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Ta có : \(\dfrac{x}{2}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x=2\left(1-x\right)\)
\(\Leftrightarrow3x=2-2x\)
\(\Leftrightarrow5x=2\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy ...
Ta có: \(\dfrac{x}{2}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x=2\left(1-x\right)\)
\(\Leftrightarrow3x=2-2x\)
\(\Leftrightarrow3x+2x=2\)
\(\Leftrightarrow5x=2\)
hay \(x=\dfrac{2}{5}\)
Vậy: \(x=\dfrac{2}{5}\)
A(x)=-2 rồi thì A(x):B(x) dư 6 sao được bạn? Bạn xem lại đề.
a)\(f\left(1\right)=2.1^2+5.1-3=2+5-3=4\)
\(f\left(0\right)=0+0-3=-3\)
\(f\left(1,5\right)=2.\left(1,5\right)^2-5.1,5-3=4,5-7,5-3=-6\)
\(D=\dfrac{-x+3+1}{x-3}=-1+\dfrac{1}{x-3}\)
D min khi x-3=-1
=>x=2
\(f\left(3\right)=3a-3=9\)
\(3a=12\Rightarrow a=4\)
\(f\left(5\right)=5a-3=11\)
\(5a=14\Rightarrow a=\dfrac{14}{5}\)
\(f\left(-1\right)=-a-3=6\)
\(-a=9\Rightarrow a=9\)
ĐKXĐ: x<>0
2x-y=3
=>\(y=2x-3\)
\(\dfrac{2}{x}=\dfrac{y}{5}\)
=>\(\dfrac{2}{x}=\dfrac{2x-3}{5}\)
=>x(2x-3)=10
=>\(2x^2-3x-10=0\)
=>\(\left[{}\begin{matrix}x=\dfrac{3+\sqrt{89}}{4}\left(nhận\right)\\x=\dfrac{3-\sqrt{89}}{4}\left(nhận\right)\end{matrix}\right.\)
Khi \(x=\dfrac{3+\sqrt{89}}{4}\) thì \(y=2\cdot\dfrac{3+\sqrt{89}}{4}-3=\dfrac{-3+\sqrt{89}}{2}\)
Khi \(x=\dfrac{3-\sqrt{89}}{4}\) thì \(y=2\cdot\dfrac{3-\sqrt{89}}{4}-3=\dfrac{-3-\sqrt{89}}{2}\)
\(2x^2-5x-3=0\)
=>\(2x^2+x-6x-3=0\)
<=> \(x.\left(2x+1\right)-3.\left(2x+1\right)=0\)
<=>\(\left(x-3\right)\left(2x+1\right)=0\)
=> x-3=0 <=> x=3
hoặc 2x+1=0 => x=\(\dfrac{-1}{2}\)
`2x^2-5x-3=0`
`(2x^2+x)-(6x+3)=0`
`x(2x+1)-3(2x+1)=0`
`(x-3)(2x+1)=0`
\(\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
`a)`
`@f(1)=2.1^2+5.1-3=2.1+5-3=2+5-3=4`
`@f(0)=2.0^2+5.0-3=-3`
`@f(1,5)=2.(1,5)^2+5.1,5-3=4,5+7,5-3=9`
_____________________________________________________
`b)`
`***f(3)=9`
`=>3a-3=9`
`=>3a=12=>a=4`
`***f(5)=11`
`=>5a-3=11`
`=>5a=14=>a=14/5`
`***f(-1)=6`
`=>-a-3=6`
`=>-a=9=>a=-9`
a: f(1)=2+5-3=4
f(0)=-3
f(1,5)=4,5+7,5-3=9
b: f(3)=9 nên 3a-3=9
hay a=4
f(5)=11 nên 5a-3=11
hay a=14/5
f(-1)=6 nên -a-3=6
=>-a=9
hay a=-9
\(\left|x\right|+\frac{1}{6}=\frac{2}{3}\)
\(\left|x\right|=\frac{2}{3}-\frac{1}{6}\)
\(\left|x\right|=\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{1}{2}\right\}\)
\(|x|=\frac{2}{3}-\frac{1}{6}\)
\(|x|=\frac{1}{2}\)
\(\Rightarrow\)\(x\in\left\{\frac{1}{2};\frac{-1}{2}\right\}\)