đề kiểm tra thì mỗi giáo viên mỗi khác, giống nhau đâu mà có

http://123doc.org/document/3890871-de-kiem-tra-15-phut-mon-toan-lop-7.htm

|3x-2|=10

=>3x-2=10 hoặc 3x-2=-10

=>3x=12 hoặc 3x=-8

=>x=4 hoặc x=-8/3

k nhé những người tốt bụng

|3x-2|=10

=>3x-2=10

=>3x=12

=>x=4

\(\left|2,5-x\right|=7,5\)

\(\orbr{\begin{cases}2,5-x=7,5\\2,5-x=-7,5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2,5-7,5\\x=2,5+7,5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-5\\x=10\end{cases}}\)

\(Vayx\in\left\{-5;10\right\}\)

\(1,6-\left|x-0,2\right|=0\)

\(\left|x-0,2\right|=1,6\)

\(\orbr{\begin{cases}x-0,2=1,6\\x-0,2=-1,6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1,6+0,2\\x=-1,6+0,2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1,8\\x=-1,4\end{cases}}\)

giúp mk với

đợi tý đc ko

e. \(x=\frac{22}{15}-\frac{8}{27}\)

\(x=\frac{158}{135}\)

Vậy \(x\in\left\{\frac{158}{135}\right\}\)

\(\frac{22}{15}-x=\frac{8}{27}\)

             \(x=\frac{22}{15}-\frac{8}{27}\)

               \(x=\frac{198}{135}-\frac{40}{135}\)

              \(x=\frac{158}{135}\)

 Vậy \(x=\frac{158}{135}\)

Chúc bạn học tốt !!!

\(=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)

5/1.3 + 5/3.5 + ... + 5/99.101

= 5/2.(1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)

= 5/2.(1 - 1/101)

=5/2.100/101

= 250/101

=1/1.2+5/2.3+11/3.4+19/4.5+29/5.6+41/6.7

=1-1/2+5/2-5/3+11/3-11/4+19/4-19/5+29/5-29/6+41/6-41/7

=3+2+2+2+2-41/7

=77/7-41/7

=36/7

k nhé

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}\)

\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)\)

\(=\left(1+1+1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(=6-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}\right)\)

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

\(=6-\left(1-\frac{1}{7}\right)=6-\frac{6}{7}=\frac{36}{7}\)

Ta có: |x+1|+|x+4|=3x

=> x+1+x+4=3x ( do trị tuyệt đối không âm)

=> 2x+5=3x

=> 3x-2x=5

=> x=5

bạn làm thiếu bước rồi:

Ta có:\(|x+1|\ge0\)với mọi x

      \(|x+4|\ge0\)

\(\Rightarrow|x+1|+|x+4|\ge0\)

\(\Rightarrow3x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\hept{\begin{cases}x+1>0\Rightarrow|x+1|=x+1\\x+4>0\Rightarrow|x+4|=x+4\end{cases}}\)

Khi đó ta có

(x+1)+(x+4)=3x

2x+5=3x

x=5

Đây là cách giải chi tiết

|x+1|+|x+4|=3x      (1)

ĐK: 3x>0=> x>0

Với  x>0=> x+1>0;x+4>0

từ (1) ta có:x+1+x+4=3x

              => 2x+5=3x

             => 5=3x-2x

               => 5=x

Vậy x=5