\(a,\Leftrightarrow5\left(x-2\right)-15x\le9+10\left(x+1\right)\)

\(\Leftrightarrow5x-10-15x\le9+10x+10\)

\(\Leftrightarrow-20x\le29\)

\(\Leftrightarrow x\ge-1,45\)

Vậy ...........

\(b,\Rightarrow\left(x+2\right)-3\left(x-3\right)=5\left(x-2\right)\)

\(\Leftrightarrow x+2-3x+9-5x+10=0\)

\(\Leftrightarrow-7x+21=0\)

\(\Leftrightarrow x=3\)

Vậy ..............

 \(\frac{x-2}{6}-\frac{x}{2}\le\frac{3}{10}+\frac{x+1}{3}\Leftrightarrow\frac{5\left(x-2\right)}{30}-\frac{15x}{30}\le\frac{9}{30}+\frac{10\left(x+1\right)}{30}\)

\(\Leftrightarrow5x-10-15x-9-10x-10\le0\) 

 \(\Leftrightarrow-20x-29\le0\Leftrightarrow\left(-20x\right)\cdot\frac{-1}{20}\ge29\cdot-\frac{1}{20}\)

 \(\Leftrightarrow x\ge-\frac{29}{20}\)

ĐKXĐ : \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)

\(\frac{x+2}{x^2-5x+6}-\frac{3}{x-2}=\frac{5}{x-3}\) 

 \(\Rightarrow\frac{x+2}{x-2x-3x+6}-\frac{3}{x-2}=\frac{5}{x-3}\)

 \(\Rightarrow\frac{x+2}{\left(x-2\right)\left(x-3\right)}-\frac{3}{x-2}=\frac{5}{x-3}\)

 \(\Rightarrow\frac{x+2}{\left(x-2\right)\left(x-3\right)}-\frac{3\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

 \(\Rightarrow x+2-3x+9-5x+10=0\)

\(\Leftrightarrow-7x+21=0\Leftrightarrow x=3\) (nhân)

 tập nghiệm của phương trình là S= 3

Bài làm:

PT:

đkxđ: \(x\ne0;x\ne2\)

Ta có: \(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)

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

\(\Rightarrow x^2+2x=2+x-2\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(vl\right)\\x+1=0\end{cases}}\Rightarrow x=-1\)

BPT:

Ta có: \(\frac{x+1}{2}-x\le\frac{1}{2}\)

\(\Leftrightarrow\frac{x+1}{2}-x-\frac{1}{2}\le0\)

\(\Leftrightarrow\frac{x+1-2x-1}{2}\le0\)

\(\Leftrightarrow\frac{-x}{2}\le0\)

\(\Rightarrow-x\le0\)

\(\Rightarrow x\ge0\)

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)

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

\(\Leftrightarrow\frac{2}{x\left(x-2\right)}+\frac{1}{x}-\frac{x+2}{x-2}=0\)

\(\Leftrightarrow\frac{2+x-2-x^2-2x}{x\left(x-2\right)}=0\)

\(\Leftrightarrow-x^2-x=0\)

\(\Leftrightarrow-x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}}\)

Vậy \(S=\left\{-1\right\}\)

b) \(\frac{x+1}{2}-x\le\frac{1}{2}\)

\(\Leftrightarrow x+1-2x-1\le0\)

\(\Leftrightarrow-x\le0\)

\(\Leftrightarrow x\ge0\)

Vậy \(x\ge0\)

ĐKXĐ : \(x\ne0;2\)

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

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

\(\Leftrightarrow x^2+2x=2+x-2\)

\(\Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}\)

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 !!!

\(a,\left(\dfrac{1}{3}\right)^{2x+1}\le9\\ \Leftrightarrow2x+1\ge-2\\ \Leftrightarrow2x\ge-3\\ \Leftrightarrow x\ge-\dfrac{3}{2}\)

\(b,4^x>2^{x-2}\\ \Leftrightarrow2^{2x}>2^{x-2}\\ \Leftrightarrow2x>x-2\\ \Leftrightarrow x>-2\)

a) \({2^x} > 16 \Leftrightarrow {2^x} > {2^4} \Leftrightarrow x > 4\) (do \(2 > 1\)) .

b) \(0,{1^x} \le 0,001 \Leftrightarrow 0,{1^x} \le 0,{1^3} \Leftrightarrow x \ge 3\) (do \(0 < 0,1 < 1\)).

c) \({\left( {\frac{1}{5}} \right)^{x - 2}} \ge {\left( {\frac{1}{{25}}} \right)^x} \Leftrightarrow {\left( {\frac{1}{5}} \right)^{x - 2}} \ge {\left( {{{\left( {\frac{1}{5}} \right)}^2}} \right)^x} \Leftrightarrow {\left( {\frac{1}{5}} \right)^{x - 2}} \ge {\left( {\frac{1}{5}} \right)^{2x}} \Leftrightarrow x - 2 \le 2{\rm{x}}\) (do \(0 < \frac{1}{5} < 1\))

\( \Leftrightarrow x \ge  - 2\).

Mình giải thử thôi nha

\(\frac{\left(2x-1\right)^2}{2}-\frac{\left(1-3x\right)^2}{3}\le x\left(2-x\right)\)

\(\Leftrightarrow3\left(2x-1\right)^2-2\left(1-3x\right)^2\le6x\left(2-x\right)\)

\(\Leftrightarrow12x^2-12x+3-2+12x-18x^2\le12x-6x^2\)

\(\Leftrightarrow-6x^2+1\le12x-6x^2\)

\(\Leftrightarrow1\le12x\)

\(\Leftrightarrow\frac{1}{12}\le x\)

\(\Rightarrow x\ge\frac{1}{12}\)

bạn là nam hay nữ zở

bn nhìn tên rồi đoán nha bn