\(cho:ab+bc+ac=2006\left(a,b,c\in Z\right)\)
\(CM:P=\left(a^2+2006\right)\left(b^2+2006\right)\left(c^2+2006\right)\)là số chính phương
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh:
a) \(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
b)\(\frac{4a^4+5b^4}{4c^4+5d^4}=\frac{a^2b^2}{c^2d^2}\)
c)\(\left(\frac{a-b}{c-d}\right)^{2005}=\frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}\)
d)\(\frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}=\frac{\left(a+b\right)^{2005}}{\left(c+d\right)^{2005}}\)
e)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
f)\(\frac{\left(20a^{2007}-11c^{2007}\right)^{2006}}{\left(20a^{2006}+11c^{2006}\right)^{2007}}=\frac{\left(20b^{2007}-11d^{2007}\right)^{2006}}{\left(20b^{2006}+11d^{2006}\right)^{2007}}\)
ừ, bạn bik làm thì giúp mình nha ^^
Cho a+b+c=0 và ab+bc+ca=0
Tính M = \(\left(a-2005\right)^{2006}+\left(b-2005\right)^{2006}+ \left(c-2005\right)^{2006}\)
Ta có a+b+c=0\(\Rightarrow\)\(\left(a+b+c\right)^2=0\)\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)\(\Rightarrow a^2+b^2+c^2=0\).Mặt khác ta có :\(a^2\ge0\forall a;b^2\ge0\forall b;c^2\ge0\forall c\)\(\Rightarrow a=b=c=0\)\(\Rightarrow\)\(M=\left(a-2005\right)^{2006}+\left(b-2005\right)^{2006}+\left(c-2005\right)^{2006}\)=\(\left(-2005\right)^{2006}+\left(-2005\right)^{2006}+\left(-2005\right)^{2006}\)=\(3.2005^{2006}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)Chứng minh:
a)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}^{ }}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
b)\(\left(4a+5b\right)\left(7c-11d\right)=\left(7a-11b\right)\left(4c+5d\right)\)
Cho \(\left(a+\sqrt{a^2+2006}\right)\left(b+\sqrt{b^2+2006}\right)=2006\) hãy tính tổng a+b
bài 1: cho abc=2006
tính A=
\(\dfrac{a}{ab+a+2006}+\dfrac{b}{bc+b+1}+\dfrac{2006c}{ac+2006c+2006}\)
bài 2:a,b,c thỏa mãn \(a^3+b^3+c^3\)=3abc
tính N=\(\left(1+\dfrac{a}{b}\right).\left(1+\dfrac{b}{c}\right).\left(1+\dfrac{c}{a}\right)\)
2)
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-cb-ac\right)\)
\(\Rightarrow a+b+c=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(\Rightarrow N=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(\Rightarrow N=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\)
\(\Rightarrow N=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\)
\(\Rightarrow N=-1\)
Bài 1:
Thay 2006 = abc vào biểu thức A ,có :
\(\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{abc^2}{ac+abc^2+abc}\)
\(=\dfrac{a}{a+ab+abc}+\dfrac{ab}{a\left(1+b+bc\right)}+\dfrac{c.abc}{c\left(a+ab+abc\right)}\)
\(=\dfrac{a}{a+ab+abc}+\dfrac{ab}{a+ab+abc}+\dfrac{abc}{a+ab+abc}\)
\(=\dfrac{a+ab+abc}{a+ab+abc}=1\)
Vậy tại abc = 2006 giá trị biểu thức A là 1
E xin ủng hộ cách khác cho bài 2 :(
Áp dụng hđt mở rộng ta có:\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc=3abc\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2=ab+bc+ac\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
\(N=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Với \(a=b=c\) ta có: \(N=\dfrac{2a.2a.2a}{a^3}=\dfrac{8a^3}{a^3}=8\)
Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\) ta có: \(N=\dfrac{-abc}{abc}=-1\)
Cho \(\left(a+\sqrt{a^2+2006}\right)\left(b+\sqrt{b^2+2006}\right)=2006\). Hãy tính tổng a+b
Ta có:
\(\left(a+\sqrt{a^2+2006}\right)\left(b+\sqrt{b^2+2006}\right)=2006\)
Dễ thấy \(\left\{{}\begin{matrix}\sqrt{a^2+2006}-a\ne0\\\sqrt{b^2+2006}-b\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+\sqrt{a^2+2006}\right)\left(\sqrt{a^2+2006}-a\right)\left(b+\sqrt{b^2+2006}\right)=2006\left(\sqrt{a^2+2006}-a\right)\\\left(a+\sqrt{a^2+2006}\right)\left(b+\sqrt{b^2+2006}\right)\left(\sqrt{b^2+2006}-b\right)=2006\left(\sqrt{b^2+2006}-b\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2006\left(b+\sqrt{b^2+2006}\right)=2006\left(\sqrt{a^2+2006}-a\right)\\2006\left(a+\sqrt{a^2+2006}\right)=2006\left(\sqrt{b^2+2006}-b\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b+\sqrt{b^2+2006}=\sqrt{a^2+2006}-a\\a+\sqrt{a^2+2006}=\sqrt{b^2+2006}-b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b+a=\sqrt{a^2+2006}-\sqrt{b^2+2006}\left(1\right)\\a+b=\sqrt{b^2+2006}-\sqrt{a^2+2006}\left(2\right)\end{matrix}\right.\)
Lấy (1) + (2) ta được
\(a+b=0\)
Ta có : \(\left(a+\sqrt{a^2+2006}\right)\left(b+\sqrt{b^2+2006}\right)=2006\) (*)
Nhân liên hợp ta được :
(*)\(\Leftrightarrow\dfrac{\left(a+\sqrt{a^2+2006}\right)\left(a-\sqrt{a^2+2006}\right)}{a-\sqrt{a^2+2006}}.\)\(\dfrac{\left(b+\sqrt{b^2+2006}\right)\left(b-\sqrt{b^2+2006}\right)}{b-\sqrt{b^2-2006}}=2006\)
\(\Leftrightarrow\dfrac{a^2-a^2-2006}{a-\sqrt{a^2+2006}}.\dfrac{b^2-b-2006}{b-\sqrt{b^2+2006}}=2006\)
\(\Leftrightarrow\left(-2006\right).\left(-2006\right)\dfrac{1}{\left(a-\sqrt{a^2+2006}\right)\left(b-\sqrt{b^2+2006}\right)}=2006\)
\(\Leftrightarrow\)\(\Leftrightarrow\dfrac{1}{\left(a-\sqrt{a^2+2006}\right)\left(b-\sqrt{b^2+2006}\right)}=\dfrac{1}{2006}\)
=> \(\left(a-\sqrt{a^2+2006}\right)\left(b-\sqrt{b^2+2006}\right)=2006\) (**)
Từ (*) và (**) ta suy ra :
\(\dfrac{\left(a-\sqrt{a^2+2006}\right)\left(b-\sqrt{b^2+2006}\right)}{\left(a+\sqrt{a^2+2006}\right)\left(b+\sqrt{b^2+2006}\right)}=1\)
Và \(\dfrac{a-\sqrt{a^2+2006}}{a+\sqrt{a^2+2006}}=\dfrac{b+\sqrt{b^2+2006}}{b-\sqrt{b^2+2006}}\)
=> \(\dfrac{a-\sqrt{a^2+2006}}{a+\sqrt{a^2+2006}}=\dfrac{b+\sqrt{b^2+2006}}{b-\sqrt{b^2+2006}}=\dfrac{1}{2}\)
+ , \(\dfrac{a-\sqrt{a^2+2006}}{a+\sqrt{a^2+2006}}=\dfrac{1}{2}\Rightarrow2a-2\sqrt{a^2+2006}=a+\sqrt{a^2+2006}\Rightarrow a=3\sqrt{a^2+2006}\)
Tương tự : b = \(3\sqrt{b^2+2006}\)
=> a+b = \(3\left(\sqrt{a^2+2006}+\sqrt{b^2+2006}\right)\)
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không biết hướng làm này có đúng không nữa ... tại còn dính ẩn ...
Tìm x biết : \(\frac{\left(2006-x\right)^2+\left(2006-x\right)\left(x-2007\right)+\left(x-2007\right)^2}{\left(2006-x\right)^2-\left(2006-x\right)\left(x-2007\right)+\left(x-2007\right)^2}=\frac{19}{49}\)
Đặt x -2006 = y
pt <=> \(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)
<=> \(\frac{y^2-y^2+y+y^2-2y+1}{y^2+y^2-y+y^2-2y+1}=\frac{19}{49}\)
<=> \(\frac{y^2-y+1}{3y^2-3y+1}=\frac{19}{49}\)
<=> \(49y^2-49y+49=57y^2-57y+19\)
<=> \(8y^2-8y-30=0\)
<=> \(4y^2-4y+15=0\)
Giải tiếp nha
Bài 1 : cho a , b ,c là ba số khác 0 thỏa mãn điều kiện a^3 + b^3 +c^3 = 3abc . và a + b + c = 0 . Tính gtbt :
\(M=\left(1+\frac{a}{b}\right)\left(a+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Bài 2 :
Cho 3 số x , y , z \(\ge0\) và x2006 + y2006 + z2006 = 3 . Tìm GTLN của A = x^2 + y ^2 + x ^2
Ai làm được mk link cho nha
Sửa đề \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Ta có: \(a^3+b^3+c^3=3ab\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
TH1: a+b+c=0
=> \(\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}}\)
Thay vào M ta được M=\(\left(1-\frac{b+c}{b}\right)\left(1-\frac{a+c}{c}\right)\left(1-\frac{a+b}{a}\right)\)
\(\Rightarrow M=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)
TH2: \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow M=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Bài 1 : cho a , b ,c là ba số khác 0 thỏa mãn điều kiện a^3 + b^3 +c^3 = 3abc . và a + b + c = 0 . Tính gtbt :
\(M=\left(1+\frac{a}{b}\right)\left(a+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Bài 2 :
Cho 3 số x , y , z \(\ge0\) và x2006 + y2006 + z2006 = 3 . Tìm GTLN của A = x^2 + y ^2 + x ^2
Ai làm được mk link cho nha