bỏ số 1 ở đầu thì giải dc á, còn có số 1 thì chịu

\(\dfrac{1}{x+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)

\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+8\right)}=\dfrac{4}{105}\)

\(\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)

\(\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\dfrac{x+8-x}{x\left(x+8\right)}=\dfrac{8}{105}\)

\(\dfrac{8}{x.\left(x+8\right)}=\dfrac{8}{105}\)

\(\Rightarrow x\left(x+8\right)=105\)

\(x^2+8x-105=0\)

\(\left(x-7\right)\left(x+15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)

\(ĐKXĐ:x\ne0;-2;-4;-6;-8\)\(\frac{1}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+8\right)}=\frac{4}{105}\)

\(\Leftrightarrow\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+6\right)\left(x+8\right)}=\frac{8}{105}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+...+\frac{1}{x+6}-\frac{1}{x+8}=\frac{8}{105}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+8}=\frac{8}{105}\)

Quy đồng làm nốt

tks nha

phân tích mẫu thành nhân tử

VD:x2+6x+8=x2+2x+4x+8=(x+2)(x+4)

x2+10x+24=x2+4x+6x+24=(x+6)(x+4).....

kết quả ra1/x-1/x+8=4/105

chuyển vế rồi tính

PT<=> \(\frac{1}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+8\right)}=\frac{4}{105}\)

<=> \(\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+6\right)\left(x+8\right)}=\frac{8}{105}\)

<=> \(\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-...+\frac{1}{x+6}-\frac{1}{x+8}=\frac{8}{105}\)

<=> \(\frac{1}{x}-\frac{1}{x+8}=\frac{8}{105}\)

<=> \(\frac{8}{x\left(x+8\right)}=\frac{8}{105}\)

<=> x(x+8) = 105

<=> x = 7

\(A=\dfrac{2}{x^2+2x}+\dfrac{2}{x^2+6x+8}+\dfrac{2}{x^2+10x+24}+\dfrac{2}{x^2+14x+48}\)

\(A=\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+8}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{x+8}{x\left(x+8\right)}-\dfrac{x}{\left(x+8\right)}=\dfrac{8}{x\left(x+8\right)}\)

\(B=\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{32}{1-x^{32}}\)

Sai đề bạn ơi

Sai chỗ \(\frac{1}{x^2+6x+18}\)

Bạn sửa lại rồi mk giải cho..

\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)

\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow x\left(x+8\right)=105\)

\(\Leftrightarrow x^2+8x-105=0\)

\(\Leftrightarrow x^2-7x+15x-105=0\)

\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)

Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)

2: \(x^2-2xy+y^2-2x+2y\)

\(=\left(x-y\right)^2-2\left(x-y\right)\)

=(x-y)(x-y-2)

3: \(3x^2-2x-5\)

\(=3x^2-5x+3x-5\)

=x(3x-5)+(3x-5)

=(3x-5)(x+1)

4: \(16-x^2+4xy-4y^2\)

\(=16-\left(x^2-4xy+4y^2\right)\)

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

=(4-x+2y)(4+x-2y)

5: \(x^2-2x+1-y^2\)

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

=(x-1-y)(x-1+y)

6: \(x^2+8x+15\)

\(=x^2+3x+5x+15\)

=x(x+3)+5(x+3)

=(x+3)(x+5)

7: \(\left(x^2+6x+8\right)\left(x^2+14x+48\right)-9\)

=(x+2)(x+4)(x+6)(x+8)-9

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)-9\)

\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+384-9\)

\(=\left(x^2+10x\right)^2+15\left(x^2+10x\right)+25\left(x^2+10x\right)+375\)

\(=\left(x^2+10x+25\right)\left(x^2+10x+15\right)=\left(x+5\right)^2\cdot\left(x^2+10x+15\right)\)

8: \(\left(x^2-8x+15\right)\left(x^2-16x+60\right)-24x^2\)

\(=\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)

\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)

\(=\left(x^2+30\right)^2-24x\left(x^2+30\right)+143x^2-24x^2\)

\(=\left(x^2+30\right)^2-24x\left(x^2+30\right)+119x^2\)

\(=\left(x^2-7x+30\right)\left(x^2-17x+30\right)\)

\(=\left(x^2-7x+30\right)\left(x-2\right)\left(x-15\right)\)