.

.

.

Đặt x + 29 = a (a \(\ne-29;-30\))

Đề trở thành: \(\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{5}{4}\)

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

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

\(\Leftrightarrow\frac{2a^2+2a+1}{a^4+2a^3+a^2}=\frac{5}{4}\)

\(\Leftrightarrow8a^2+8a+4=5a^4+10a^3+5a^2\)

\(\Leftrightarrow5a^4+10a^3-3a^2-8a-4=0\)

\(\Leftrightarrow5a^4+10a^3-3a^2-6a-2a-4=0\)

\(\Leftrightarrow5a^3\left(a+2\right)-3a\left(a+2\right)-2\left(a+2\right)=0\)

\(\Leftrightarrow\left(a+2\right)\left(5a^3-3a-2\right)=0\)

\(\Leftrightarrow\left(a+2\right)\left(5a^3-5a+2a-2\right)=0\)

\(\Leftrightarrow\left(a+2\right)\left(a-1\right)\left(5a^2+5a+2\right)=0\)

tới đây dễ r`

x = -28, đoán bừa.

=\(\frac{1}{x^2}\)+ \(\frac{1}{^{29^2}}\)+\(\frac{1}{x^2}\)+\(\frac{1}{30^2}\)=5/4

=\(\frac{1}{x^2}\)x2 + \(\frac{1}{29^2}\)+\(\frac{1}{30^2}\)=5/4

=\(\frac{2}{x^2}\)+\(\frac{1}{841}\)+\(\frac{1}{900}\)=5/4

\(\frac{2}{x^2}\)=\(\frac{5}{4}\)-\(\frac{1}{841}\)-\(\frac{1}{900}\)bạn tự tính tiếp đi

\(\frac{1}{\left(x+29\right)^2}+\frac{1}{\left(x+30\right)^2}=\frac{\left(x+30\right)^2}{\left(x+29\right)^2\left(x+30\right)^2}+\frac{\left(x+29\right)^2}{\left(x+29\right)^2\left(x+30\right)^2}\)

\(=\frac{x^2+60x+900+x^2+58x+841}{\left(x+29\right)^2\left(x+30\right)^2}\)

\(=\frac{2x^2+118x+1741}{\left(x+29\right)^2\left(x+30\right)^2}\)

TK MÌNH ĐI RỒI MÌNH GIẢI CHO.

cau a: 8x^3 -12x^2 + 6x + 1 =29

<=>8x^3 - 12x^2 + 6x - 28 =0

<=>(8x^3 - 16x^2)+(4x^2 - 8x)+(14x-28)=0

<=>8x^2 ( x-2) + 4x(x-2) + 14(x-2)=0

<=>(x-2)(8x^2 + 4x +14)=0

<=>8x^2 +4x +14 =0 <=> 8(x^2 +1/2 x +7/4)=0<=>(x^2 +2* x*1/4  + 1/16) +27/16 =0 <=>(x+ 1/4)^2=-27/16 (0xay ra) (loai)

=>(x-2)(8x^2 +4x+14)=0 <=> x-2=0 <=>x=2

Vay tap nghiem phuong trinh S={2}

b) x+3/x+2 +x+6/x+5=x+5/x+4 + x+4/x+3

=(x+3/x+2 -1) +(x+6/x+5 - 1)=(x+5/x+4 -1) +(x+4/x+3 -1)

 =1/x+2 +1/x+5 -1/x+4 -1/x+3 =0

x=1000

em moi hoc lop 7 thoi a doi xong ki 2 nha

em mới học lớp 7 thôi

\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\frac{x^2-10x-29}{1971}-1+\frac{x^2-10x-27}{1973}-1=\frac{x^2-10x-1971}{29}-1+\frac{x^2-10x-1973}{27}-1\)

\(\Leftrightarrow\frac{x^2-10x-29-1971}{1971}+\frac{x^2-10x-27-1973}{1973}=\frac{x^2-10x-1971-29}{29}+\frac{x^2-10x-193-27}{27}\)

\(\Leftrightarrow\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}-\frac{x^2-10x-2000}{29}-\frac{x^2-10x-2000}{27}=0\)

\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\right)=0\)

\(\Leftrightarrow x^2-10x-2000=0\)

\(\Leftrightarrow x^2-50x+40x-2000=0\)

\(\Leftrightarrow x\left(x-50\right)+40\left(x-50\right)=0\)

\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)

     Th1  \(x-50=0\Leftrightarrow x=50\)

    Th2  \(x+40=0\Leftrightarrow x=-40\)

Vậy tập nghiệm của phương trình là   \(S=\left\{50;-40\right\}\)