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8 tháng 11 2017
\(\text{a) }\dfrac{x^2+2}{x^3-1}+\dfrac{x}{x^2+x+1}+\dfrac{1}{1-x}\\ =\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x}{x^2+x+1}-\dfrac{1}{x-1}\\ =\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}-\dfrac{\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{x^2+2+x\left(x-1\right)-\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x-1\right)}\\ =\dfrac{x^2+2+x^2-x-x^2-x-1}{\left(x^2+x+1\right)\left(x-1\right)}\\ =\dfrac{x^2-2x+1}{\left(x^2+x+1\right)\left(x-1\right)}\\ =\dfrac{\left(x-1\right)^2}{\left(x^2+x+1\right)\left(x-1\right)}\\ =\dfrac{x-1}{x^2+x+1}\\ \)
\(\text{b) }\dfrac{x}{xy-y^2}+\dfrac{2x-y}{xy-x^2}\\ =\dfrac{x}{y\left(x-y\right)}+\dfrac{2x-y}{x\left(y-x\right)}\\ =\dfrac{x}{y\left(x-y\right)}-\dfrac{2x-y}{x\left(x-y\right)}\\ =\dfrac{x^2}{y\left(x-y\right)x}-\dfrac{\left(2x-y\right)y}{x\left(x-y\right)y}\\ =\dfrac{x^2-\left(2x-y\right)y}{xy\left(x-y\right)}\\ =\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)}\\ =\dfrac{\left(x-y\right)^2}{xy\left(x-y\right)}\\ =\dfrac{x-y}{xy}\)
21 tháng 4 2017
a) −x2+21−5x−x2+21−5x =x2+2−(1−5x)=x2+2−(1−5x) =x2+25x−1=x2+25x−1;
b) −4x+15−x−4x+15−x =4x+1−(5−x)=4x+1−(5−x) =4x+1x−5
TH
20 tháng 12 2018
Bài 2:
a) \(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{1}{abc}\left(a^3+b^3+c^3\right)\)
\(A=\dfrac{1}{abc}\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]\)
Vì \(a+b+c=0\)
Nên a + b = -c (1)
Thay (1) vào A, ta được:
\(A=\dfrac{1}{abc}\left[\left(-c\right)^3-3ab\left(-c\right)+c^3\right]\)
\(A=\dfrac{1}{abc}.3abc\)
\(A=3\)
b) \(B=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
\(B=\dfrac{a^2}{a^2-\left(b^2+c^2\right)}+\dfrac{b^2}{b^2-\left(c^2+a^2\right)}+\dfrac{c^2}{c^2-\left(a^2+b^2\right)}\)
Vì \(a+b+c=0\)
Nên b + c = -a
=> ( b + c )2 = (-a)2
=> b2 + c2 + 2bc = a2
=> b2 + c2 = a2 - 2bc (1)
Tương tự ta có: c2 + a2 = b2 - 2ac (2)
a2 + b2 = c - 2ab (3)
Thay (1), (2) và (3) vào B, ta được:
\(B=\dfrac{a^2}{a^2-\left(a^2-2bc\right)}+\dfrac{b^2}{b^2-\left(b^2-2ac\right)}+\dfrac{c^2}{c^2-\left(c^2-2ab\right)}\)
\(B=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ac}+\dfrac{c^2}{c^2-c^2+2ab}\)
\(B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)
\(B=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\)
\(B=\dfrac{1}{2abc}\left(a^3+b^3+c^3\right)\)
Mà \(a^3+b^3+c^3=3abc\) ( câu a )
\(\Rightarrow B=\dfrac{1}{2abc}.3abc\)
\(\Rightarrow B=\dfrac{3}{2}\)
TH
20 tháng 12 2018
Bài 1:
a) GT: abc = 2
\(M=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)
\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{abc+2cb+2b}\)
\(M=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2+2cb+2b}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2\left(1+cb+b\right)}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(M=\dfrac{1+b+bc}{bc+b+1}\)
\(M=1\)
b) GT: abc = 1
\(N=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)
\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)
\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(N=\dfrac{1+b+bc}{bc+b+1}\)
\(N=1\)
16 tháng 7 2022
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a-c}{c}=\dfrac{bk-dk}{dk}=\dfrac{b-d}{d}\)
b: \(\dfrac{a+b}{c+d}=\dfrac{bk+b}{dk+d}=\dfrac{b}{d}\)
\(\dfrac{a-b}{c-d}=\dfrac{bk-b}{dk-d}=\dfrac{b}{d}\)
Do đó: \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
28 tháng 4 2017
a)ĐK: a>0 b>0 nhé bạn đề thiếu
(a-b)2\(\ge\)0
<=>a2+b2\(\ge\)2ab
<=>a2+2ab+b2\(\ge\)4ab
<=>(a+b)2\(\ge\)4ab
<=>\(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)
<=>\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
Dấu "=" xảy ra <=> (a-b)2=0<=>a=b
=>A\(\ge\)\(\left(a+b\right)\dfrac{4}{a+b}=4\)(đpcm)
b)\(B=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{a+c}{b}=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
Áp dụng bất đẳng thức cosi x+y\(\ge\)2\(\sqrt{xy}\)cho 2 số dương x;y ta có:
\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{ac}{ca}}=2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\)
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\)
Dấu "=" xảy ra khi và chỉ khi:\(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{c}{a}\\\dfrac{b}{c}=\dfrac{c}{b}\\\dfrac{a}{b}=\dfrac{b}{a}\end{matrix}\right.\)\(\Leftrightarrow\)a=b=c
=>B\(\ge2+2+2=6\)(đpcm)
RD
27 tháng 2 2018
a)\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{a^3+b^3+c^3}{abc}\)
\(A=\dfrac{3abc}{abc}=3\)(vì a+b+c=0)
b)Ta có: a+b+c=0
\(\Rightarrow\left\{{}\begin{matrix}a=-b-c\\b=-c-a\\c=-a-b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(c+a\right)^2\\c^2=\left(a+b\right)^2\end{matrix}\right.\)
\(\Rightarrow B=\dfrac{a^2}{\left(b+c\right)^2-b^2-c^2}+\dfrac{b^2}{\left(a+c\right)^2-c^2-a^2}+\dfrac{c^2}{\left(a+b\right)^2-a^2-b^2}\)
\(\Rightarrow B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}\)
\(\Rightarrow B=\dfrac{a^3+b^3+c^3}{2abc}\)
\(\Rightarrow B=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)(vì a+b+c=0)
27 tháng 2 2018
cm:nếu a+b+c=0 thì a^3+b^3+c^3=3abc
a^3+b^3+c^3=3abc
=>a^3+b^3+c^3-3abc=0
=>(a+b)^3-3ab(a+b)+c^3-3abc=0
=>[(a+b)^3+c^3]-3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2] -3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0
vì a+b+c=0 nên a^3+b^3+c^3=3abc
thay kết quả vừa chúng minh vào đề bài ta đc
\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)
chúc bạn học tốt ^ ^
NN
29 tháng 11 2017
a) \(\dfrac{4x-1}{3x^2y}-\dfrac{7x-1}{3x^2y}\)
\(=\dfrac{\left(4x-1\right)-\left(7x-1\right)}{3x^2y}\)
\(=\dfrac{4x-1-7x+1}{3x^2y}\)
\(=\dfrac{-3x}{3x^2y}\)
\(=\dfrac{-1}{xy}\)
b) \(\dfrac{4x+5}{2x-1}-\dfrac{5-9x}{2x-1}\)
\(=\dfrac{\left(4x+5\right)-\left(5-9x\right)}{2x-1}\)
\(=\dfrac{4x+5-5+9x}{2x-1}\)
\(=\dfrac{13x}{2x-1}\)
c) \(\dfrac{11x}{2x-3}-\dfrac{x-18}{3-2x}\)
\(=\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\)
\(=\dfrac{11x+\left(x-18\right)}{2x-3}\)
\(=\dfrac{11x+x-18}{2x-3}\)
\(=\dfrac{12x-18}{2x-3}\)
\(=\dfrac{6\left(2x-3\right)}{2x-3}\)
\(=\dfrac{6}{1}\)
\(=6\)
d) \(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)
\(=\dfrac{2x-7}{10x-4}+\dfrac{3x+5}{10x-4}\)
\(=\dfrac{\left(2x-7\right)+\left(3x+5\right)}{10x-4}\)
\(=\dfrac{2x-7+3x+5}{10x-4}\)
\(=\dfrac{5x-2}{10x-4}\)
\(=\dfrac{5x-2}{2\left(5x-2\right)}\)
\(=\dfrac{1}{2}\)
`#3107.101107`
`b/(a - b) + a/(b - a)`
`= b/(a - b) - a/(a - b)`
`= (b - a)/(a - b)`
`= -(a - b)/(a - b)`
`= -1`