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a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
a: \(\left(3x-1\right)^2-\left(x+3\right)^3=\left(2-x\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow9x^2-6x+1-x^3-9x^2-27x-27=8-x^3\)
\(\Leftrightarrow-x^3-33x-26-8+x^3=0\)
=>-33x=34
hay x=-34/33
b: \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow x^4-1-x^4+2x^2-1=2\)
\(\Leftrightarrow2x^2=4\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
c: \(x^2-2\sqrt{3}x+3=0\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)^2=0\)
hay \(x=\sqrt{3}\)
d: \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-x+\sqrt{2}\right)=0\)
\(\Leftrightarrow x-\sqrt{2}=0\)
hay \(x=\sqrt{2}\)
a) ĐKXĐ: x khác 0
\(x+\dfrac{5}{x}>0\)
\(\Leftrightarrow x^2+5>0\) ( luôn đúng)
Vậy bất pt vô số nghiệm ( loại x = 0)
d)
\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)
\(\Leftrightarrow2x+2-4x+4>-15\)
\(\Leftrightarrow-2x>-21\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
Vậy....................
a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)
Mà \(x^2+5>0\)
\(\Rightarrow x>0\)
d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)
\(\Leftrightarrow-x>-\dfrac{21}{2}\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
a.
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow\left(x^2+5x\right)-\left(2x+10\right)=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b.
\(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=1\end{matrix}\right.\)
bài 2:
ĐKXĐ: x khác -1
\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\)
\(\Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x+1}\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow0x=-1\)
\(\Leftrightarrow x\in\varnothing\)
Suy ra pt vô nghiệm
b.
ĐKXĐ: x khác \(\dfrac{3}{2}\)
\(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow\dfrac{x^2+4x+4}{2x-3}-\dfrac{2x-3}{2x-3}=\dfrac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\) ( loại)
a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)
\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1
b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)
\(\Leftrightarrow x< 5\)
c) \(\dfrac{-1}{2x+3}< 0\)
dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp
a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)
\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)
\(\Leftrightarrow6x+6+12x-8=x-7\)
\(\Leftrightarrow6x+12x-x=-7-6+8\)
\(\Leftrightarrow17x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy .........................
b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)
\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)
\(\Leftrightarrow2x^2-x^2+x+15-21=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy .........................
P/s: các câu còn lại tương tự, bn tự giải nha
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
a) \(\dfrac{2x-9}{2x-5}+\dfrac{3x}{3x-2}=2\)
\(\Rightarrow\dfrac{2x-5-4}{2x-5}+\dfrac{3x-2+2}{3x-2}=2\)
\(\Rightarrow1-\dfrac{4}{2x-5}+1+\dfrac{2}{3x-2}=2\)
\(\Rightarrow\dfrac{4}{2x-5}+\dfrac{2}{3x-2}=0\)
\(\Rightarrow\dfrac{12x-8}{\left(2x-5\right)\left(3x-2\right)}+\dfrac{4x-10}{\left(2x-5\right)\left(3x-2\right)}=0\)
\(\Rightarrow\dfrac{16x-18}{\left(2x-5\right)\left(3x-2\right)}=0\)
\(\Rightarrow16x-18=0\)
\(\Rightarrow x=\dfrac{18}{16}=\dfrac{9}{8}\)
b) \(\dfrac{x+2}{2002}+\dfrac{x+5}{1999}+\dfrac{x+201}{1803}=-3\)
\(\Rightarrow\dfrac{x+2}{2002}+1+\dfrac{x+5}{1999}+1+\dfrac{x+201}{1803}+1=0\)
\(\Rightarrow\dfrac{x+2004}{2002}+\dfrac{x+2004}{1999}+\dfrac{x+2004}{1809}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2002}+\dfrac{1}{1999}+\dfrac{1}{1809}\right)=0\)
Vì \(\left(\dfrac{1}{2002}+\dfrac{1}{1999}+\dfrac{1}{1809}\right)\ne0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
c) \(\left(2x^2+3x+1\right)\left(2x^2+5x+3\right)=18\)
\(\Rightarrow\left(2x+1\right)\left(x+1\right)\left(x+1\right)\left(2x+3\right)=18\)
\(\Rightarrow\left(x+1\right)^2\left(2x+1\right)=18\)
Thôi, xử tiếp đi nhé :)))))
a) Đặt \(2x^2-3x-1=a\)
Bt \(\Leftrightarrow a^2-3\left(a-4\right)-16=0\)
\(\Leftrightarrow a^2-3a+12-16=0\)
\(\Leftrightarrow\left(a-4\right)\left(a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-1=4\\2x^2-3x-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-5=0\\2x^2-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)\left(2x-5\right)=0\\x\left(2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=-1\\x=2.5\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=1.5\end{matrix}\right.\end{matrix}\right.\)