Ôn tập cuối năm phần số học
Các câu hỏi tương tự
Cho a,b,c khác 0 thỏa mãn \(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)=8\)
CMR \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}=\dfrac{3}{4}+\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{ca}{\left(c+a\right)\left(a+b\right)}\)
Tìm GTNN của :
a) \(A=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)với a, b > 0
b) \(B=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)với a, b, c > 0
c) \(C=\left(a+b+c+d\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\)với a, b, c, d > 0
CM CÁC BẤT ĐẲNG THỨC SAU
A) \(\left(A+B\right)\left(\dfrac{1}{A}+\dfrac{1}{B}\right)\ge4\)
B) \(\left(A+B+C\right)\left(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\right)\ge9\)
C) \(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\ge\dfrac{9}{A+B+C}\)
cho a, b, c thỏa mãn
\(\dfrac{a+b-c}{a}=\dfrac{b+c-a}{c}=\dfrac{a+c-b}{b}\)
tính B = \(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{b}\right)\)
Rút gọn biểu thức:
a) \(A=\dfrac{bc}{\left(a-b\right)\left(a-c\right)}+\dfrac{ca}{\left(b-c\right)\left(b-a\right)}+\dfrac{ab}{\left(c-a\right)\left(c-b\right)}\)
b) \(B=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+x^3+\dfrac{1}{x^3}}\)
Chứng minh
\(\dfrac{\left(a-b\right)^2}{ab}+\dfrac{\left(b-c\right)^2}{bc}+\dfrac{\left(c-a\right)^2}{ca}=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Cho 3 số a, b, c thỏa mãn a # -b, b # -c, c # -a.
Chứng minh rằng : \(\dfrac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^2-ab}{\left(c+a\right)\left(c+b\right)}=0\)
Cho \(\dfrac{a-\left(c-b\right)}{b-c}+\dfrac{b-\left(a-c\right)}{c-a}+\dfrac{c-\left(b-a\right)}{a-b}=3\)
Chứng minh rằng: \(\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}=0\)
chứng minh
\(2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)-3=\dfrac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\dfrac{\left(b-c\right)^2}{\left(b+a\right)\left(c+a\right)}+\dfrac{\left(c-a\right)^2}{\left(c+b\right)\left(a+b\right)}\)