cac ban giup minh voi minh rat can dap an nay
so a>0 thoan man 15/a=a-1/28
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\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{3}+\dfrac{3}{15}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)
tu ve hinh :
xet tamgiac AMB va tamgiac AMC co : goc BAM = goc CAM do AM la phan giac cua goc BAC (gt)
AB = AC va goc ABC = goc ACB do tamgiac ABC can tai A (gt)
=> tamgiac AMB = tamgiac AMC (c - g - c) (1)
b, (1) => goc AMB = goc AMC
goc AMB + goc AMC = 180 (ke bu)
=> goc AMB = 90
=> AM | BC (dn)
Bài 35:
b) ĐKXĐ: \(x\notin\left\{5;2\right\}\)
Ta có: \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\)
\(\Leftrightarrow\dfrac{x+2}{x-5}+3-\dfrac{6}{2-x}=0\)
\(\Leftrightarrow\dfrac{x+2}{x-5}+3+\dfrac{6}{x-2}=0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\dfrac{3\left(x-5\right)\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\dfrac{6\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=0\)
Suy ra: \(x^2-4+3\left(x^2-7x+10\right)+6x-30=0\)
\(\Leftrightarrow x^2-4+3x^2-21x+30+6x-30=0\)
\(\Leftrightarrow4x^2-15x-4=0\)
\(\Leftrightarrow4x^2-16x+x-4=0\)
\(\Leftrightarrow4x\left(x-4\right)+\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\4x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-\dfrac{1}{4}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{4;-\dfrac{1}{4}\right\}\)
Bài 36:
a) Ta có: \(\left(3x^2-5x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(3x^2-5x+1\right)=0\)
mà \(3x^2-5x+1>0\forall x\)
nên (x-2)(x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: S={2;-2}
Bà con ai đang gặp các vấn đề về còi hãy gọi ngay tổng đài 190000 không biết :))