cho abc\(\ne\)0,1
và \(\dfrac{ab+1}{b}=\dfrac{bc+1}{c}=\dfrac{ac+1}{a}\)
CMR a=b=c
BT
Những câu hỏi liên quan
Cho a,b,c thỏa mãn ab+ac+bc=a+b+c+abc ; 3+ab ≠ 2a+b; 3+bc ≠ 2b+c;3+ac ≠2c+a.
C/M: \(\dfrac{1}{3+ab-\left(2a+b\right)}+\dfrac{1}{3+bc-\left(2b+c\right)}+\dfrac{1}{3+ac-\left(2c+a\right)}=1\)
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 abc=1. CMR \(\dfrac{a}{ab}+a+1+\dfrac{b}{bc}+b+1+\dfrac{c}{ac}+c+1=1\)
mình nghĩ đề thế này, do bạn ko viết a+1,b+1,c+1 dưới mẫu
Cho abc = 1 . CMR : \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}=1\)
GIẢI
Ta có : \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\)
\(=\frac{ab+a+1}{ab+a+1}=1\)
Đúng 0
Bình luận (0)
Cho abc=1. CMR \(\dfrac{a}{ab}+a+1+\dfrac{b}{bc}+b+1+\dfrac{c}{ac}+c+1=1\)
Chắc bạn viết nhầm đề, cho \(a=b=c=1\) đâu có đúng
Sửa lại đề: cho \(abc=1\) chứng minh \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\)
Ta có
\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=\dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{c}{ac+c+abc}\)
\(=\dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{c}{c\left(a+1+ab\right)}\)
\(=\dfrac{a}{ab+a+1}+\dfrac{ab}{ab+a+1}+\dfrac{1}{ab+a+1}\)
\(=\dfrac{a+ab+1}{ab+a+1}=1\) (đpcm)
Đúng 0
Bình luận (0)
Cho a, b, c \(\in\)\([0,1]\). Chứng minh:
\(\dfrac{a}{1+bc}+\dfrac{b}{1+ac}+\dfrac{c}{1+ab}\le2\)
Lời giải:
Do $0\leq a,b,c\le1 1$ nên: \(\text{VT}\leq \frac{a+b+c}{1+abc}\)
Giờ ta cần cm: $a+b+c\leq 2(1+abc)(*)$
Thật vậy:
$c(a-1)(b-1)\geq 0$
$\Leftrightarrow c(ab-a-b+1)\geq 0$
$\Leftrightarrow abc\geq ac+bc-c$
$\Leftrightarrow 2(abc+1)\geq ac+bc-c+abc+2$
Mà:
$ac+bc-c+abc+2-(a+b+c)=abc+(a+b)(c-1)-2(c-1)$
$=abc+(a+b-2)(c-1)\geq 0$ với mọi $0\leq a,b,c\leq 1$
$\Rightarrow ac+bc-c+abc+2\geq a+b+c$
$\Rightarrow 2(abc+1)\geq a+b+c$
Do đó BĐT $(*)$ đúng nên ta có đpcm.
Đúng 0
Bình luận (0)
Cho abc=1.CMR:
\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+}=1\)
\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac}{abc+ac+1}+\dfrac{ab}{abc+ab+1}+\dfrac{bc}{abc+bc+1}\)
\(=\dfrac{ac}{ac+2}+\dfrac{ab}{ab+2}+\dfrac{bc}{bc+2}\)
\(=abc\left(\dfrac{b}{abc+2}+\dfrac{c}{abc+2}+\dfrac{a}{abc+2}\right)\)
\(=1.1=1\)(đpcm).
Vậy \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\).
Đúng 0
Bình luận (0)
Cho tan giác ABC có: \(\widehat{C}=2\widehat{B}=4\widehat{A}\). CMR: \(\dfrac{1}{AB}+\dfrac{1}{AC}=\dfrac{1}{BC}\)
Gọi ACAC sao cho EE là điểm trên cạnh CE=ABCE=ABˆDBC=ˆDCB=12ˆABC=ˆABDDBC^=DCB^=12ABC^=ABD^⇒ABAC=BDCB⇒ABAC=BDCB (1)⇒ˆDEC=ˆDAB=4ˆC⇒DEC^=DAB^=4C^⇒ˆEDC=ˆADB=2ˆC⇒EDC^=ADB^=2C^⇒DB=EB⇒DB=EB (5)từ (1, 5)⇒ABAC+ABBC=1⇒ABAC+ABBC=1
Đúng 0
Bình luận (0)
Cho tam giác ABC có \(\widehat{C}=2\widehat{B}=4\widehat{A}\). CMR: \(\dfrac{1}{AB}+\dfrac{1}{AC}=\dfrac{1}{BC}\)
Cho abc=1 .CMR: \(\dfrac{1}{1+a+ab}+\dfrac{1}{1+b+bc}+\dfrac{1}{1+c+ac}\)
\(\dfrac{1}{1+a+ab}+\dfrac{a}{a+ab+abc}+\dfrac{1}{abc+ac+c}\)
\(=\dfrac{1}{1+a+ab}+\dfrac{a}{1+a+ab}+\dfrac{1}{c\left(1+a+ab\right)}\)
\(=\dfrac{ac+c+1}{c\left(1+a+ab\right)}=\dfrac{c\left(a+1+ab\right)}{c\left(1+a+ab\right)}=1\)
Đúng 0
Bình luận (0)