$ĐKXĐ:x \neq -4;-5;-6;-7$

$pt⇔\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}$

$⇔\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$

$⇔\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}$

$⇔\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}$

$⇔\dfrac{3}{(x+4)(x+7)}=\dfrac{1}{18}$

$⇔x^2+11x+28=54$

$⇔x^2+11x-26=0$

$⇔x^2-2x+13x-26=0$

$⇔(x-2)(x+13)=0$

$⇔$ \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)(t/m)

Vậy phương trình đã cho có tập nghiệm $S=(2;-13)$

phân tích mẫu thành nhân tử r áp dụng \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) sau đó rút gọn quy đồng

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\) \(\left(ĐKXĐ:x\ne0;x\ne-4;x\ne-5;x\ne-6;x\ne-7\right)\)

\(\Leftrightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{\left(x+6\right)\left(x+7\right)+\left(x+4\right)\left(x+7\right)+\left(x+4\right)\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{\left(x^2+13x+42\right)+\left(x^2+11x+28\right)+\left(x^2+9x+20\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{x^2+13x+42+x^2+11x+28+x^2+9x+20}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3x^2+33x+90}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3\left(x^2+11x+30\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=18.3\left(x^2+11x+30\right)\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=54\left(x+5\right)\left(x+6\right)\)

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)

\(\Leftrightarrow x^2+11x+28-54=0\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow x^2+13x-2x-26=0\)

\(\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\)

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+13=0\\x-2=0\end{cases}}\) 

\(\Leftrightarrow\orbr{\begin{cases}x=-13\left(tm\right)\\x=2\left(tm\right)\end{cases}}\)

rần Duy Thanh:gà dữ đến dấu <=> thứ 3 r` mà còn quy đồng. What is your favourite food :v

câu a)
bạn lập bảng xét dấu
x          -3/2               0 
x        -   ||      -          0 +
2x+3   -   0     +         ||  +
từ đó bạn xét từng trường hợp x< -3/2 và -3/2<x<0 và 0<x và bạn sẽ tìm được từng kết quả x
b)1/(x^2 + 13x + 42) = 1/((x+7)(x+6))
1/(x^2 + 11x + 30) = 1/((x+ 5)(x +6))
1/(x^2 + 9x + 20) = 1/((x + 5)(x+4))
chuyển 1/18 sang bạn sẽ có 1/((x+7)(x+6)) + 1/((x+ 5)(x +6)) + 1/((x + 5)(x+4)) - 1/18 = 0
mẫu số chung sẽ là 18(x+4)(x+5)(x+6)(x+7). quy đồng và rút gọn bạn sẽ có 1 biểu thức khá đẹp:
-(x^2 + 11x - 26)/(18(x+4)(x+7)) = 0.
giải phương trình -x^2 - 11x + 26 bạn sẽ có nghiệm là x = -13 và x = 2.
 

1/x²+9x+20=1/(x+4)(x+5)=1/x+4 -1/x+5 

CMTT=>1/x+4-1/x+7-1/18=0=>x=2;-13

Câu hỏi của Phạm Tiến Dũng new - Toán lớp 9 - Học toán với OnlineMath

Ta có:

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}\)  \(=\frac{1}{18}\)

\(\Leftrightarrow\)\(\frac{1}{\left(x+4\right)\left(x+5\right)}\) \(+\frac{1}{\left(x+5\right)\left(x+6\right)}\) \(+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}\) \(=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow x^2+11x-26=0\Leftrightarrow\hept{\begin{cases}x_1=2\\x_2=-13\end{cases}}\)

Vậy nghiệm của phương trình là {2;-13}

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)(điều kiện: \(x\ne\left\{-4;-5;-6;-7\right\}\) )

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow54=\left(x+4\right)\left(x+7\right)\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=-13\\x=2\end{cases}}\)(thỏa mãn ĐKXĐ)

Vậy tập nghiệm của pt là: \(S=\left\{-13;2\right\}\)

Lâu lắm không làm nhể

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x.\left(x+4\right)+5.\left(x+4\right)}+\frac{1}{x.\left(x+5\right)+6.\left(x+5\right)}+\frac{1}{x.\left(x+6\right)+7.\left(x+6\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}\)

Dùng công thứ \(\frac{1}{x.\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)

Khi đó \(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{x+7}{\left(x+4\right).\left(x+7\right)}-\frac{\left(x+4\right)}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{3}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow\left(x+4\right).\left(x+7\right)=54\)

\(\Rightarrow\hept{\begin{cases}x+4=6\\x+7=9\end{cases}}\)hoặc \(\hept{\begin{cases}x+4=-6\\x+7=-9\end{cases}}\)

Suy ra \(x=3\)hoặc \(x=-3\)

Thêm đk: \(x\in Z\)

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\frac{1}{\left(x^2+4x\right)+\left(4x+20\right)}+\frac{1}{\left(x^2+5x\right)+\left(6x+30\right)}+\frac{1}{\left(x^2+6x\right)+\left(7x+42\right)}=\frac{1}{18}\)

\(\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)

\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\frac{\left(x+7\right)-\left(x+4\right)}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)

đến đây b tự lập bảng nhé~

ĐKXĐ: \(x\ne-4;-5;-6;-7\)

\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

Ta có: 

\(x^2+9x+2x=\left(x+4\right)\left(x+5\right)\)

\(x^2+11x+30=\left(x+6\right)\left(x+5\right)\)

\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)

ĐK: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)

pt \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)

\(\Rightarrow18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)\)

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)  (tm)

\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\) ĐKXĐ:x\(\ne\)-4,-5,-6,-7

\(\Leftrightarrow\)\(\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{1}{x.\left(x+4\right)+5.\left(x+4\right)}+\dfrac{1}{x.\left(x+5\right)+6.\left(x+5\right)}+\dfrac{1}{x.\left(x+6\right)+7.\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{1}{\left(x+4\right).\left(x+5\right)}+\dfrac{1}{\left(x+5\right).\left(x+6\right)}+\dfrac{1}{\left(x+6\right).\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{x+7-x-4}{\left(x+4\right).\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\)\(\dfrac{3}{\left(x+4\right).\left(x+7\right)}=\dfrac{3}{54}\)

\(\Leftrightarrow\)(x+4).(x+7)=54

\(\Leftrightarrow\)x²+11x+28=54

\(\Leftrightarrow\)x²+11x-26=0

\(\Leftrightarrow\)x²+13x-2x-26=0

\(\Leftrightarrow\)x.(x+13)-2.(x+13)=0

\(\Leftrightarrow\)(x-2).(x+13)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\left(TM\right)\\x=-13\left(TM\right)\end{matrix}\right.\)

Vậy tập nghiệm của pt trên là S={-13;2}

ĐKXĐ: \(x\ne-4;x\ne-5;x\ne-6;x\ne-7\)

\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow54=x^2+11x+28\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-13\left(TM\right)\end{matrix}\right.\)