tìm x:
2/5.x-1/5.x=3/4
giúp em vs ạ
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a) ĐKXĐ: \(x\ne2\)
\(\Rightarrow\left(x+2\right)\left(x-2\right)=5.1\)
\(\Rightarrow x^2-4=5\Rightarrow x^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)
b) ĐKXĐ: \(x\ne-1\)
\(\Rightarrow\left(x+1\right)^2=2.8=16\)
\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)
c) giống câu a
d) ĐKXĐ: \(x\ne5,x\ne-1\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)=\left(x-3\right)\left(x-5\right)\)
\(\Rightarrow x^2+3x+2=x^2-8x+15\)
\(\Rightarrow11x=13\)
\(\Rightarrow x=\dfrac{13}{11}\left(tm\right)\)
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Ta có :
\(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)
\(\Leftrightarrow\)\(4x^2-4x-3x^2+15=x-3-x-4\)
\(\Leftrightarrow\)\(x^2-4x+15=-7\)
\(\Leftrightarrow\)\(\left(x^2-2.x.2+2^2\right)+11=-7\)
\(\Leftrightarrow\)\(\left(x-2\right)^2=-18\)
Mà \(\left(x-2\right)^2\ge0\) \(\left(\forall x\inℝ\right)\)
\(\Rightarrow\)\(x\in\left\{\varnothing\right\}\)
Vậy không có giá trị nào của x thoã mãn đề bài
Chúc bạn học tốt ~
x/y=3/4
=>x/3=y/4
=>x/15=y/20
y/z=5/7
=>y/5=z/7
=>y/20=z/28
=>x/15=y/20=z/28=(2x+3y-z)/(2*15+3*20-28)=186/62=3
=>x=45; y=60; z=84
\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\Leftrightarrow\left(4-x\right)\left(4-x\right)=-5\times-5\)
\(\Rightarrow\left(4-x\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\Rightarrow\left(4-x\right).\left(4-x\right)=\left(-5\right).\left(-5\right)\)
\(\Rightarrow\left(4-x\right)^2=25\)
\(\Rightarrow\left(4-x\right)^2=5^2\)
\(\Rightarrow4-x=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4-5\\x=4-\left(-5\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
`@` `\text {Ans}`
`\downarrow`
`2+(x+3)=7`
`\Rightarrow x+3=7-2`
`\Rightarrow x+3=5`
`\Rightarrow x=5-3`
`\Rightarrow x=2`
`5+(3+x)=10`
`\Rightarrow 3+x=10-5`
`\Rightarrow 3+x=5`
`\Rightarrow x=5-3`
`\Rightarrow x=2`
`(4+x)+1=7`
`\Rightarrow 4+x=7-1`
`\Rightarrow 4+x=6`
`\Rightarrow x=6-4`
`\Rightarrow x=2`
`(x+5)+3=9`
`\Rightarrow x+5=9-3`
`\Rightarrow x+5=6`
`\Rightarrow x=6-5`
`\Rightarrow x=1`
`(x-1)-4=7`
`\Rightarrow x-1=7+4`
`\Rightarrow x-1=11`
`\Rightarrow x=11+1`
`\Rightarrow x=12`
`4-(6-x)=1`
`\Rightarrow 6-x=4-1`
`\Rightarrow 6-x=3`
`\Rightarrow x=6-3`
`\Rightarrow x=3`
\(2+\left(x+3\right)=7\)
\(\Rightarrow2+x+3=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(5+\left(3+x\right)=10\)
\(\Rightarrow5+3+x=10\)
\(\Rightarrow x+8=10\)
\(\Rightarrow x=2\)
\(\left(4+x\right)+1=7\)
\(\Rightarrow4+x+1=7\)
\(\Rightarrow x+5=7\)
\(\Rightarrow x=2\)
\(\left(x+5\right)+3=9\)
\(=x+5+3=9\)
\(\Rightarrow x+8=9\)
\(\Rightarrow x=1\)
\(\left(x-1\right)-4=7\)
\(\Rightarrow x-1-4=7\)
\(\Rightarrow x-5=7\)
\(\Rightarrow x=12\)
\(4-\left(6-x\right)=1\)
\(\Rightarrow4-6-x=1\)
\(\Rightarrow-2-x=1\)
\(\Rightarrow x=-3\)
\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
\(\dfrac{2}{5}x-\dfrac{1}{5}x=\dfrac{3}{4}\)
\(\left(\dfrac{2}{5}-\dfrac{1}{5}\right)x=\dfrac{3}{4}\)
\(\dfrac{1}{5}x=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}:\dfrac{1}{5}=\dfrac{15}{4}\)
\(\dfrac{2}{5}.x-\dfrac{1}{5}.x=\dfrac{3}{4}\)
\(x.\left(\dfrac{2}{5}-\dfrac{1}{5}\right)=\dfrac{3}{4}\)
\(x.\dfrac{1}{5}\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}:\dfrac{1}{5}\)
\(x\) \(\dfrac{15}{4}\)