x phần 3 = 0
x phần 3 < 0
1<x phần 3 bằng hoặc bé hơn 2
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a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)
a)13,2 - x = -4,3
=> x = 13,2 - ( -4,3)
=> x = 17,5
b)x phần 3 = 20 phần 15
\(\dfrac{x}{3}=\dfrac{20}{15}=>20.3=x.15=>60=x.15=>x=60:15=>x=4\)
c) 2 phần 3 - 1 phần 3 x = -5 phần 3 + 1 phần 2
Mong bạn viết lại đề giúp mình
d) x - 3 phần 5 = -7 phần 10
\(\dfrac{x-3}{5}=\dfrac{-7}{10}=>5.-7=\left(x-3\right).10=>-35=\left(x-3\right).10=>x-3=-35:10=>x-3=\text{-3,5}=>x=\text{-3,5}+3=>x=6,5\)
a)
\(\dfrac{x-2}{4}+\dfrac{2x-3}{3}=\dfrac{x-18}{6}\)
`<=> 3x-6+8x-12=2x-36`
`<=> 3x+8x-2x=-36+6+12`
`<=> 9x=-18`
`<=> x=-2`
b)
\(\dfrac{x+3}{x-3}+\dfrac{3-x}{x+3}=\dfrac{36}{x^2-9}\left(x\ne3;x\ne-3\right)\)
suy ra
`(x+3)^2 +(3-x)(x-3)=36`
`<=>x^2 +6x+9+3x-9-x^2 +3x=36`
`<=> x^2 -x^2 +6x+3x+3x+9-9-36=0`
`<=> 12x-36=0`
`<=> 12x=36`
`<=> x=3 (KTMĐK)
\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)
\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)
\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)
\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)
\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)
\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)
\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)
\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)
\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)
\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)
\(< =>5x-35=32x-8< =>32x-5x=-35+8\)
\(< =>27x=-27< =>x=-1\)
\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)
\(< =>12=6x-3< =>6x=12+3\)
\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)
\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)
\(< =>8x-4=9x+3< =>9x-8x=-4-3\)
\(< =>9x-8x=-7< =>x=-7\)
\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)
\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)
\(< =>12x-7x=14-4< =>5x=10\)
\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)
\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)
\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)
\(< =>6x-4x=4+6< =>2x=10\)
\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)
\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)
\(< =>\left(x+1\right)\left(x+1\right)=3.3\)
\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)
\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)
\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)
\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)
a) \(\frac{x-5}{7}=\frac{5}{3}\)
\(\Rightarrow\left(x-5\right)\cdot3=7\cdot5\)
\(\Rightarrow3x-15=35\)
\(\Rightarrow3x=15+35\)
\(\Rightarrow3x=50\)
\(\Rightarrow x=\frac{50}{3}\)
b) \(\frac{x-3}{3}=\frac{12}{x-3}\)
\(\Rightarrow\left(x-3\right)\cdot\left(x-3\right)=3\cdot12\)
\(\Rightarrow\left(x-3\right)^2=36\)
\(\Rightarrow\left(x-3\right)^2=6^2\)hoặc \(\left(x-3\right)^2=\left(-6\right)^2\)
\(\Rightarrow x-3=6\) \(x-3=-6\)
\(\Rightarrow x=6+3\) \(x=-6+3\)
\(\Rightarrow x=9\) hoặc \(x=-3\)
c) \(\frac{5-2x}{4}=\frac{7}{3}\)
\(\Rightarrow\left(5-2x\right)\cdot3=4\cdot7\)
\(\Rightarrow15-6x=28\)
\(\Rightarrow6x=15-28\)
\(\Rightarrow6x=-13\)
\(\Rightarrow x=-\frac{13}{6}\)
d) \(\frac{x+1}{4}=\frac{7}{3}\)
\(\Rightarrow\left(x+1\right)\cdot3=2\cdot7\)
\(\Rightarrow3x+3=28\)
\(\Rightarrow3x=28-3\)
\(\Rightarrow3x=25\)
\(\Rightarrow x=\frac{25}{3}\)
Chúc bạn học tốt !!!
+)\(\frac{x}{3}=0\) \(\Rightarrow x=0\)
+) \(\frac{x}{3}< 0\) \(\Rightarrow x< 0\)
+)\(1< \frac{x}{3}< 2\) \(\Rightarrow3< x< 6\)
x/3=0 thì x=0