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\(a,A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(=x-0,2-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(=\left(-0,2-2+2\right)+\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)\)
\(=-0,2\)
\(b,B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(=x^3-8y^3-x^3+8y^3-10\)
\(=-10\)
\(c,C=4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)-4x\)
\(=4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=13\)
a) \(A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(A=x-\dfrac{1}{5}-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(A=\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)-\left(\dfrac{1}{5}+2-2\right)\)
\(A=-\dfrac{1}{5}\)
Vậy: ...
b) \(B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(B=\left[x^3-\left(2y\right)^3\right]-\left[x^3-\left(2y\right)^3\right]-10\)
\(B=-10\)
Vậy: ...
c) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)-4x\)
\(=4\left(x^2+2x+4\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=\left(4x^2+4x^2-8x^2\right)+\left(8x-4x-4x\right)+\left(4+1+8\right)\)
\(=13\)
Vậy:...
Hai tam giác vuông CAB và CFE đồng dạng (chung góc C)
\(\Rightarrow\dfrac{CF}{CA}=\dfrac{EF}{AB}=\dfrac{AD}{AB}=\dfrac{AD}{3}\)
\(\Rightarrow\dfrac{AC-AF}{AC}=\dfrac{AD}{3}\Leftrightarrow\dfrac{AC-2}{AC}=\dfrac{AD}{3}\Rightarrow AD=3\left(\dfrac{AC-2}{AC}\right)\)
\(S_{ABC}=\dfrac{1}{2}AB.AC=\dfrac{3}{2}AC\)
\(S_{ADEF}=AD.AF=2AD=6\left(\dfrac{AC-2}{AC}\right)\)
Theo đề bài: \(S_{ADEF}=\dfrac{1}{2}S_{ABC}\Rightarrow6\left(\dfrac{AC-2}{AC}\right)=\dfrac{1}{2}.\dfrac{3}{2}AC\)
\(\Leftrightarrow8\left(AC-2\right)=AC^2\Leftrightarrow AC^2-8AC+16=0\)
\(\Leftrightarrow\left(AC-4\right)^2=0\Leftrightarrow AC=4\)
Vậy \(S_{ABC}=\dfrac{1}{2}AB.AC=\dfrac{1}{2}.3.4=6\left(cm^2\right)\) \(\Rightarrow S_{ADEF}=3\)
\(=\dfrac{1}{\left(5x-1\right)\left(25x^2+5x+1\right)}\)
\(\dfrac{1}{125}x^3-1=\left(\dfrac{1}{5}x-1\right)\left(\dfrac{1}{25}x^2+\dfrac{1}{5}x+1\right)\)
CM:(n-1)^2(n+1)+(n-1)(n+1) chia hết cho 6 với 1 số nguyên n. Mng giúp mình vs ạ. Mình c.on nhiều ạaa
\(\left(n-1\right)^2\left(n+1\right)+\left(n-1\right)\left(n+1\right)\)
\(=\left(n-1\right)\left(n+1\right)\left[\left(n-1\right)+1\right]\)
\(=\left(n-1\right)\left(n+1\right)\left(n-1+1\right)\)
\(=n\left(n-1\right)\left(n+1\right)\)
Xét:
\(n\left(n-1\right)\) là hai số tự nhiên liên tiếp nên sẽ có số chẵn nên sẽ chia hết cho 2
\(n\left(n-1\right)\left(n+1\right)\) là 3 số tự nhiên liên tiếp nên sẽ chia hết cho 3
Mà: (2;3)=1 nên
\(n\left(n-1\right)\left(n+1\right)\) sẽ chia hết cho 2 x 3 = 6 (đpcm)
\(\left(n-1\right)^2\left(n+1\right)+\left(n-1\right)\left(n+1\right)\)
\(=\left(n-1\right)\left(n+1\right)\left(n-1+1\right)\)
\(=\left(n-1\right)n\left(n+1\right)\) là 3 số tự nhiên liên tiếp
\(\Rightarrow\left\{{}\begin{matrix}\left(n-1\right)n\left(n+1\right)⋮2\\\left(n-1\right)n\left(n+1\right)⋮3\end{matrix}\right.\)
\(\Rightarrow\left(n-1\right)n\left(n+1\right)⋮\left(2.3\right)\)
mà \(UCLN\left(2;3\right)=1\)
\(\Rightarrow\left(n-1\right)n\left(n+1\right)⋮6\)
\(\Rightarrow dpcm\)