tính a,b,c,d biết a=\(\frac{8}{9}\)b; b=\(\frac{4}{9}\)c; c=\(\frac{6}{5}\)d và (a+c)-(b+d)=3
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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 ...
\(A=\frac{11}{9}-\frac{7}{8}+-\frac{2}{3}-\frac{1}{8}+\frac{25}{9}-\frac{4}{3}\)
\(A=1\)
\(B=1\frac{3}{4}:\frac{3}{5}-\frac{2}{3}x1,75+\left(\frac{1}{2}\right)^2:\frac{1}{7}\)
\(B=3,5\)
I don't now
or no I don't
..................
sorry
a) ta có:\(\frac{a}{b}=\frac{5}{8}\Rightarrow8a=5b\)
ab=360\(\Rightarrow5ab=1800\Rightarrow8a^2=1800\Rightarrow225=a^2\Rightarrow a=15\)
đề câu b hình như sai bạn ạ
Áp dụng t/c của dãy tỉ số bằng nhau ta có:
\(\frac{a-1}{9}=\frac{b-2}{8}=\frac{c-3}{7}=....=\frac{i-9}{1}=\frac{\left(a-1\right)+\left(b-2\right)+\left(c-3\right)+...+\left(i-9\right)}{9+8+7+...+1}=\frac{\left(a+b+c+..+i\right)-\left(1+2+3+...+9\right)}{1+2+3+...+9}\)
=> \(\frac{a-1}{9}=\frac{b-2}{8}=\frac{c-3}{7}=....=\frac{i-9}{1}=\frac{90-45}{45}=1\)
=> a - 1 = 9 ; b - 2 = 8; c - 3 = 7; d- 4 = 6; e - 5 = 5; f - 6 = 4; ...; i - 9 = 1
=> a = 10; b = 10; c = 10= d = ..= i
\(\frac{a-1}{9}=\frac{b-2}{8}=\frac{c-3}{7}=...=\frac{i-9}{1}=\frac{\left(a-1\right)+\left(b-2\right)+\left(c-3\right)+...+\left(i-9\right)}{9+8+7+...+1}=\frac{\left(a+b+c+...+i\right)-\left(1+2+3+...+9\right)}{9+8+7+...+1}\)\(=\frac{90-\frac{9.10}{2}}{\frac{9.10}{2}}=\frac{90-45}{45}=\frac{45}{45}=1\)
=> a = 9 + 1 = 10
b = 8 + 2 = 10
c = 7 + 3 = 10
....
i = 1 + 9 = 10
Vậy a = b = c = ... = i = 10
Bài 1:suy ra 5*(44-x)=3*(x-12)
220-5x=3x-36
-5x-3x=-36-220
-8x =-256
x=32
Bài 2 :Đặt a/3=b/4=k
suy ra a=3k ; b=4k
Ta có a*b=48
suy ra 3k*4k=48
12k =48
k=4
suy ra a=3*4=12
b=4*4 =16
Bài 3: áp dụng tính chất dãy số bằng nhau ta được
a+b+c+d/3+5+7+9 = 12/24=0,5
suy ra a=1,5; b=2,5; c=3,5; d=4,