a+c=2b
Những câu hỏi liên quan
Cho tỉ lệ thức 3a+2b+c/a+2b-c=3a+2b+c/a-2b-c (b≠0)chứng minh a+c =0
Cho tỉ lệ thức 3a+2b+c/a+2b-c=3a+2b+c/a-2b-c (b≠0)chứng minh a+c =0
\(\dfrac{ }{ }\)
cho 3a+2b+c/a+2b-c=3a-2b+c/a-2b-c và b ko =0 CMR a=c=0 hộ mik nha!
Cho 2y + 2z - x/ a = 2z + 2x - y/ b = 2x + 2y - z/c với a,b,c khác 0; 2c +2b khác c; 2b + 2c khác a; 2c +2b khác b. Chứng minh : x/ 2b + 2c - a= y/ 2c + 2a - b= z/ 2a + 2b - c
Cho a,b,c khác 0 thỏa mãn: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
Tính \(E=\dfrac{a^2b^2c^2}{a^2b^2+b^2c^2-c^2a^2}+\dfrac{a^2b^2c^2}{b^2c^2+c^2a^2-a^2b^2}+\dfrac{a^2b^2c^2}{c^2a^2+a^2b^2-b^2c^2}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=> bc+ac+ab=0
ta có
\(bc+ac=-ab\)
<=> \(\left(bc+ac\right)^2=a^2b^2\)
<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)
<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)
tương tự
\(a^2b^2+b^2c^2-c^2a^2=-2ab^2c\)
\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)
thay vào E ta đc
\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)
=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)
Đúng 1
Bình luận (0)
Cho sao nha nhưng tui ko bít làm
Cho \(\dfrac{a}{2b}=\dfrac{2b}{c}=\dfrac{c}{a}\)và a+2b+c≠0. Tính giá trị của biểu thức M=\(\dfrac{a^3.c^2.b^{2015}}{b^{2020}}\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{a}{2b}=\dfrac{2b}{c}=\dfrac{c}{a}=\dfrac{a+2b+c}{2b+c+a}=1\)
\(\dfrac{a}{2b}=1\Rightarrow a=2b\\ \dfrac{2b}{c}=1\Rightarrow c=2b\\ \dfrac{c}{a}=1\Rightarrow a=c\\ \Rightarrow a=2b=c\)
\(M=\dfrac{a^3.c^2.b^{2015}}{b^{2020}}=\dfrac{a^3.a^2}{b^5}=\dfrac{a^5}{b^5}=\dfrac{\left(2b\right)^5}{b^5}=\dfrac{32b^5}{b^5}=32\)
Đúng 1
Bình luận (0)
Có \(\dfrac{a}{2b}=\dfrac{2b}{c}=\dfrac{c}{a}=\dfrac{a+2b+c}{2b+c+a}=1\)
=> a = 2b = c
M = \(\dfrac{a^3.c^2.b^{2015}}{b^{2020}}=\dfrac{a^3.c^2}{b^5}=\dfrac{\left(2b\right)^3.\left(2b\right)^2}{b^5}=\dfrac{32.b^5}{b^5}=32\)
Đúng 2
Bình luận (0)
A + A + A = 90
A + 2B + 2B = 230
C + 2B + A = 210
B - A + C =?
A+A+A=90 => 3A=90 => A=90:3=30
A+2B+2B=230 = 30+4B=230 => 4B=230-30=200 => B=200:4=50
C+2B+A=210 => C+100+30=210 => C=210-100-30=80
=> B-A+C=50-30+80=100
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Cho a, b, c \(\ne\)0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\). Tính : \(E=\frac{a^2b^2c^2}{a^2b^2+b^2c^2-a^2c^2}+\frac{a^2b^2c^2}{b^2c^2+c^2a^2-a^2b^2}+\frac{a^2b^2c^2}{c^2a^2+a^2b^2-b^2c^2}.\)
\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow abc\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-2abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
CHÚC BẠN HỌC TỐT
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow a.b.c\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
Vậy \(E=0\)
cho abc=36,1/a+1/b+1/c=o.
Tính Q=a^2(b^2+c^2)-b^2c^2/a^2b^2c^2*b^2(c^2+a^2)-c^2a^2/a^2b^2c^2*c^2(a^2+b^2)-a^2b^2/a^2b^2c^2
Cho a,b,c là các số thực khác 0 thỏa mãn. Tính giá trị biểu thức:
\(P=\frac{a^2c}{a^2c+c^2b+b^2a}+\frac{b^2a}{b^2a+a^2c+c^2b}+\frac{c^2b}{c^2b+b^2a+a^2c}\)
P = \(\frac{a^2c}{a^2c+c^2b+b^2a+}+\frac{b^2a}{b^2a+a^2c+c^2b}+\frac{c^2b}{c^2b+b^2a+a^2c}\)
P = \(\frac{a^2c+b^2a+c^2b}{a^2c+c^2b+b^2a}=1\)
\(P=\frac{\frac{a}{b}}{\frac{a}{b}+\frac{c}{a}+\frac{b}{c}}+\frac{\frac{b}{c}}{\frac{b}{c}+\frac{a}{b}+\frac{c}{a}}+\frac{\frac{c}{a}}{\frac{c}{a}+\frac{b}{c}+\frac{a}{b}}=\frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}=1\)