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

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

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

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

\(4.\)

\(a.A=5-8x-x^2\)

\(=-\left(16+8x+x^2\right)+21\)

\(=-\left(4+x\right)^2+21\le21\)

\(A_{max}=21\)

Dấu '='xảy ra khi \(x=-4\)

\(b.B=5-x^2+2x-4y^2-4y\)

\(=-\left(1-2x+x^2\right)-\left(4+4y+4y^2\right)+10\)

\(=-\left(1-x\right)^2-\left(2+2y\right)^2+10\le10\)

\(B_{max}=10\)

Dấu '=' xảy ra khi \(x=1;y=-1\)

\(5.\)

\(a.\) Ta có:\(a^2+b^2+c^2=ab+bc+ca\)

              \(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

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

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

              \(\Leftrightarrow a-b=0\Leftrightarrow a=b\left(1\right)\)

              hay\(b-c=0\Leftrightarrow b=c\left(2\right)\)

             hay\(c-a=0\Leftrightarrow c=a\left(3\right)\)

Từ \(\left(1\right),\left(2\right)\)và\(\left(3\right)\)suy ra:\(a=b=c\left(đpcm\right)\)

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

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

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

\(\Leftrightarrow a-1=0\Leftrightarrow a=1\)

hay\(b+2=0\Leftrightarrow b=-2\)

hay\(2c-2=0\Leftrightarrow c=1\)

V...

^^

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

\(2\left(x^2+8x+16\right)-x^2+4=0\)

\(\Leftrightarrow2x^2+16x+32-x^2+4=0\)

\(\Leftrightarrow x^2+16x+36=0\)

\(\Leftrightarrow x^2+16x+64=28\)

\(\Leftrightarrow\left(x+8\right)^2=28\)

\(\Leftrightarrow\orbr{\begin{cases}x_1=\sqrt{28}-8\\x_2=-\sqrt{28}-8\end{cases}}\)

\(2\left(x^2+8x+16\right)-x^2+4=0\)

\(2x^2+16x+32-x^2+4=0\)

\(x^2+16x+36=0\)

\(x^2+16x+64=28\)

\(\left(x+8\right)^2=28\)

bình phương thì chia lm 2 trường hợp 

lm tiếp phần sau 

a) \(A=5-8x-x^2\)

        \(=-\left(x^2+8x-5\right)\)

        \(=-\left(x^2+2.x.4+4^2-16-5\right)\)

        \(=-\left[\left(x+4\right)^2-21\right]\)

        \(=-\left(x+4\right)^2+21\le21\)

       Dấu "=" khi x + 4 = 0 => x = -4

       Vậy GTLN của A là 21 khi x = -4

b) \(B=5-x^2+2x-4y^2-4y\)

       \(=-\left(x^2-2x+4y^2+4y-5\right)\)

       \(=-\left[x^2-2x+1+\left(2y\right)^2+2.2y.1+1-7\right]\)

      \(=-\left[\left(x-1\right)^2+\left(2y+1\right)^2\right]+7\le7\)

    Dấu "=" khi \(\hept{\begin{cases}x-1=0\\2y+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=\frac{-1}{2}\end{cases}}}\)

   Vậy GTLN của B là 7 khi x = 1 và y = -1/2

c) Theo đề: \(a^2+b^2+c^2=ab+bc+ca\)

           \(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

         \(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

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

          \(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)(ĐPCM)

d) \(a^2-2a+b^2+4b+4c^2-4c+6=0\)

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

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

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

   Vậy nghiệm phương trình: a = 1; b = -2; c = 1/2

Chúc bạn học tốt ^_^

sao ko ai giúp nhỉ ;(

c. Ta có: a^2 + b^2 + c^2 = ab + bc + ca 
<=> 2.a^2 + 2.b^2 + 2.c^2 = 2.ab + 2.bc + 2.ca 
<=> ( a^2 - 2ab + b^2 ) + ( b^2 - 2bc +c^2 ) + ( c^2 - 2ac + a^2 ) =0 
<=> (a-b)^2 + (b-c)^2 + (c -a)^2 =0 (1) 
Vì (a-b)^2 ; (b-c)^2 ; (c -a)^2 ≧ 0 với mọi a,b,c. 
=> (a-b)^2 + (b-c)^2 + (c -a)^2 ≧ 0 (2) 
Từ (1) và (2) khẳng định dấu "=" khi: 
a - b = 0; b - c = 0 ; c - a = 0 => a=b=c 
Vậy a=b=c.
=> đpcm

<=>a^2-2a+b^2+4b+4c^2-4c+1+4+1=0

<=>(a^2-2a+1)+(b^2+4b+4)+(4c^2-4c+1)=0

<=>(a-1)²+(b+2)²+(2c-1)²=0

<=>(a-1)^2=0 hoặc(b+2)^2=0 hoặc (2c-1)^2=0

+,(a-1)^2=0<=>a-1=0<=>a=1

+,(b+2)^2=0<=>b+2=0<=>b=-2

+,(2c-1)^2=0<=>2c-1=0<=>2c=1<=>c=1/2