\(a,\)\(4Al\)\(+\)\(3O_2\)\(=\)\(2Al_2O_3\)

\(b,\)\(m_{Al}\)\(+\)\(mO_2\)\(=\)\(mAl_2O_3\)

\(c,\)\(Ta\)\(có\)\(:\)

\(mAl\)\(+\)\(mO_2\)\(=\)\(mAl_2O_3\)

=> 10,8 +mO2 =20,4

=> mO2   = 20,4 -10,8 = 9,6 ( g ) 

Vậy mO2 = 9,6 g

\(B=\frac{a+1}{a^2-a+1}-\frac{1}{a+1}-\frac{a-2}{a^3+1}=\frac{\left(a+1\right)^2}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a-2}{a^3+1}\\ \)

 \(=\frac{a^2+2a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a^2-a+1}{\left(a+1\right).\left(a^2-a+1\right)}-\frac{a-2}{\left(a+1\right).\left(a^2-a+1\right)}\)

\(=\frac{a^2+2a+1-\left(a^2-a+1\right)-\left(a-2\right)}{\left(a+1\right).\left(a^2-a+1\right)}=\frac{2a+2}{\left(a+1\right).\left(a^2-a+1\right)}=\frac{2}{a^2-a+1}\)

ơ bài này là bài lớp 2 mà  , sao lại ghi là lớp 8 vậy

a. Ta có: \(x^2-10x+26+y^2+2y=0\Leftrightarrow\left(x^2-10x+25\right)+\left(y^2+2y+1\right)=0\\ \)

\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\Rightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}}\)

b. \(\left(2x+5\right)^2-\left(x-7\right)^2=0\Leftrightarrow\left(2x+5+x-7\right).\left(2x+5-x+7\right)=0\)

\(\Leftrightarrow\left(3x-2\right).\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-12\end{cases}}}\)

c. \(25.\left(x-3\right)^2=49.\left(1-2x\right)^2\Leftrightarrow\left(5x-15\right)^2=\left(7-14x\right)^2\Leftrightarrow\left(5x-15\right)^2-\left(7-14x\right)^2=0\)

\(\Leftrightarrow\left(5x-15-7+14x\right).\left(5x-15+7-14x\right)=0\Leftrightarrow\left(19x-22\right).\left(-9x-8\right)=0\)

\(\Leftrightarrow\left(19x-22\right).\left(9x+8\right)=0\Leftrightarrow\orbr{\begin{cases}19x-22=0\\9x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{22}{19}\\x=-\frac{8}{9}\end{cases}}}\)

d. \(\left(x+2\right)^2=\left(3x-5\right)^2\Leftrightarrow\left(x+2\right)^2-\left(3x-5\right)^2=0\Leftrightarrow\left(x+2+3x-5\right).\left(x+3-3x+5\right)=0\)

\(\Leftrightarrow\left(4x-3\right).\left(8-2x\right)=0\Leftrightarrow\orbr{\begin{cases}4x-3=0\\8-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=4\end{cases}}}\)

e. \(x^2-2x+1=16\Leftrightarrow\left(x-1\right)^2-16=0\Leftrightarrow\left(x-1-4\right).\left(x-1+4\right)=0\)

\(\Leftrightarrow\left(x-5\right).\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)

Cảm ơn bn rất nhìu nha!!!^-^!!!

\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\frac{x^2+12}{4-x^2}=\frac{\left(x-1\right).\left(x-2\right)}{x^2-4}-\frac{\left(x+2\right)^2}{x^2-4}+\frac{x^2+12}{x^2-4}\)

  \(=\frac{x^2-3x+2}{x^2-4}-\frac{x^2+4x+4}{x^2-4}+\frac{x^2+12}{x^2-4}=\frac{x^2-7x+10}{x^2-4}=\frac{\left(x-2\right).\left(x-5\right)}{\left(x-2\right).\left(x+2\right)}=\frac{x-5}{x+2}\)

\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\)\(\frac{x^2+12}{4-x^2}\)\(ĐKXĐ\): \(x\ne\pm2\)

\(=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)\(-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)\(+\frac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2-2x-x+2-x^2-4x-4+x^2+12}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2-7x+10}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2-2x-5x+10}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x\left(x-2\right)-5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{\left(x-5\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x-5}{x+2}\)

Trên CD lấy điểm K sao cho DA = DK. Khi đó ta chứng minh AK là phân giác góc A và BK cũng là phân giác góc B.

Bạn xem lời giải ở đây nhé.

Câu hỏi của Amber Shindouya - Toán lớp 8 - Học toán với OnlineMath

2/

a, \(A=2x^2+6x-5=2\left(x^2+3x-\frac{5}{2}\right)=2\left(x^2+2x\cdot\frac{3}{2}+\frac{9}{4}-\frac{19}{4}\right)=2\left[\left(x+\frac{3}{2}\right)^2-\frac{19}{4}\right]=2\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0\Rightarrow A=\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\ge-\frac{19}{2}\)

Dấu "=" xảy ra khi x=-3/2

Vậy Amin=-19/2 khi x=-3/2

b,bài này phải tìm min 

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

Vì \(\left(x-2\right)^2\ge0\Rightarrow B=\left(x-2\right)^2+4\ge4\)

Dấu "=" xảy ra khi x = 2

Vậy Bmin=4 khi x=2

Bài 2)Ta có:

\(2x^2+6x-5\)

\(=2x^2+6x+\frac{9}{2}-\frac{19}{2}\)

\(=2\left(x^2+3x+\frac{9}{4}\right)-\frac{19}{2}\)

\(=2\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\ge-\frac{19}{2}\)

vt đề tử tế giùm -_-

nhìn dell hỉu giề :)

bn ơi cs fải đề thế này ko?

\(2xy\left(x^2y-\frac{1}{2}xy\right)-2x^2y\left(xy-\frac{1}{2}y\right)+1\)

\(=\) \(2x^3y^2-x^2y^2-2x^3y^2+x^2y^2+1\)

\(=1\)

Vậy giá trị của biểu thức trên ko phụ thuộc vào biến nên giá trị của biểu thức luôn bằng 1

1~2~3~4~5~6~7~8~9~10~11~12~13~14~15~16~17~18~19~20

:))