mọi người giúp mk với, mk đang cần gấp. Tối nay mk phải nộp rồi
B3: Cho biểu thức: C= \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a. Hãy rút gọn C
b. Tìm a để C ≥ 1/6
cho C= \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right)\div\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a) rút gọn C
b) tìm a để C= \(\dfrac{1}{4}\)
lm nhanh giúp mk nhé mk đang cần gấp
1) rút gọn và tìm A để A nguyên
A= \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
giúp mk vs ạ mk cần gấp
Bài 1: CMR:
\(a,\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(b,\dfrac{a^3}{b\left(2c+a\right)}+\dfrac{b^3}{c\left(2a+b\right)}+\dfrac{c^3}{a\left(2b+c\right)}\ge1\) với a+b+c=3
Bài 2: \(a,b,c\in N,a+b+c=2021\)
Tìm GTNN \(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
cho a,b,c>0
CMR:
1) \(a+b+\dfrac{1}{4}\ge\sqrt{a+b}\)
2) \(\left(a+b+\dfrac{1}{2}\right)^2+\left(b+c+\dfrac{1}{2}\right)^2+\left(c+a+\dfrac{1}{2}\right)^2\ge4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\right)\)
Cho a,b,c khác 0 thỏa mãn a\(\left(\dfrac{1}{c}+\dfrac{1}{b}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-2\)
a(1b+1c)+b(1c+1a)+c(1a+1b)=−2
và a3+b3+c3=1. CMR \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Cho a,b,c khác 0 thỏa mãn \(a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-2\)
và a3+b3+c3=1. CMR \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Giải pt:
\(\dfrac{\left(b-c\right)\left(1+a^2\right)}{x+a^2}+\dfrac{\left(c-a\right)\left(1+b^2\right)}{x+b^2}+\dfrac{\left(a-b\right)\left(1+c^2\right)}{x+c^2}=0\)
Cho a,b,c>0 thỏa mãn ab+bc+ca=1. CMR:
\(\left(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\right)^3\le\dfrac{3}{2}\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\)