x-2}+3-\dfrac{3-x}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3\left(x-2\right)-\left(3-x\right)}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3x-6-3+x}{x-2}=0\) ( đk \(x\ne2\) )
\(\Leftrightarrow4x-8=0\Rightarrow x=2\)
đ) \(\dfrac{8-x}{x-7}-8-\dfrac{1}{x-7}=0\)
\(\Leftrightarrow\dfrac{8-x-8\left(x-7\right)-1}{x-7}=0\) (đk \(x\ne7\))
\(\Leftrightarrow8-x-8x+56-1=0\)
\(\Leftrightarrow-9x+63=0\)
\(\Leftrightarrow x=7\)
Cái này áp dụng hằng đẳng thức 100%
a, \(\left(3x-1\right)^3=27x^3-3.9x^2+3.3x-1=27x^3-27x^2+9x-1\)
b, \(\left(4x-\dfrac{1}{2}\right)^2=16x^2-2.4x.\dfrac{1}{2}+\dfrac{1}{4}=16x^2-4x+\dfrac{1}{4}\)
c, \(\left(\dfrac{1}{3}x+1\right)^3=\dfrac{1}{27}x^3+3.\dfrac{1}{9}x^2+3.\dfrac{1}{3}x+1=\dfrac{1}{27}x^3+\dfrac{1}{3}x^2+x+1\)
d, \(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}x^2+2.\dfrac{2}{3}x.\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{4}{9}x^2+\dfrac{2}{3}x+\dfrac{1}{4}\)
e, \(x^6-1=\left(x^3\right)^2-1=\left(x^3-1\right)\left(x^3+1\right)\)
f, \(27x^3+8=\left(3x\right)^3+2^3=\left(3x+2\right)\left(9x^2-6x+4\right)\)
g, \(9x^2-4=\left(3x\right)^2-2^2=\left(3x-2\right)\left(3x+2\right)\)
a) \(\left(3x-1\right)^3=21x^3-27x^2+9x-1\)
b) \(\left(4x-\dfrac{1}{2}\right)^2=16x^2-4x+\dfrac{1}{4}\)
c) \(\left(\dfrac{1}{3}x+1\right)^3=\dfrac{1}{27}x^3+\dfrac{1}{3}x^2+x+1\)
d) \(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}x^2+\dfrac{2}{3}x+\dfrac{1}{4}\)
e) \(x^6-1=\left(x^3\right)^2-1=\left(x^3-1\right)\times\left(x^3+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
f) \(27x^3+8=\left(3x+2\right)\left(9x^2-6x+4\right)\)
g) \(9x^2-4=\left(3x-2\right)\left(3x+2\right)\)