Cái bài đầu giải BPT bn ghi cái dj ak ,mik cx k hỉu nữa

V mik giải bài 2 nghen, sửa lại đề bài đầu rồi mik giải cho

\(3x-3=|2x+1|\)

Điều kiện: \(3x-3\ge0\Leftrightarrow3x\ge3\Leftrightarrow x\ge1\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=3x-3\\2x+1=-3x+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1-3\\2x+3x=-1+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=-3\\5x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\left(n\right)\\x=\frac{2}{5}\left(l\right)\end{cases}}}\)

Vậy S={3}

Cài đề câu b ,bn xem lại nhé!

\(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}>\frac{x^2}{7}-\frac{2x-3}{5}\)

\(\Leftrightarrow\frac{2x-3}{35}+\frac{5x\left(x-2\right)}{35}-\frac{5x^2}{35}+\frac{7\left(2x-3\right)}{35}>0\)

\(\Leftrightarrow2x-3+5x\left(x-2\right)-5x^2+7\left(2x-3\right)>0\)

\(\Leftrightarrow2x-3+5x^2-10x-5x^2+14x-21>0\)

\(\Leftrightarrow6x-24>0\)

\(\Leftrightarrow x>4\)

VẬY TẬP NGHIỆM CỦA BẤT PHƯƠNG TRÌNH LÀ :  S = {  \(x\text{\x}>4\)}

\(\frac{6x+1}{18}+\frac{x+3}{12}\le\frac{5x+3}{6}+\frac{12-5x}{9}\)

\(\Leftrightarrow\frac{6\left(6x+1\right)}{108}+\frac{9\left(x+3\right)}{108}\le\frac{18\left(5x+3\right)}{108}+\frac{12\left(12-5x\right)}{108}\)

\(\Leftrightarrow36x+6+9x+27\le90x+54+144-60x\)

\(\Leftrightarrow36x+6+9x+27-90x-54-144+60x\le0\)

\(\Leftrightarrow15x-165\le0\)

\(\Leftrightarrow x\le11\)

VẬY TẬP NGHIỆM CỦA BẤT PHƯƠNG trình ..........

tk mk nka !!! chúc bạn học tốt !!!

\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)

\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)

\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)

\(\Rightarrow-1\le x< 6\)

\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)

Bài b tương tự

bạn ơi bạn giải câu b được ko. mk ko biết làm câu b

\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)

\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)

2 câu sau tự làm nha

\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)

\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)

=> \(x\in\left\{-2;-1;0;1;2\right\}\)

thank nha

1)

\(\frac{7.8^3-5.2^{10}}{\left(-16\right)^2}\)       

=  \(\frac{7.2^8.2-5.2^8.2^2}{16^2}\)

=  \(\frac{2^8.\left(2.7-5.2^2\right)}{2^8}\)

=   \(\frac{2^8.\left(-6\right)}{2^8}\)

=   \(-6\)

2)

b)\(\frac{-28}{4}\le x\le\frac{-21}{7}\)

\(\Rightarrow-7\le x\le-3\)

\(\Rightarrow x=\left\{-7;-6;-5;-4;-3\right\}\)

2)

a)\(\frac{108}{12}\le x\le\frac{91}{7}\)

\(\Rightarrow9\le x\le13\)

\(\Rightarrow x=\left\{9;10;11;12;13\right\}\)

\(\frac{2}{3}\) .\(\frac{3}{4}\)\(\le\)\(\frac{x}{18}\) \(\le\)\(\frac{7}{3}\).\(\frac{1}{3}\)

\(\frac{1}{2}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{9}{18}\le\frac{x}{18}\le\frac{14}{18}\)

\(\Rightarrow x\in\){9:10;11;12;13;14}

\(\frac{2}{3}.\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\)

\(\frac{2}{3}.\left(\frac{5}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\frac{1}{3}\)

\(\frac{2}{3}.\frac{11}{12}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{11}{18}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{11}{18}\le\frac{x}{18}\le\frac{14}{18}\)

Vậy \(x\in\left\{11;12;13\right\}\)

\(\frac{2}{3}\cdot\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}\cdot\left(\frac{1}{2}-\frac{1}{6}\right)\)

\(\frac{2}{3}\cdot\left(\frac{2}{4}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}\cdot\left(\frac{3}{6}-\frac{1}{6}\right)\)

\(\frac{2}{3}\cdot\left(\frac{5}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}\cdot\frac{1}{3}\)

\(\frac{2}{3}\cdot\left(\frac{15}{12}-\frac{4}{12}\right)\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{2}{3}\cdot\frac{11}{12}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{11}{18}\le\frac{x}{18}\le\frac{14}{18}\)

Để \(x\)phải nhỏ hơn hoặc bằng thì x lần lượt bằng \(\left\{11;12;13;14\right\}\)

mình cần gấp nhé