câu B bạn nhé

câu b nha

HÈ RỒI ÍT  NGƯỜI LÀM LẮM

VỚI LẠI LÀ KO BIẾT ĐANG HỌC LỚP 5 LÊN LỚP 6

1) Tìm x:

a) \(\frac{11}{12}-\frac{5}{12}.\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\Leftrightarrow\frac{5}{12}.\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)

\(\Leftrightarrow\frac{2}{5}+x=\frac{1}{4}:\frac{5}{12}=\frac{3}{5}\)

\(\Leftrightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)

b) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)

\(\Leftrightarrow x=-\frac{7}{20}:\frac{1}{4}=\frac{-7}{5}\)

a) \(\frac{11}{12}-\frac{5}{12}\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{5}{12}.\frac{2}{5}-\frac{5}{12}x=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{1}{6}-\frac{5}{12}x=\frac{2}{3}\)

\(\Leftrightarrow\frac{-5}{12}x=\frac{2}{3}-\frac{11}{12}+\frac{1}{6}\)

\(\Leftrightarrow-\frac{5}{12}x=\frac{8}{12}-\frac{11}{12}+\frac{2}{12}=-\frac{1}{12}\)

\(\Leftrightarrow x=\frac{-1}{12}:\left(-\frac{5}{12}\right)=-\frac{1}{12}.\left(-\frac{12}{5}\right)=\frac{1}{5}\)

Vậy x = 1/5

b) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=\frac{8}{20}-\frac{15}{20}=-\frac{7}{20}\)

\(\Leftrightarrow x=\frac{1}{4}:\left(-\frac{7}{20}\right)=\frac{1}{4}.\left(-\frac{20}{7}\right)=-\frac{5}{7}\)

Vậy x = -5/7

c) \(2x\left(x-\frac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\frac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\)

d) \(\left(x+1\right)\left(x-2\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\end{matrix}\right.\)

Ta thấy x <-1 và x >2 vô lí

Do đó: x >-1 và x <2

Vậy -1 < x <2

e) \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\)

Vậy x > 2 hoặc x < -2/3

A=1+(2-3-3+5)+(6-7-8+9)+....+(98-99-100+101)+102

=1+0+0+....+102=103

b) |1-2x|>7

=> 1-2x>7 hoặc 1-2x<-7

=> 2x<-6 hoặc 2x>8

=> x<-3 hoặc x>4

Câu 1.

ta có: 63/72 =7/8

=>x +3 =7

=>x=...

Câu 2.

=>x +4<11

=>x=0;1;2;3;4;5;6

=>x +4>9

=>x=6;7;8;9;...

=> x=6

\(a)\) \(\frac{-11}{12}< \frac{x}{12}< \frac{-3}{4}\)

\(\Leftrightarrow\)\(\frac{-11}{12}< \frac{x}{12}< \frac{-9}{12}\)

\(\Leftrightarrow\)\(-11< x< -9\)

\(\Rightarrow\)\(x=-10\)