Cho \(abc\ne0\) và \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
Tính \(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)
Cho \(abc\ne0\) và \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}.\) Tính \(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)
Giúp ik
Lời giải:
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Kết quả $P=8$ hoặc $P=-1$
Cho \(a^3+b^3+c^3=3abc,abc\ne0,a+b+c\ne0\)
Chứng minh:
\(B=\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(\dfrac{1}{c}+\dfrac{1}{b}\right)\left(\dfrac{1}{c}+\dfrac{1}{a}\right)=\dfrac{8}{abc}\)
\(a^3+b^3+c^3=3abc\\ \Rightarrow a^3+b^3+c^3-3abc=0\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\\ \Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\left(a+b+c\ne0\right)\\ \Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\\ \Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\\ \Rightarrow a=b=c\\ \Rightarrow B=\dfrac{2}{a}.\dfrac{2}{b}.\dfrac{2}{c}=\dfrac{8}{abc}\)
Cho \(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{1}{a+b+c}\left(a,b,c\ne0,a+b+c\ne0\right)\)
Tính \(\left(a-3b\right)\left(b-c\right)\left(3c-a\right)\)
Ai giúp mik đi, mik cho 5 coin
\(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{2a+2b+2c}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}b+c-5=2a\\a+c+2=2b\\a+b+3=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c=a+5\\a+b+c=b-2\\a+b+c=c-3\end{matrix}\right.\)
Lại có \(\dfrac{1}{a+b+c}=2\Rightarrow a+b+c=\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}a+5=\dfrac{1}{2}\\b-2=\dfrac{1}{2}\\c-3=\dfrac{1}{2}\end{matrix}\right.\)
Từ đó tự giải ra
Áp dụng t/c dtsbn:
\(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{b+c-5+a+c+2+a+b+3}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}b+c-5=2a\\a+c+2=2b\\a+b+3=2c\end{matrix}\right.\)\(\left(1\right)\)
Mặt khác \(\dfrac{1}{a+b+c}=\dfrac{b+c-5}{a}=2\)\(\Rightarrow a+b+c=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{1}{2}-c\\a+c=\dfrac{1}{2}-b\\b+c=\dfrac{1}{2}-a\end{matrix}\right.\)\(\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-a-5=2a\\\dfrac{1}{2}-b+2=2b\\\dfrac{1}{2}-c+3=2c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{3}{2}\\b=\dfrac{5}{6}\\c=\dfrac{7}{6}\end{matrix}\right.\)
\(\left(a-3b\right)\left(b-c\right)\left(3c-a\right)=\left(-\dfrac{3}{2}-3.\dfrac{5}{6}\right)\left(\dfrac{5}{6}-\dfrac{7}{6}\right)\left(3.\dfrac{7}{6}+\dfrac{3}{2}\right)=\dfrac{20}{3}\)
Cho \(a,b,c\ne0\) và \(a+b+c=\dfrac{a+2b-c}{c}=\dfrac{b+2c-a}{a}=\dfrac{c+2a-b}{b}\)
Tính \(P=\left(2+\dfrac{a}{b}\right)\left(2+\dfrac{b}{c}\right)\left(2+\dfrac{c}{a}\right)\)
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Cho \(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(a;b;c\ne0;b\ne c\right).\) Chứng minh: \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
Cho \(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ac\right)\left(a-abc\right);abc\ne0;a\ne b\)
CMR:\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c\)
Cho đa thức \(P\left(x\right)=ax^2+bx+c\). Trong đó \(a,b,c\) là các hằng số thỏa mãn \(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}\) và \(a\ne0\). Tính \(\dfrac{P\left(-2\right)-3P\left(1\right)}{a}\).
P(x)=\(ax^2+bx+c\) (1)(a\(\ne0\) )
Ta có :
\(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}\)\(\Rightarrow\left\{{}\begin{matrix}b=2a\\c=3a\end{matrix}\right.\)(2)
Thay(2) vào (1)\(\Rightarrow P\left(x\right)=ax^2+2ax+3a\)
\(\Rightarrow\dfrac{P\left(-2\right)-3P\left(-1\right)}{a}=\dfrac{4a-4a+3a-3\left(a-2a+3a\right)}{a}\)=\(\dfrac{3a-3a+6a-9a}{a}=\dfrac{-3a}{a}=-3\)
Cho \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\).Tính P =\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)
Cho a,b,c thỏa mãn \(\dfrac{a b-c}{c}=\dfrac{b c-a}{a}=\dfrac{c a-b}{b}\) Tính giá trị M = \(\left(1 \dfrac{b}{a}\right... - Hoc24
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Cho \(\dfrac{a+b-2017c}{c}=\dfrac{b+c-2017a}{a}=\dfrac{c+a-2017b}{b}\)
Với a, b, c \(\ne0\). TínhP = \(\left(1+\dfrac{a}{b}\right).\left(1+\dfrac{b}{c}\right).\left(1+\dfrac{c}{a}\right)\)
Trần Thọ Đạt ông giải dùm đi!Bn ý k bk tag nên tui tag dùm!
TH1 : Nếu a + c+ b \(\ne\) 0
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{a+b-2017c}{c}=\dfrac{b+c-2017a}{a}=\dfrac{c+a-2017b}{b}=\dfrac{a+b-2017c+b+c-2017a+c+a-2017b}{c+a+b}=\dfrac{-2015a-2015b-2015c}{a+b+c}=\dfrac{-2015.\left(a+b+c\right)}{a+b+c}=-2015\)( vì a + b + c \(\ne\)0 )
\(\Rightarrow\left\{{}\begin{matrix}a+b-2017c=-2015c\\b+c-2017a=-2015a\\c+a-2017b=-2015b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\a+c=2b\end{matrix}\right.\)
Mặt khác , P = \(\left(\dfrac{b+a}{b}\right).\left(\dfrac{c+b}{c}\right).\left(\dfrac{a+c}{a}\right)\)
\(\Rightarrow\)\(\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}=\dfrac{2.âbc}{abc}=2\)
TH2 : Nếu a + b+ c = 0
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\)=-1
Vây ...
Mk ko chắc là TH2 đúng ko nữa