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Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
\(\frac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)
\(=\frac{1}{2}x^2.6x+\frac{1}{2}x^2.\left(-3\right)+\left(-x\right).x^2+\left(-x\right).\frac{1}{2}+\frac{1}{2}.x+\frac{1}{2}.4\)
\(=3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2\)
\(=\left(3x^3-x^3\right)-\frac{3}{2}x^2+\left(-\frac{1}{2}x+\frac{1}{2}x\right)+2\)
\(=2x^3-\frac{3}{2}x^2+2\)
\(a,\)\(\frac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)
\(=3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2\)
\(=2x^3-\frac{3}{2}x^2+2\)
\(b,\)\(2x\left(3x^3-x\right)-4x^2\left(x-x^2+1\right)+\left(x-3x^2\right)x\)
\(=6x^4-2x^2-4x^3+4x^4-4x^2+x^2-3x^3\)
\(=10x^4-7x^3-5x^2\)
b/ \(3-100x+8x^2=8x^2+x-300\)
\(\Leftrightarrow-101x=-303\)
\(\Rightarrow x=3\)
c/ \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-79x=-158\)
\(\Rightarrow x=2\)
d/ \(3\left(3x+2\right)-\left(3x+1\right)=12x+10\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
\(\Leftrightarrow-6x=5\)
\(\Rightarrow x=-\frac{5}{6}\)
e/ \(30x-6\left(2x-5\right)+5\left(x+8\right)=210+10\left(x-1\right)\)
\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)
\(\Leftrightarrow13x=130\)
\(\Rightarrow x=10\)
\(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)
\(\Rightarrow A_{min}=-3\) khi \(x=2\)
\(B=4x^2+4x+11=\left(2x+1\right)^2+10\ge10\)
\(\Rightarrow B_{min}=10\) khi \(x=-\frac{1}{2}\)
\(C=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
\(\Rightarrow C_{min}=-36\) khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(D=-x^2-8x-16+21=21-\left(x+4\right)^2\le21\)
\(\Rightarrow C_{max}=21\) khi \(x=-4\)
\(E=-x^2+4x-4+5=5-\left(x-2\right)^2\le5\)
\(\Rightarrow E_{max}=5\) khi \(x=2\)
a) \(\frac{7x}{8}-5\left(x-9\right)=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40\left(x-9\right)}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{7x}{8}-\frac{40x-360}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow\frac{360-33x}{8}=\frac{20x+1,5}{6}\)
\(\Leftrightarrow2160-198x=160x+12\)
\(\Leftrightarrow358x=2148\)
\(\Leftrightarrow x=6\)
Vậy nghiệm của pt x=6
b) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
\(\Leftrightarrow\frac{10\left(x-1\right)+4}{12}-\frac{21x-3}{12}=\frac{4x+2}{7}-\frac{35}{7}\)
\(\Leftrightarrow\frac{-11x-3}{12}=\frac{4x-33}{7}\)
\(\Leftrightarrow-77x-21=48x-396\)
\(\Leftrightarrow125x=375\)
\(\Leftrightarrow3\)
Vậy nghiệm của pt x=3