Sửa đề: x² + 13x + 41 --> x² + 13x + 42

Giải:

\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+41}=\frac{1}{2}\)

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

(ĐKXĐ: \(x\ne\left\{-1;-2;-3;-4;-5;-6;-7\right\}\))

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{2}\)

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

\(\Leftrightarrow\frac{x+7-x-1}{\left(x+1\right)\left(x+7\right)}=\frac{1}{2}\)

\(\Leftrightarrow\left(x+1\right)\left(x+7\right)=12\)

\(\Leftrightarrow x^2+8x+7=12\)

⇔$x^2-8x=5$

⇔ $8x+{(-4)}^2=5+{(-4)}^2$
⇔ $x^2-8x+16=21$
⇔ ${(x-4)}^2=21$
⇔ $x=\pm \sqrt{21}+4$

Vậy...

Chúc bạn học tốt@@

vabh ơi cho mk hỏi bạn có ghi sai đề k ạ?

sr mik viết sai đề 42 ko phải 41

\(\left(1-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}\right)\cdot X=\frac{11}{6}\)

\(< =>\left(\frac{1}{2}-\frac{1}{12}-\frac{1}{60}\right)\cdot X=\frac{11}{6}\)

\(< =>\left(\frac{30}{60}-\frac{5}{60}-\frac{1}{60}\right)\cdot X=\frac{11}{6}\)

\(< =>\left(\frac{30-5-1}{60}\right)\cdot X=\frac{11}{6}\)

\(< =>\frac{2}{5}\cdot X=\frac{11}{6}\)

\(< =>X=\frac{11}{6}:\frac{2}{5}\)

\(< =>X=\frac{55}{12}\)

CHUC BAN HOC TOT >.<

a, 10+15+20+....+295+x.300+x=67

10+15+20+...+295+x(300+1)=67

10+15+20+...+295+x.301=67

8845+x.301=67

67-8845=x.301

-8878=x.301

x=-29/149/301

b, 

\(\frac{1}{7.6}+\frac{1}{6.5}+\frac{1}{5.4}+\frac{1}{4.3}+\frac{1}{3.2}+\frac{1}{2.1}-\frac{1}{x+1}=\frac{59}{77}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}-\frac{1}{x+1}=\frac{59}{77}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)\(1-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)

\(\frac{6}{7}-\frac{1}{x+1}=\frac{59}{77}\)

\(\frac{1}{x+1}=\frac{6}{7}-\frac{59}{77}\)

\(\frac{1}{x+1}=\frac{1}{11}\)

suy ra x+1=11

suy ra x=10

ket bn nha cau

mk dng ko co bn olm

\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}.....\frac{30}{62}.\frac{31}{64}=4^x\)

\(\Leftrightarrow\)\(\frac{1.2.3.4.5.....30.31}{4.6.8.10.12.....62.64}=4^x\)

\(\Leftrightarrow\)\(\frac{2.3.4.5.....30.31}{2\left(2.3.4.5.....30.31\right).64}=4^x\)

\(\Leftrightarrow\)\(\frac{1}{128}=4^x\)

\(\Leftrightarrow\)\(2^{2x}=2^{-7}\) ( trong sgk có phần đọc thêm nói về cái này nhé ) 

\(\Leftrightarrow\)\(2x=-7\)

\(\Leftrightarrow\)\(x=\frac{-7}{2}\)

Vậy \(x=\frac{-7}{2}\)

Chúc bạn học tốt ~ 

\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}.....\frac{30}{62}.\frac{31}{64}=4^x\)

\(=\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}....\frac{30}{2.31}.\frac{31}{2.32}=4^x\)

\(=\frac{1}{2^{31}.2^5}=4^x\)

\(=\frac{1}{2^{36}}=4^x\)

\(=\frac{1}{2^{36}}=2^{2x}\)

\(=2^{-36}=2^{2x}\)

\(=-36=2.x\)

\(x.2=-36\)

\(x=-36:2\)

\(x=-18\)

bn cứ quy đồng lần lượt 2 hạng tử đầu tiên là đc thôi

mk giải phần a k ra

a, Đề có vẻ sai sai nhé :v

b, \(\left|\frac{1}{2}x-\frac{2}{3}\right|-1=\frac{1}{6}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{2}{3}\right|=\frac{7}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{2}{3}=\frac{7}{6}\\\frac{1}{2}x-\frac{2}{3}=-\frac{7}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{3}\\x=-1\end{cases}}\)

Vậy : ....

c, \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x(x+1)}=\frac{4}{5}\)

\(\Leftrightarrow\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\cdot(x+1)}=\frac{4}{5}\)

\(\Leftrightarrow2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{4}{5}\)

\(\Leftrightarrow2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{4}{5}\)

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

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{10}\)

\(\Leftrightarrow x+1=10\Leftrightarrow x=9\)

Vậy x = 9

a) x=0 nhé

\(\Leftrightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\frac{5}{14}\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\frac{x}{3}=\frac{2}{3}\)

\(\Leftrightarrow x=2\)

\(\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{7}{14}-\frac{2}{14}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{5}{14}.\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{x}{3}=\frac{5}{21}:\frac{5}{14}\)

\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

Đặt \(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(\Rightarrow A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

\(\Leftrightarrow\frac{5}{14}.\frac{4}{3}x=\frac{5}{14}\)

\(\Rightarrow\frac{4}{3}.x=\frac{5}{21}:\frac{5}{14}\)

\(\Rightarrow x=2\)

x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{5}{6}\) -\(\frac{3}{4}\) + \(\frac{2}{3}\) -\(\frac{1}{2}\)

x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{10}{12}\)-\(\frac{9}{12}\)+\(\frac{8}{12}\)-\(\frac{6}{12}\)

=>x.(1/2-2/3+3/4)=1/4

=>x.7/12=1/4

=>x=1/4:7/12

=>x=1/4.12/7

=>x=3/7