cho\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\) va a+b+c khac 0
a] so sanh ac so a,b,c
cho a=2017. tinh b,c
cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)va a+b+c khac 0. tinh M=\(\frac{a^{10}b^7c^{2000}}{b^{2017}}\)
Ta có:M=\(\frac{a^{10}b^7c^{2000}}{b^{2017}}\)=\(\frac{a^{10}}{b^{10}}\)x\(\frac{b^7}{b^7}\)x\(\frac{c^{2000}}{b^{2000}}\)=\(\left(\frac{a}{b}\right)^{10}\)x\(\left(\frac{c}{b}\right)^{2000}\)=\(\left(\frac{a}{b}\right)^{10}\)x\(\left(\frac{b}{c}\right)^{-2000}\)
Mà \(\frac{a}{b}\)=\(\frac{b}{c}\)nên M=\(\left(\frac{a}{b}\right)^{10}\)x\(\left(\frac{a}{b}\right)^{-2000}\)=\(\left(\frac{a}{b}\right)^{-1990}\)
cho a,b,c,d la 4 so khac 0 thoa man b^2 =ac va c^2=bd.Cmr\(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)=\(\frac{a}{b}\)
Cho ba so a , b ,c \(\in\) Q khac nhau tung doi mot va khac 0 thoa \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\). Chung minh \(\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}\)khong phu thuoc vao gia tri cua a,b,c.
M.n oi, giup mik voi ngay mai mik phai nop roi....
Đề sửa lại là: Chứng minh \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\) nhé.
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}=\frac{a+b+c}{2.\left(a+b+c\right)}.\)
Xét 2 trường hợp:
TH1: \(a+b+c=0\) thì \(\left\{{}\begin{matrix}b+c=-a\\a+c=-b\\a+b=-c\end{matrix}\right.\)
Có: \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\), không phụ thuộc vào các giá trị \(a;b;c\) (1)
TH2: \(a+b+c\ne0\) thì \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2.\left(a+b+c\right)}=\frac{1}{2}.\)
\(\Rightarrow\left\{{}\begin{matrix}2a=b+c\\2b=a+c\\2c=a+b\end{matrix}\right.\)
Có: \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}=2+2+2=6\), không phụ thuộc vào các giá trị \(a;b;c\) (2)
Từ (1) và (2) => \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\) không phụ thuộc vào các giá trị của \(a;b;c.\)
Chúc bạn học tốt!
cho 3 so a,b,c khac 0 va (a+b+c)^2=a^2+b^2+c^2 . chung minh \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3abc\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\)
\(\Rightarrow2\left(ab+bc+ac\right)=0\)
\(\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{\left(a+b+c\right)}{abc}=0\)
\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)
\(\Rightarrow\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{-1}{c}\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(\frac{-1}{c}\right)^3\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}.\left(-\frac{1}{c}\right)=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{ab}=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\left(đpcm\right)\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{ab+bc+ac}{abc}=0\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\left(\frac{1}{a}\right)^3+\left(\frac{1}{b}\right)^3+\left(\frac{1}{c}\right)^3=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Cho a,b,c la 3 so doi mot khac nhau va \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
CMR\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
Ta có:\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=\frac{b}{a-c}+\frac{c}{b-a}=\frac{b^2-ab+ac-c^2}{\left(c-a\right)\left(a-b\right)}\)
\(\frac{\Leftrightarrow a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(1\right)\) Nhân hai vế với \(\frac{1}{b-c}\)
Tương tự ta có:\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ba-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(2\right);\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(3\right)\)
Cộng (1),(2),(3) ta được đpcm
Cho cac so a,b,c va thoa man\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=2\)Tinh gia tri bieu thuc \(P=\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\)
cho a,b,c khac 0 va a+b+c=0 . tinh Q=\(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}\)
a + b + c = 0 => c = -a - b ; b= -a - c ; a = - b - c
Thay vào Q ta có :
\(Q=\frac{1}{a^2+b^2-\left(a+b\right)^2}+\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{a^2+c^2-\left(a+c\right)^2}\)
\(Q=\frac{1}{a^2+b^2-a^2-b^2-2ab}+\frac{1}{b^2+c^2-b^2-c^2-2bc}+\frac{1}{c^2+a^2-c^2-a^2-2ac}\)
\(Q=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}=\frac{c+a+b}{-2abc}=0\)
cau 1: Cho A= \(\frac{100^{2014}+2}{3}-\frac{100^{2015}+17}{9}.\)tong cac cua so cua B=-9A
Cau 2: So sanh A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{100.101}\)voi 1 ta doc A....1
Cau 3 : Cho bon so a,b,c,d sao cho a+b+c+d khac 0 . Biet \(\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{d+a+b}{c}=\frac{a+b+c}{d}=k\). Vay k=
Cau 4 : so cac so nguyen am x thoa man \(x^{2015}=\left(-2\right)^{2014}\)
Cau 5; tim x,y,z biet \(\frac{x}{y}=\frac{10}{9},\frac{y}{z}=\frac{3}{4}\)va x-y+z=78
Cau 6: tap hop cac so co ba chu so chia het cho 18 va tong cac chu so ti le voi 1;2;3 la
Cau 7: gia tri cua tong S=1.2+2.3+.....+49.50 la S=
Cho a , b , c khac 0 va \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\) Tinh C=\(\frac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}\)