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Ta có: \(a+b+c=0\)
\(=>\left(a+b+c\right)^2=0\)
\(=>a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(=>a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(=>a^2+b^2+c^2=0\)
\(=>a^2+b^2+c^2=ab+bc+ac\)
\(=>2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(=>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)(nhân phân phối, đổi qua bên kia dấu bằng, tách thành hằng đẳng thức)
\(=>\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(=>\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\)
\(=>a=b=c=0\)
***\(A=\left(a-1\right)^{22}+b^{12}+\left(c-1\right)^{2014}\)
\(A=\left(-1\right)^{22}+1+\left(-1\right)^{2014}\)
\(A=1+1+1\)
\(A=3\)
Ta có
a + b + c = 0
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow\)a2 + b2 + c2 = ab + bc + ca
Mà ta có a2 + b2 + c2 \(\ge\) ab + bc + ca
Dấu = xảy ra khi a = b = c = 0
\(\Rightarrow\)(a - 1)22 + b12 + (c - 1)2014 = 1 + 0 + 1 = 2
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{3}{a^2b}+\frac{3}{ab^2}+\frac{1}{b^3}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{b^3}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{-3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{-3}{ab}\cdot\frac{-1}{c}=\frac{3}{abc}\)
Ta có: \(M=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)
Ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+3\frac{1}{a}.\frac{1}{b}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3\frac{1}{a}\frac{1}{b}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3\frac{1}{a}\frac{1}{b}\left(-\frac{1}{c}\right)\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\frac{1}{abc}=\frac{3}{abc}\)
Ta lại có :
\(P=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{bca}{b^3}+\frac{cab}{c^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)
\(\)
Bài làm:
Ta có: \(P=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
CM HĐT phụ:
Ta có: \(a^3+b^3+c^3=\left(a^3+b^3+c^3-3abc\right)+3abc\)
\(=\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\right]+3abc\)
\(=\left[\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\right]+3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc\)
Áp dụng vào trên ta được:
\(abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\left[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{bc}-\frac{1}{ca}\right)+\frac{3}{abc}\right]\)
Mà \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(P=abc.\frac{3}{abc}=3\)
Vậy P = 3
\(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=0\)
Mà \(a^2;b^2;c^2\ge0\forall a;b;c\) nên điều này xảy ra \(\Leftrightarrow a=b=c=0\)
\(\Rightarrow M=2018^{2014}+2018^{2014}-2018^{2014}=2018^{2014}\)
Câu hỏi của Hà Văn Minh Hiếu - Toán lớp 8 - Học toán với OnlineMath
Ta có : \(a+b+c=6\)
\(\Rightarrow\left(a+b+c\right)^2=36\)
\(\Rightarrow a^2+b^2+c^2+2.\left(ab+bc+ca\right)=36\)
\(\Rightarrow a^2+b^2+c^2=36-2.12=12\)
Do đó : \(a^2+b^2+c^2=ab+bc+ca\left(=12\right)\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Khi đó biểu thức :
\(\left(a-b\right)^{2012}+\left(b-c\right)^{2013}+\left(c-a\right)^{2014}=0+0+0=0\)
Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(=2a^2+2b^2+2c^2-2ab-2bc-2ac\)
\(=2\left(a^2+b^2+c^2+2ab+2ac+2bc\right)-6ab-6bc-6ac\)
\(=2\left(a+b+c\right)^2-6\left(ab+bc+ac\right)\)
\(=2.6^2-6.12=0\)
Mà : \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)
nên \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Do đó: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
<=> \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Leftrightarrow a=b=c\)
Vậy \(\left(a-b\right)^{2012}+\left(b-c\right)^{2013}+\left(c-a\right)^{2014}=0\)
Lời giải:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
$\Rightarrow ab+bc+ac=0$
Đặt $ab=x, bc=y, ac=z$ thì $x+y+z=0$
Có:
$M=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}$
$=\frac{b^3c^3+a^3c^3+a^3b^3}{(abc)^2}$
$=\frac{x^3+y^3+z^3}{xyz}=\frac{(x+y)^3-3xy(x+y)+z^3}{xyz}$
$=\frac{(-z)^3-3xy(-z)+z^3}{xyz}$
$+\frac{-z^3+3xyz+z^3}{xyz}=\frac{3xyz}{xyz}=3$
\(N=\dfrac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{\left(ab\right)\left(bc\right)\left(ca\right)}\)
Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\Rightarrow N=\dfrac{x^3+y^3+z^3}{xyz}\)
\(N=\dfrac{x^3+y^3+z^3-3xyz+3xyz}{xyz}=\dfrac{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]+3xyz}{xyz}=\dfrac{3xyz}{xyz}=3\)
Lời giải:
Ta có: \(a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=0^2-2.0=0\)
Vì $a^2\geq 0; b^2\geq 0; c^2\geq 0$ với mọi $a,b,c\in\mathbb{R}$ nên để tổng của chúng bằng $0$ thì $a^2=b^2=c^2=0$
$\Rightarrow a=b=c=0$
Khi đó:
$A=(a-1)^{22}+b^{12}+(c-1)^{2014}=(-1)^{22}+0^{12}+(-1)^{2014}=2$
a2+b2+c2=(a+b+c)2−2(ab+bc+ac)=02−2.0=0a2+b2+c2=(a+b+c)2−2(ab+bc+ac)=02−2.0=0
Vì a2≥0;b2≥0;c2≥0a2≥0;b2≥0;c2≥0 với mọi a,b,c∈Ra,b,c∈R nên để tổng của chúng bằng 00 thì a2=b2=c2=0a2=b2=c2=0
⇒a=b=c=0⇒a=b=c=0
Khi đó:
A=(a−1)22+b12+(c−1)2014=(−1)22+012+(−1)2014=2