Biết \(a^3.b^3+b^3.c^3+c^3.a^3=3a^2b^2c^2\).
Tính giá trị của bt :
\(A=(\frac{a}{b}+1).(\frac{b}{c}+1)+(\frac{c}{a}+1)\)
Ae làm ơn giúp mk đi ạ mk ~ Mk sẽ k câu này và 2 câu khác nữa đó !!
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
1) \(D=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+....+\frac{10}{1400}\)
\(D=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+.....+\frac{5}{700}\)
\(D=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+......+\frac{5}{25.28}\)
\(D=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+.....+\frac{3}{25.28}\right)\)
\(D=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{25}-\frac{1}{28}\right)\)
\(D=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}.\frac{6}{28}=\frac{5}{14}\)
\(E=\frac{1}{1+2}+\frac{1}{1+2+3}+.......+\frac{1}{1+2+3+....+24}\)
Ta có: \(1+2=\)\(\frac{2.\left(2+1\right)}{2}=3\);\(1+2+3=\frac{3.\left(3+1\right)}{2}=6\);\(1+2+3+...+24=\frac{24.\left(24+1\right)}{2}=300\)
\(E=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{300}\)
=>\(\frac{1}{2}E=\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{600}=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{24.25}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{24}-\frac{1}{25}=\frac{1}{2}-\frac{1}{25}=\frac{23}{50}\)
=>\(E=\frac{46}{50}\)
Vậy \(\frac{D}{E}=\frac{5}{14}:\frac{46}{50}=\frac{250}{644}=\frac{125}{322}\)
2) Theo t/c dãy tỉ số=nhau:
\(\frac{a+b}{a+c}=\frac{a-b}{a-c}=\frac{a+b-\left(a-b\right)}{a+c-\left(a-c\right)}=\frac{a+b-a+b}{a+c-a+c}=\frac{2b}{2c}=1\)
=>b=c
do đó \(A=\frac{10b^2+9bc+c^2}{2b^2+bc+2c^2}=\frac{10b^2+9b^2+b^2}{2b^2+b^2+2b^2}=\frac{\left(10+9+1\right).b^2}{\left(2+1+2\right).b^2}=4\)
1/ Đặt
\(\frac{a}{b^2}=x,\frac{b}{c^2}=y,\frac{c}{a^2}=z,xyz=1\)thì ta có
\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow xy+yz+zx=x+y+z\)
\(\Leftrightarrow xyz-xy-yz-zx+x+y+z-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)
\(\Leftrightarrow x=1;y=1;z=1\)
\(\Rightarrow\frac{a}{b^2}=1;\frac{b}{c^2}=1;\frac{c}{a^2}=1\)
\(\Leftrightarrow a=b^2;b=c^2;c=a^2\)
2/ Đặt
\(ab=x,bc=y,ca=z\) cần tính
\(P=\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\left(1+\frac{y}{x}\right)\)
\(\Rightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{cases}}\)
Xét \(x+y+z=0\)
\(\Rightarrow P=\frac{x+y}{x}.\frac{y+z}{y}.\frac{z+x}{z}=\frac{\left(-x\right)\left(-y\right)\left(-z\right)}{xyz}=-1\)
Xét \(x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
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,
vì a,b,c tỉ lệ nghịch với 1/2;1/5;1/7 nên a/2=b/5=c/7. Hay a/2=b/5=2c/14
ADTCCDTSBN TA CÓ
a/2=b/5=2c/14=a+b-2c/2+5-14=70/-7=-10
Suy ra a/2=-10 nên a=-20
b/5=-10 nên b=-50
2c/14=-10 nên c=-70
Biết 3 số a,b,c chúng tỉ lệ nghịch với 1/2 ; 1/5 ; 1/7
=> a/2 = b/5 = c/7
=> a/2 = b/5 = -2c/-14
Áp dụng tc dãy tỉ số = nhau ta đc :
a/2 = b/5 = -2c/-14 = (a+b-2c)/(2+5-14) = 70/-7 = -10
=>a= -20 ; b= -50 ; c = -70
=> a+b-c = 0
https://olm.vn/hoi-dap/detail/211794512831.html
Tham khảo ở link này (mình gửi cho)
Học tốt!!!!!!!!!!
Đặt \(ab=x;\)\(bc=y;\)\(ca=z\)
Khi đó: \(a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
<=> \(x^3+y^3+z^3=3xyz\)
<=> \(x^3+y^3+z^3-3xyz=0\)
<=> \(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)
Nếu: \(x+y+z=0\)thì: \(ab+bc+ca=0\)
\(A=\left(\frac{a}{b}+1\right)\left(\frac{b}{c}+1\right)+\left(\frac{c}{a}+1\right)\)
\(=\frac{\left(a+b\right)\left(b+c\right)}{bc}+\frac{c}{a}+1=\frac{ab+ac+bc+b^2}{bc}+\frac{c}{a}+1\)
\(=\frac{b}{c}+\frac{c}{a}+1=\frac{ab+c^2+ac}{ac}=\frac{c^2-bc}{ac}=\frac{c-b}{a}\)
Nếu: \(x^2+y^2+z^2-xy-yz-zx=0\)<=> \(x=y=z\)
<=> \(ab=bc=ca\)<=> \(a=b=c\)
\(A=\left(\frac{a}{b}+1\right)\left(\frac{b}{c}+1\right)+\left(\frac{c}{a}+1\right)=2.2+2=6\)
p/s: trg hợp 1 mk lm đc đến có z thôi, bn tham khảo