Bài 1: Tìm a , b , c , d \(\in\) N , biết :
\(\dfrac{30}{43}\) = \(\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\)
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Có \(\dfrac{30}{43}=\dfrac{1}{\dfrac{43}{30}}=\dfrac{1}{1+\dfrac{13}{30}}=\dfrac{1}{1+\dfrac{1}{\dfrac{30}{13}}}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{4}{13}}}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{13}{4}}}}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4}}}}\)
Vậy a=1; b=2 ; c=3 ; d=4
ta thấy : \(\dfrac{a}{b}\) = \(\dfrac{1}{\dfrac{b}{a}}\)
\(\Rightarrow\) \(\dfrac{30}{43}\) = \(\dfrac{1}{\dfrac{43}{30}}\)
= \(\dfrac{1}{1+\dfrac{13}{30}}\)
= \(\dfrac{1}{1+\dfrac{1}{\dfrac{30}{13}}}\)
= \(\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{15}}}\)
= \(\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{15}{2}}}}\)
=\(\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{7+\dfrac{1}{2}}}}\)
Vậy a = 1; b = 2; c = 7; d = 4
Phân tích phân số \(\dfrac{30}{43}\) ta có:
\(\dfrac{30}{43}=\dfrac{1}{\dfrac{43}{30}}=\dfrac{1}{1+\dfrac{13}{30}}=\dfrac{1}{1+\dfrac{1}{\dfrac{30}{13}}}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{4}{13}}}\)
\(=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{13}{4}}}}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4}}}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\)
Vậy: \(\left\{{}\begin{matrix}a=1\\b=2\\c=3\\d=4\end{matrix}\right.\)
b/ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\dfrac{a}{d}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
=> \(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{c+d+b}\right)^3\) (2)Từ (1) và (2)=>đpcm
2:
a: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{24}{9}=\dfrac{8}{3}\)
=>x=16/3; y=8; z=32/3
A=3x+2y-6z
=3*16/3+2*8-6*32/3
=16+16-64
=-32
b: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y+z}{5-6+7}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
=>x=5căn 2; y=6căn 2; y=7căn 2
B=xy-yz
=y(x-z)
=6căn 2(5căn 2-7căn 2)
=-6căn 2*2căn 2
=-24
a)4/5+x=2/3
x=2/3-4/5
x=-2/15
b)-5/6-x=2/3
x=-5/6-2/3
x=-3/2
c)1/2x+3/4=-3/10
1/2x=-3/10-3/4
1/2x=-21/20
x=-21/20:1/2
x=-21/10
d)x/3-1/2=1/5
x/3=1/5+1/2
x/3=7/10
10x/30=21/30
10x=21
x=21:10
x=21/10
bz-cy/a = cx- az /b = ay-bx /c => bxz-cxy / ax = cxy-azy / b = azy-bxz/c = bxz-cxy + cxy-azy+azy-bxz / a+b+c = 0/ a+b+c = 0
Suy ra : bz -cy/a = 0 => bz-cy=0 => bz = cy => z/c = b/y
cx-az/b = 0 => cx-az=0 => cx=az => x/a = z/c
ay-bx/c = 0 => ay-bx = 0 => ay=bx=> y/b = x/a
Vậy x/a=y/b=c/z
\(\dfrac{30}{43}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\)
\(\Leftrightarrow\dfrac{1}{\dfrac{43}{30}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\\ \Leftrightarrow\dfrac{1}{1+\dfrac{13}{30}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\\ \Leftrightarrow\dfrac{1}{1+\dfrac{1}{\dfrac{30}{13}}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\\ \Leftrightarrow\dfrac{1}{1+\dfrac{1}{2+\dfrac{4}{13}}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\)
\(\\ \Leftrightarrow\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{13}{4}}}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\\\Leftrightarrow\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4}}}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\)
\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\\c=3\\d=4\end{matrix}\right.\)
Vậy............
bấm máy tính casio, ta được:
a=1; b=2; c=3; d=4