\(\frac{7x-12}{5x-2}=\frac{9-7x}{-5x+1}\)
Những câu hỏi liên quan
\(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
\(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
\(\Leftrightarrow\frac{x}{840}-\frac{17}{210}=\frac{8x}{15}+\frac{19}{60}+\frac{2}{3}\)
\(\Leftrightarrow\frac{x}{840}.840-\frac{17}{210}.840=\frac{8x}{15}.840+\frac{19}{60}.840+\frac{2}{3}.840\)
\(\Leftrightarrow x-68=448x+226+560\)
\(\Leftrightarrow x-68=448x+826\)
\(\Leftrightarrow x=448x+826+68\)
\(\Leftrightarrow x=448x+894\)
\(\Leftrightarrow-447x=894\)
=> x = -2
Đúng 0
Bình luận (0)
g) |9-7x|5x-3Vì |9-7x|ge0;forall xRightarrow5x-3ge0Rightarrow xgefrac{3}{5}Ta có: |9-7x|5x-3Leftrightarroworbr{begin{cases}9-7x5x-39-7x3-5xend{cases}}Leftrightarroworbr{begin{cases}-7x-5x-3-9-7x+5x3-9end{cases}}Leftrightarroworbr{begin{cases}-12x-12-2x-6end{cases}}Leftrightarroworbr{begin{cases}x1frac{3}{5}left(chonright)x3frac{3}{5}left(chonright)end{cases}}Vậy xinleft{1;3right}h) 8x-|4x+1|x+2Leftrightarrow|4x+1|7x+2Vì |4x+1|ge0;forall xRightarrow7x+2ge0Rightarrow xgefrac{-2}{7}Ta có: |4x+...
Đọc tiếp
g) \(|9-7x|=5x-3\)
Vì \(|9-7x|\ge0;\forall x\)
\(\Rightarrow5x-3\ge0\)
\(\Rightarrow x\ge\frac{3}{5}\)
Ta có: \(|9-7x|=5x-3\)
\(\Leftrightarrow\orbr{\begin{cases}9-7x=5x-3\\9-7x=3-5x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-7x-5x=-3-9\\-7x+5x=3-9\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-12x=-12\\-2x=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1>\frac{3}{5}\left(chon\right)\\x=3>\frac{3}{5}\left(chon\right)\end{cases}}\)
Vậy \(x\in\left\{1;3\right\}\)
h) \(8x-|4x+1|=x+2\)
\(\Leftrightarrow|4x+1|=7x+2\)
Vì \(|4x+1|\ge0;\forall x\)
\(\Rightarrow7x+2\ge0\)
\(\Rightarrow x\ge\frac{-2}{7}\)
Ta có: \(|4x+1|=7x+2\)
\(\Leftrightarrow\orbr{\begin{cases}4x+1=7x+2\\4x+1=-7x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-3x=1\\11x=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}< \frac{-2}{7}\left(loai\right)\\x=\frac{-3}{11}>\frac{-2}{7}\left(chon\right)\end{cases}}\)
Vậy \(x=\frac{-3}{11}\)
\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}\)
Tìm x :
\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}\)
Ta có :
\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}\)
\(=>\left(7x+2\right)\left(5x+1\right)=\left(7x-1\right)\left(5x+7\right)\)
\(=>35x^2+7x+10x+2=35x^2+49x-5x-7\)
\(=>35x^2+17x+2=35x^2+44x-7\)
\(=>17x+2=44x-7\)
\(=>44x-17x=2+7\)
\(=>27x=9\)
\(=>x=\frac{9}{27}=\frac{1}{3}\)
Đúng 0
Bình luận (0)
\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}\)là x=...............
\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}\) tìm x
=>(7x+2)*(5x+1)=(5x+7)(7x-1)(t/c 2 p/s bằng nhau)
7x(5x+1)+2(5x+1)=5x(7x-1)+7(7x-1)
35x2+7x+10x+2=35x2-5x+49x-7
17x+2=44x-7
17x-44x=-7-2
-27x=-9
x=-9/-27=1/3
Vậy x=1/3
Đúng 0
Bình luận (0)
Rút gọn
A=\(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9-10}\)
giải phương trình
a)\(\frac{7x+10}{x+1}\left(x^2-x-2\right)=\frac{7x+10}{x+1}\left(2x^2-3x-5\right)\)
b)\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)
c)\(x^2+\frac{1}{x^2}+\frac{9x}{2}-\frac{9}{2x}+7=0\)
\(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
Giải phương trình.