\(a^2+2a+b^2+4b+4c^2-4c+6=0\)

\(\Leftrightarrow\left(a^2+2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)

\(\Leftrightarrow\left(a+1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)

Mà \(\begin{cases}\left(a+1\right)^2\ge0\\\left(b+2\right)^2\ge0\\\left(2c-1\right)^2\ge0\end{cases}\)

\(\Rightarrow\left(a+1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\)

\(\Rightarrow\begin{cases}a+1=0\\b+2=0\\2c-1=0\end{cases}\)\(\Rightarrow\begin{cases}a=-1\\b=-2\\c=\frac{1}{2}\end{cases}\)

\(a^2-2a+b^2+4b+4c^2-4c+6=0\)'

\(\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)

\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)

b tự làm nốt nhé~

\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54-x\right)\)

\(M=x^3+3^3-x^3-54+x\)

\(M=x+27-54\)

\(M=x+27-54\)

\(M=7-27\)

\(M=-20\)

\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54-x\right)\)

\(M=x^3+3^3-x^3-54+x\)

\(M=27-54+x\)

\(M=-27+x\)

thay x =7 vào M ta có :

\(M=-27+7=-20\)

b. Câu hỏi của Phạm Thị Thùy Linh - Toán lớp 8 - Học toán với OnlineMath

a² -2a+b²+4b+4c²-4c+6=0

<=>(a²-2a+1)+(b²+4b+4)+(4c²-4c+1)=0

<=>(a-1)²+(b+2)²+(2c-1)²=0

\(\left[{}\begin{matrix}a-1=0\\b+2=0\\2c-1=0\end{matrix}\right.\left[{}\begin{matrix}a=1\\b=-2\\c=\dfrac{1}{2}\end{matrix}\right.\)

Ta có: \(a^2-2a+b^2+4b+4c^2-4c+6=\)

\(=\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)\)= 0

\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)

Mà \(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\left(\forall a;b;c\right)\)

Dấu "=" xảy ra \(\Leftrightarrow\left(a-1\right)^2=0;\left(b+2\right)^2=0;\left(2c-1\right)^2=0\)

\(\Leftrightarrow a=1;b=-2;c=\dfrac{1}{2}\)

a) Ta có: \(a^2+b^2+c^2=ab+bc+ca\)

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)(1)

Mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)nên:

(1) xảy ra\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\left(đpcm\right)\)

ai làm giúp em phép tính này với em làm mãi ko dc ạ 

bài 5 tính nhanh

a 100 -99 +98 - 97 + 96 - 95 + ... + 4 -3 +2 

b 100 -5 -5 -...-5 ( có 20 chữ số 5 )

c 99- 9 -9 - ... -9 ( có 11 chữ số 9 ) 

d 2011 + 2011 + 2011 + 2011 -2008 x 4

i 14968+ 9035-968-35

k 72 x 55 + 216 x 15 

l 2010 x 125 + 1010 / 126 x 2010 -1010

e 1946 x 131 + 1000 / 132 x 1946 -946

g 45 x 16 -17 / 45 x 15 + 28 

h 253 x 75 -161 x 37 + 253 x 25 - 161 x 63 / 100 x 47 -12 x 3,5 - 5,8 : 0,1

Bài 1: 

\(=\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}+\dfrac{1}{\left(x+13\right)\left(x+16\right)}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+13\right)}+\dfrac{3}{\left(x+13\right)\cdot\left(x+16\right)}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+13}+\dfrac{1}{x+13}-\dfrac{1}{x+16}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{x+1}-\dfrac{1}{x+16}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{x+16-x-1}{\left(x+1\right)\left(x+16\right)}=\dfrac{5}{\left(x+1\right)\left(x+16\right)}\)

Bài 2: 

\(\Leftrightarrow a^2-2a+1+b^2+4b+4+4c^2-4c+1=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b+4\right)^2+\left(2c-1\right)^2=0\)

Dấu '=' xảy ra khi a=1; b=-4; c=1/2

\(a^2-2a+b^2+4b+4c^2-4c+6=0\\ \Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\\ \Leftrightarrow\left(a+1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(a+1\right)^2=0\\\left(b+2\right)^2=0\\\left(2c-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+1=0\\b+2=0\\2c-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=-2\\c=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(\left\{a;b;c\right\}=\left\{-1;-2;\dfrac{1}{2}\right\}\)

a nhân 2 vào 2 vế ta có

2a²+2b²+2c²=2ab +2bc+2ca

=> 2a²+2b²+2c²-2ab-2bc-2ca=0

=>(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²)=0

=>(a-b)²+(b-c)²+(c-a)²=0

=>(a-b)=(b-c)=(c-a)=0

=>a-b=0 =>a=b (1)

b-c=0=>b=c (2)

từ (1) và (2)

=>a=b=c (đpcm)