Tìm \(\frac{a}{b}\) và \(\frac{c}{d}\) biết :
\(\frac{a}{b}.\frac{c}{d}=\frac{32}{30}\)và \(\frac{a}{b}+\frac{c}{d}=\frac{32}{15}\)
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Đặt a/b = m ; c/d bằng n ta có :
m.n = 32/30 ; m+ n = 32/15
( m + n )^2 = ( 32/15)^2
=> ( m + n )( m + n ) = 1024/225
=> m^2 + mn + mn + n^2 = 1024/225
=> m^2 + n^2 + 2mn = 1024/225
=> m^2 + n^2 + 2mn - 4mn = 1024/225 - 4mn
=> m^2 + n^2 - 2mn = 1024/225 - 4.32/30 = 64/225
=> ( m - n)^2 = 64/225 = (8/15)^2 = ( -8/15 )^2
(+) m- n = 8/15 và m + n = 32/15 ( dây là dạng tổng hiệu tự lmaf )
(+) m -n = -8/15 và m+ n = 32/15 ( ......................)
Đặt \(\frac{a}{b}=x;\text{ }\frac{c}{d}=y\)
Ta có: \(xy=\frac{32}{30}\left(1\right)\text{ và }x+y=\frac{32}{15}\left(2\right)\)
\(\left(2\right)\Leftrightarrow y=\frac{32}{15}-x,\text{ thay vào (1), ta được:}\)
\(x\left(\frac{32}{15}-x\right)=\frac{32}{30}\Leftrightarrow x\left(32-15x\right)=16\)
\(\Leftrightarrow15x^2-32x+16=0\Leftrightarrow\left(3x-4\right)\left(5x-4\right)=0\)
\(\Leftrightarrow x=\frac{4}{3}\text{ hoặc }x=\frac{4}{5}\)
\(+x=\frac{4}{3}\text{ thì }y=\frac{32}{15}-\frac{4}{3}=\frac{4}{5}\)
\(+x=\frac{4}{5}\text{ thì }y=\frac{32}{15}-\frac{4}{5}=\frac{4}{3}\)
Vậy \(\frac{a}{b}\text{ và }\frac{c}{d}\text{ là }\frac{4}{3}\text{ và }\frac{4}{5}\text{ (hoặc ngược lại)}\)
1, \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
Do đó \(\left\{{}\begin{matrix}3a=b+c+d\left(1\right)\\3b=a+c+d\left(2\right)\\3c=a+b+d\left(3\right)\\3d=a+b+c\left(4\right)\end{matrix}\right.\)
Từ (1) và (2) \(\Rightarrow3\left(a+b\right)=a+b+2c+2d\Leftrightarrow2\left(a+b\right)=2\left(c+d\right)\Leftrightarrow a+b=c+d\Leftrightarrow\dfrac{a+b}{c+d}=1\)
Tương tự cũng có: \(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)
\(\Rightarrow A=4\)
2, Có \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)\(\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
Do đó \(\dfrac{x^2}{4}=\dfrac{1}{4};\dfrac{y^2}{16}=\dfrac{1}{4};\dfrac{z^2}{36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(1;2;3\right),\left(-1;-2;-3\right)\)
Bài 2 :
a, Ta có : \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=4\\z^2=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\\z=\pm3\end{matrix}\right.\)
Vậy ...
b, Ta có : \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{5+7}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\)
\(\Rightarrow x=2\)
\(\Rightarrow y=3\)
Vậy ...
\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\)\(\frac{d}{a+b+c}\)
\(\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)
Mà: \(a+b+c+d\ne0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}\)
\(\Rightarrow A=1+1+1+1=4\)
số đo slaf
4
nhe sbn
bài dài
lắm mình
vhir tiện ghi
thế này thôi
Vì \(\frac{a}{b+c+d}\)= \(\frac{b}{a+c+d}\)= \(\frac{c}{a+b+d}\)= \(\frac{d}{a+b+c}\)nên
\(\frac{a}{b+c+d}\)+1 = \(\frac{b}{a+c+d}\)+1 = \(\frac{c}{a+b+d}\)+1 = \(\frac{d}{a+b+c}\) +1
hay\(\frac{a+b+c+d}{b+c+d}\) = \(\frac{a+b+c+d}{a+c+d}\)= \(\frac{a+b+c+d}{a+b+d}\)= \(\frac{a+b+c+d}{a+b+c}\)
Mà a + b + c + d \(\ne\)0 \(\Rightarrow\) \(b+c+d=a+c+d=a+b+d=a+b+c\)
\(\Rightarrow\) \(a=b=c=d\)
\(\Rightarrow\) \(M=4\)
a) \(\frac{2}{5}+\frac{9}{15}=\frac{2}{5}+\frac{3}{5}=\frac{5}{5}=1\)
b) \(\frac{15}{45}+\frac{25}{30}=\frac{1}{3}+\frac{5}{6}=\frac{2}{6}+\frac{5}{6}=\frac{7}{6}\)
c) \(\frac{28}{32}+\frac{45}{72}=\frac{7}{8}+\frac{5}{8}=\frac{12}{8}=\frac{3}{2}\)
d) \(\frac{8}{28}+\frac{5}{30}=\frac{2}{7}+\frac{1}{6}=\frac{12}{42}+\frac{7}{42}=\frac{19}{42}\)
a) \(\frac{2}{5}+\frac{9}{15}\)=\(\frac{2}{5}+\frac{3}{5}=\frac{2+3}{5}=1\)
b)\(\frac{15}{45}+\frac{25}{30}=\frac{1}{3}+\frac{5}{6}=\frac{2}{6}+\frac{5}{6}=\frac{2+5}{6}=\frac{7}{6}\)
c)\(\frac{28}{32}+\frac{45}{72}=\frac{7}{8}+\frac{5}{8}=\frac{7+5}{8}=\frac{12}{8}=\frac{3}{2}\)
d)\(\frac{8}{28}+\frac{5}{30}=\frac{2}{7}+\frac{1}{6}=\frac{12}{42}+\frac{7}{42}=\frac{12+7}{42}=\frac{19}{42}\)
\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
Vay a=1; b=2; c=3; d=4
Ta có : \(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
Vậy a = 1,b = 2,c = 3,d = 4
Đặt \(\frac{a}{b}=x;\text{ }\frac{c}{d}=y\)
Ta có: \(xy=\frac{32}{30}\left(1\right)\text{ và }x+y=\frac{32}{15}\left(2\right)\)
\(\left(2\right)\Leftrightarrow y=\frac{32}{15}-x,\text{ thay vào (1), ta được:}\)
\(x\left(\frac{32}{15}-x\right)=\frac{32}{30}\Leftrightarrow x\left(32-15x\right)=16\)
\(\Leftrightarrow15x^2-32x+16=0\Leftrightarrow\left(3x-4\right)\left(5x-4\right)=0\)
\(\Leftrightarrow x=\frac{4}{3}\text{ hoặc }x=\frac{4}{5}\)
\(+x=\frac{4}{3}\text{ thì }y=\frac{32}{15}-\frac{4}{3}=\frac{4}{5}\)
\(+x=\frac{4}{5}\text{ thì }y=\frac{32}{15}-\frac{4}{5}=\frac{4}{3}\)
Vậy \(\frac{a}{b}\text{ và }\frac{c}{d}\text{ là }\frac{4}{3}\text{ và }\frac{4}{5}\text{ (hoặc ngược lại)}\)