Tìm x biết
a) x + 30 % x = − 1 , 3
b) 1 3 x + 2 5 x − 1 = 0
c) 3 x − 1 2 − 5 x + 3 5 = − x + 1 5
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b) \(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)
\(\Leftrightarrow x-1=0\) hoặc \(x+1=0\)
\(\Leftrightarrow x=1\) hoặc \(x=-1\)
c) \(x^2-6x+8=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
a) \(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
(do \(x^2+1\ge1>0\))
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
Lời giải:
a. $2x^2+3(x-1)(x+1)=5x(x+1)$
$\Leftrightarrow 2x^2+3x^2-3=5x^2+5x$
$\Leftrightarrow 5x^2-3=5x^2+5x$
$\Leftrightarrow 5x=-3$
$\Leftrightarrow x=\frac{-3}{5}$
b.
PT $\Leftrightarrow (-5x^2-2x+16)+4(x^2-x-2)=4-x^2$
$\Leftrightarrow -x^2-6x+8=4-x^2$
$\Leftrightarrow -6x+8=4$
$\Leftrightarrow -6x=-4$
$\Leftrightarrow x=\frac{2}{3}$
c.
PT $\Leftrightarrow 4(x^2+4x-5)-(x^2+7x+10)=3(x^2+x-2)$
$\Leftrightarrow 4x^2+16x-20-x^2-7x-10=3x^2+3x-6$
$\Leftrightarrow 3x^2+9x-30=3x^2+3x-6$
$\Leftrightarrow 6x=24$
$\Leftrightarrow x=4$
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)
\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)
a) x + 30 % x = − 1 , 3
x 1 + 3 10 = − 13 10 13 10 x = − 13 10 x = − 1
b) 1 3 x + 2 5 x − 1 = 0
1 3 x + 2 5 x − 2 5 = 0 11 15 x = 2 5 x = 2 5 : 11 15 x = 6 11
c) 3 x − 1 2 − 5 x + 3 5 = − x + 1 5
3 x − 3 2 − 5 x − 3 = − x + 1 5 x = − 3 2 − 3 − 1 5 x = − 47 10