Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Từ \(\frac{a}{b}\) = \(\frac{132}{143}\) = \(\frac{12}{13}\) => \(\frac{a}{12}\) = \(\frac{b}{13}\)
Đặt \(\frac{a}{12}\) = \(\frac{b}{13}\) = k => a= 12k ; b=13k
BCNN(a,b)=BCNN(12k;13k) = 12.13.k = 1092
=> k =7 =>a= 7.12 = 84
Vậy a = 84
CHÚC BẠN HK TỐT!!!
TICK MK NHA
a=84 (mk k biết cách giải), bài này ở trong violympic toán cũng có
Bạn vào link này có câu trả lời rồi nhé !
https://olm.vn/hoi-dap/question/90515.html
Rút gọn \(\frac{132}{143}=\frac{12}{13}\)
\(\Rightarrow a=12k;b=13k\) với \(k\in N\)
Ta có:
\(a.b=1092\Leftrightarrow12k.13k=1092\)
\(\Leftrightarrow\left(12.13\right)k=1092\)
\(\Leftrightarrow156k=1092\)
\(\Leftrightarrow k=7\)
\(\Rightarrow\frac{a}{b}=\frac{12k}{13k}=\frac{12.7}{13.7}=\frac{84}{91}\)
Vậy \(a=84;b=91\)
theo bai ra, ta co:
\(\frac{a}{b}=\frac{132}{143}\Leftrightarrow\frac{a}{132}=\frac{b}{143}=k\)
\(\Rightarrow a=132k;b=143k\)
ta co: BCNN(a,b)=BCNN(132k;143k)=156k
\(\Rightarrow\)156k=1092\(\Leftrightarrow\)k=7
\(\Rightarrow\)a=132.k=924
Ta có: \(A=a\left(a^2-bc\right)+b\left(b^2-ac\right)+c\left(c^2-ab\right)=0\)
\(\Rightarrow A=a^3+b^3+c^3-3abc=0\) \(\Rightarrow A=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow A=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow A=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)
Xét \(M=a^2+b^2+c^2-ab-ac-bc=0\)
\(\Rightarrow2M=2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Rightarrow2M=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Rightarrow a-b=0;b-c=0;c-a=0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)
Lời giải:
Ta có
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=\left ( \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \right )(a+b+c)-\frac{a(b+c)}{b+c}-\frac{b(c+a)}{c+a}-\frac{c(a+b)}{a+b}\)
\(=a+b+c-(a+b+c)=0\)
Ta có đpcm
a)\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{a^3+b^3+c^3}{abc}\)
\(A=\dfrac{3abc}{abc}=3\)(vì a+b+c=0)
b)Ta có: a+b+c=0
\(\Rightarrow\left\{{}\begin{matrix}a=-b-c\\b=-c-a\\c=-a-b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(c+a\right)^2\\c^2=\left(a+b\right)^2\end{matrix}\right.\)
\(\Rightarrow B=\dfrac{a^2}{\left(b+c\right)^2-b^2-c^2}+\dfrac{b^2}{\left(a+c\right)^2-c^2-a^2}+\dfrac{c^2}{\left(a+b\right)^2-a^2-b^2}\)
\(\Rightarrow B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}\)
\(\Rightarrow B=\dfrac{a^3+b^3+c^3}{2abc}\)
\(\Rightarrow B=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)(vì a+b+c=0)
cm:nếu a+b+c=0 thì a^3+b^3+c^3=3abc
a^3+b^3+c^3=3abc
=>a^3+b^3+c^3-3abc=0
=>(a+b)^3-3ab(a+b)+c^3-3abc=0
=>[(a+b)^3+c^3]-3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2] -3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0
vì a+b+c=0 nên a^3+b^3+c^3=3abc
thay kết quả vừa chúng minh vào đề bài ta đc
\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)
chúc bạn học tốt ^ ^
\(A=\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}=\frac{a^3b^3+b^3c^3+c^3a^3}{abc}=\frac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)
\(\Rightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3=3.ab.bc.ca=3a^2b^2c^2\)
Vậy \(A=\frac{3a^2b^2c^2}{abc}=3abc\left(a,b,c\ne0\right)\)
ta có: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\)
<=>\(\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
<=>\(\dfrac{a^2}{b+c}+\dfrac{ab}{b+c}+\dfrac{ac}{b+c}+\dfrac{b^2}{a+c}+\dfrac{ab}{a+c}+\dfrac{bc}{a+c}+\dfrac{c^2}{a+b}+\dfrac{ac}{a+b}+\dfrac{cb}{a+b}=a+c+b\)
<=>\(\dfrac{a^2}{b+c}+\dfrac{ab+ac}{b+c}+\dfrac{b^2}{a+c}+\dfrac{ab+bc}{a+c}+\dfrac{c^2}{a+b}+\dfrac{ac+cb}{a+b}=a+c+b\)
<=>\(\dfrac{a^2}{b+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{b^2}{a+c}+\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c^2}{a+b}+\dfrac{c\left(a+b\right)}{a+b}=a+c+b\)
<=>\(\dfrac{a^2}{b+c}+a+\dfrac{b^2}{a+c}+b+\dfrac{c^2}{a+b}+c=a+c+b\)
<=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=a+c+b-a-c-b=0\) (đpcm)
chúc bạn học tốt ^ ^
Ta có: \(\dfrac{a}{b}=\dfrac{132}{143}=\dfrac{12}{13}\)
nên a=12k; b=13k với k∈N (1)
Ta có: ƯCLN (12;13) = 1 => ƯCLN (12k;13k)=1 =>BCNN(12k;13k)=12.13k (2)
Theo đề bài thì BCNN(a,b)=1092 (3)
Từ (1); (2) và (3) , ta có:
12.13k=1092 ⇔ 156k=1092 ⇔ k=7
Khi đó a=12.7=84; b=13.7=91
Vậy a=84; b=91