\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)

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

\(=\frac{1}{ab}\)

\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+14xy+y^2}{16x}\)

\(=\frac{\left(2x+y\right)^2+2\left(2x+y\right)\left(2x-y\right)+\left(2x-y\right)^2}{\left(2x+y\right)^2.\left(2x-y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)

\(=\frac{\left(2x+y+2x-y\right)^2}{\left(2x+y\right)^2.\left(2x-y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)

\(=\frac{x}{\left(2x-y\right)^2}\)

\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{1}{a}+\frac{1}{b}\right)\right].\frac{ab}{\left(a+b\right)^2}\)

ĐK: a, b khác 0, a khác -b

\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}.\left(\frac{a+b}{ab}\right)\right].\frac{ab}{\left(a+b\right)^2}\)

\(A=\left[\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}\right].\frac{ab}{\left(a+b\right)^2}=\left(\frac{1}{a}+\frac{1}{b}\right)^2.\frac{ab}{\left(a+b\right)^2}\)

\(A=\frac{\left(a+b\right)^2}{ab}.\frac{ab}{\left(a+b\right)^2}=1\)

 \(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{\left(4x^2-y^2\right)}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16xy}\)

ĐK: xy khác 0, y  \(\ne\pm\)2x

\(B=\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{\left(2x-y\right).\left(2x+y\right)}+\frac{1}{\left(2x+y\right)^2}\right].\frac{\left(2x+y\right)^2}{16xy}\)

\(B=\left[\frac{1}{\left(2x-y\right)}+\frac{1}{\left(2x+y\right)}\right]^2.\frac{\left(2x+y\right)^2}{16xy}\)

\(B=\left(\frac{2x+y+2x-y}{\left(2x-y\right).\left(2x+y\right)}\right)^2.\frac{\left(2x+y\right)^2}{16xy}\)

\(B=\frac{16x^2}{\left(2x-y\right)^2.\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16xy}\)

\(B=\frac{x}{\left(2x-y\right)^2.y}\)

Ta thấy \(1+x+x^2=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow1+x+x^2+x^3>x^3\)

\(+,\left(x+2\right)^2-\left(1+x+x^2+x^3\right)=x^2+3.x^2.2+3.x.4+8-1-x-x^2-x^3\)

\(=5x^2+11x+7=5\left(x^2+2.\frac{11}{10}x+\frac{121}{100}\right)+\frac{19}{20}=5.\left(x+\frac{11}{10}\right)^2+\frac{19}{20}>0\)

\(\Rightarrow\left(x+2\right)^2>1+x+x^2+x^3\)

\(\Rightarrow x^3< 1+x+x^2+x^3< \left(x+2\right)^3\)

Vậy \(1+x+x^2+x^3=\left(x+1\right)^3\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\left(tm\right)\)

Với x=-1 => y=0

      x=0   =>y=1 

(thay vào là đc nha)

Vậy ....

Bước 2 là sao cậu

\(1+x+x^2+x^3\)>\(x^3\)+.....(chỗ này mik k hiểu lắm)

làm hộ tớ với

\(ĐKXĐ:x\ne\frac{-1}{2};x\ne-1\)

\(pt\Leftrightarrow\frac{x^2-5x+1}{2x+1}+\frac{x^2-4x+1}{x+1}=-2\)

\(\Leftrightarrow\frac{\left(x^2-5x+1\right)\left(x+1\right)+\left(2x+1\right)\left(x^2-4x+1\right)}{\left(2x+1\right)\left(x+1\right)}=-2\)

\(\Leftrightarrow\frac{x^3-4x^2-4x+1+2x^3-7x^2-2x+1}{2x^2+3x+1}=-2\)

\(\Leftrightarrow\frac{3x^3-11x^2-6x+2}{2x^2+3x+1}=-2\)

\(\Leftrightarrow3x^3-11x^2-6x+2=-4x^2-6x-2\)

\(\Leftrightarrow3x^3-7x^2+4=0\)

Tính được: \(\Delta=b^2-3ac\)

\(m=\frac{9abc-2b^3-27a^2d}{2\sqrt{\left|\Delta\right|^3}}\)rồi thay vào giải pt bậc ba như thường

tách:

\(\frac{\left(t-x\right)\left(t-y\right)}{\left(t-a\right)\left(t-b\right)\left(t-c\right)}=\frac{A}{t-a}+\frac{B}{t-b}+\frac{C}{t-c}\left(1\right)\)

khi đó:

\(\left(t-x\right)\left(t-y\right)=A\left(t-b\right)\left(t-c\right)+B\left(t-c\right)\left(t-a\right)+C\left(t-a\right)\left(t-b\right)\)

Cho t=a; t=b; t=c

=> \(A=\frac{\left(a-x\right)\left(a-y\right)}{\left(a-b\right)\left(a-c\right)};B=\frac{\left(b-x\right)\left(b-y\right)}{\left(b-c\right)\left(b-a\right)};C=\frac{\left(c-x\right)\left(c-y\right)}{\left(c-a\right)\left(c-b\right)}\)

trong đẳng thức (1) ta cho t=0 ta được \(P=\frac{xy}{abc}\)

Ta xét hiệu :

\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ac+a^2}\right)\)

\(=a-b+b-c+c-a=0\)

Do đó : \(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ac+a^2}=1006\)

Khi đó \(M=2\cdot1006=2012\)

Chỉ ra được : \(M=2\cdot1006=2012\)

Gợi ý : Xét hiệu .

A B C D I K O

\(1,\hept{\begin{cases}OI//AB\Rightarrow\frac{OI}{AB}=\frac{OD}{BD}\\OI//CD\Rightarrow\frac{OI}{CD}=\frac{OA}{AC}\\AB//CD\Rightarrow\frac{OA}{AC}=\frac{OB}{BD}\end{cases}}\Rightarrow\frac{OI}{AB}+\frac{OI}{CD}=\frac{OD}{BD}+\frac{OA}{AC}=\frac{OD}{BD}+\frac{OB}{BD}=\frac{BD}{BD}=1\)

\(\hept{\begin{cases}OK//AB\Rightarrow\frac{OC}{AC}=\frac{OK}{AB}\\OK//CD\Rightarrow\frac{OK}{CD}=\frac{OB}{BD}\\\frac{CB}{BD}=\frac{OA}{AC}\end{cases}}\Rightarrow\frac{OK}{AB}+\frac{OK}{CD}=\frac{OC}{AC}+\frac{OB}{BD}=\frac{OC}{AC}+\frac{OA}{AC}=\frac{AC}{AC}=1\)

\(2,\hept{\begin{cases}\frac{OI}{AB}+\frac{OI}{CD}=1\\\frac{OK}{AB}+\frac{OK}{CD}=1\end{cases}}\Rightarrow\frac{OI}{AB}+\frac{OI}{CD}+\frac{OK}{AB}+\frac{OK}{CD}=2\)

\(\Leftrightarrow\frac{OI+OK}{AB}+\frac{OI+OK}{CD}=2\)

\(\Leftrightarrow\frac{IK}{AB}+\frac{IK}{CD}=2\)

\(\Leftrightarrow\frac{1}{AB}+\frac{1}{CD}=\frac{2}{IK}\left(đpcm\right)\)

Giúp mik bài này với: https://olm.vn/hoi-dap/detail/244594379058.html

a) 3x - 2 = 2x-3

<=> 3x-2 -2x +3 = 0

<=> x +1 = 0

<=> x = -1

c) 3 - 4y+24+6y=y+27+3y

<=> 3 - 4y+24+6y - y - 27 - 3y = 0

<=> -2y =0

<=> y = 0

b,7-2x = 22 - 3x

<=> 7-2x -22 +3x = 0

<=> -15 +x = 0

<=> x = 15

d) x-12+4x = 25+2x-1

<=> x-12+4x -25-2x+1=0

<=> 3x -36 = 0

<=> 3x = 36

<=> x = 12

còn câu e bạn tự làm nha

\(a,3x-2=2x-3\)

\(3x-2x=-3+2\)

\(x=-1\)

Vậy pt cs nghiệm là  { -1 }

\(b,7-2x=22-3x\)

\(-2x+3x=22-7\)

\(x=15\)

Vậy pt cs nghiệm là { 15 }

bn lm nốt nha ... 

Cho hợp chất \(B\) tác dụng hết với kim loại \(Al\) thu đc \(AlCl_3\) và \(H_2\)

\(\Rightarrow B\)là \(HCl\) đó có n.tố H, Cl ở sp

Thử lại thấy thoả mãn yêu cầu

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

(Không chắc lắm @@)