''TẮM LÀ GÌ?''

b) \(2y-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)

\(=\dfrac{2y\left(3x+2y\right)}{3x+2y}-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)

\(=\dfrac{2y\left(3x+2y\right)-\left(6xy+2y\right)+\left(2y-9x^2\right)}{3x+2y}\)

\(=\dfrac{6xy+4y^2-6xy-2y+2y-9x^2}{3x+2y}\)

\(=\dfrac{4y^2-9x^2}{3x+2y}\)

\(=\dfrac{-\left(9x^2-4y^2\right)}{3x+2y}\)

\(=\dfrac{-\left[\left(3x\right)^2-\left(2y\right)^2\right]}{3x+2y}\)

\(=\dfrac{-\left(3x-2y\right)\left(3x+2y\right)}{3x+2y}\)

\(=-\left(3x-2y\right)\)

\(=-3x+2y\)

Theo đề ta có:

\(n+5⋮n+1\)

\(\Rightarrow\left(n+1\right)+4⋮n+1\)

\(\Rightarrow4⋮n+1\)

\(\Rightarrow n+1\inƯ\left(4\right)=\left\{1;2;4\right\}\) ( vì \(n\in N\) )

\(\Rightarrow\left\{{}\begin{matrix}n+1=1\Rightarrow n=0\\n+1=2\Rightarrow n=1\\n+1=4\Rightarrow n=3\end{matrix}\right.\)

Vậy \(n\in\left\{0;1;3\right\}\)

a) Vì tam giác ABC vuông cân tại A nên: \(\widehat{A}=90^o\) và \(AB=AC\).

M là trung điểm của BC \(\Rightarrow BM=MC=\dfrac{BC}{2}\)

Tam giác ABC có \(\widehat{A}=90^o\)

\(\Rightarrow AM=BM=MC=\dfrac{BC}{2}\) ( định lý cạnh huyền và đường trung tuyến của tam giác vuông )

Xét \(\Delta AMB\) và \(\Delta AMC\) có:

\(AB=AC\left(cmt\right)\)

\(BM=MC\left(cmt\right)\)

\(AM:\) cạnh chung

\(\Rightarrow\Delta AMB=\Delta AMC\left(c.c.c\right)\)

\(\Rightarrow\widehat{AMC}=\widehat{AMB}\) ( hai góc tương ứng )

Ta có: \(\widehat{AMC}+\widehat{AMB}=180^o\) ( kề bù )

Mà \(\widehat{AMC}=\widehat{AMB}\left(cmt\right)\)

\(\widehat{\Rightarrow AMC}=\widehat{AMB}=\dfrac{180^o}{2}=90^o\)

Ta có: \(\widehat{AMC}=90^o\) và \(AM=MC\)

\(\Rightarrow\Delta AMC\) vuông cân tại M

tự sát là tốt nhất

\(100^2-99^2+98^2-97^2+......+4^2-3^2+2^2-1^2\)

\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+......+\left(4^2-3^2\right)+\left(2^2-1^2\right)\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+......+\left(4-3\right)\left(4+3\right)+\left(2-1\right)\left(2+1\right)\)

\(=1.\left(100+99\right)+1.\left(98+97\right)+......+1.\left(4+3\right)+1.\left(2+1\right)\)

\(=199+195+......+7+3\)

\(=\dfrac{\left(199+3\right)\left[\left(199-3\right):4+1\right]}{2}\)

\(=\dfrac{202.\left(196:4+1\right)}{2}\)

\(=\dfrac{202.\left(49+1\right)}{2}\)

\(=\dfrac{202.50}{2}\)

\(=\dfrac{10100}{2}\)

\(=5050\)

\(\left(2-\dfrac{5}{6}x\right):\left(-2\right)+\dfrac{1}{3}x=-0,91\)

\(\Leftrightarrow\dfrac{2}{-2}-\dfrac{\dfrac{5}{6}x}{-2}+\dfrac{1}{3}x=-0,91\)

\(\Leftrightarrow-1+\dfrac{5}{12}x+\dfrac{1}{3}x=-0,91\)

\(\Leftrightarrow-1+x\left(\dfrac{5}{12}+\dfrac{1}{3}\right)=-0,91\)

\(\Leftrightarrow-1+\dfrac{3}{4}x=-0,91\)

\(\Leftrightarrow\dfrac{3}{4}x=-0,91-\left(-1\right)\)

\(\Leftrightarrow\dfrac{3}{4}x=-0,91+1\)

\(\Leftrightarrow\dfrac{3}{4}x=-\left(0,91-1\right)\)

\(\Leftrightarrow\dfrac{3}{4}x=-\left(-0,09\right)\)

\(\Leftrightarrow\dfrac{3}{4}x=0,09\)

\(\Leftrightarrow x=0,09:\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{3}{25}\)

Vậy \(x=\dfrac{3}{25}\)

c) Ta có:

\(10x+23⋮2x+1\)

\(\Rightarrow\left(10x+5\right)+18⋮2x+1\)

\(\Rightarrow5\left(2x+1\right)+18⋮2x+1\)

\(\Rightarrow18⋮2x+1\)

\(\Rightarrow2x+1\inƯ\left(18\right)=\left\{1;2;3;6;9;18\right\}\) ( vì \(x\in N\) )

+) \(2x+1=1\Rightarrow x=0\left(thoa\right)\)

+) \(2x+1=2\Rightarrow x=0,5\left(loai\right)\)

+) \(2x+1=3\Rightarrow x=1\left(thoa\right)\)

+) \(2x+1=6\Rightarrow x=2,5\left(loai\right)\)

+) \(2x+1=1\Rightarrow x=4\left(thoa\right)\)

+) \(2x+1=1\Rightarrow x=8,5\left(loai\right)\)

Vậy \(x\in\left\{0;1;4\right\}\)

Bài 8: Phân tích đa thức thành nhân tử

1) \(5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)

\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=5\left(x-y-2x\right)\left(x-y+2x\right)\)

2 \(16x-5x^2-3\)

\(=-5x^2+16x-3\)

\(=-5x^2+15x+x-3\)

\(=-\left(5x^2-15x\right)+\left(x-3\right)\)

\(=-5x\left(x-3\right)+\left(x-3\right)\)

\(=\left(x-3\right)\left(-5x+1\right)\)

3) \(x^2-5x+5y-y^2\)

\(=\left(x^2-y^2\right)-\left(5x-5y\right)\)

\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-5\right)\)

4) \(3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)

\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=3\left(x-y-2x\right)\left(x-y+2x\right)\)

5) \(x^2+4x+3\)

\(=x^2+x+3x+3\)

\(=\left(x^2+x\right)+\left(3x+3\right)\)

\(=x\left(x+1\right)+3\left(x+1\right)\)

\(=\left(x+1\right)\left(x+3\right)\)

6) \(\left(x^2+1\right)^2-4x^2\)

\(=\left(x^2+1\right)^2-\left(2x\right)^2\)

\(=\left(x^2+1-2x\right)\left(x^2+1+2x\right)\)

\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)

\(=\left(x-1\right)^2\left(x+1\right)^2\)

7) \(x^2-4x-5\)

\(=x^2+x-5x-5\)

\(=\left(x^2+x\right)-\left(5x+5\right)\)

\(=x\left(x+1\right)-5\left(x+1\right)\)

\(=\left(x+1\right)\left(x-5\right)\)

x = 8

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