Tìm x, biết:
5x2 = 2 + 3x
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(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2=x.5x^2+x.\left(-3x\right)+x.2+\left(-1\right).5x^2+\left(-1\right)\left(-3x\right)+\left(-1\right).2=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
\(\left(5x^2-7x^2y^3+3y^4\right)-K=3x^2-7x^2y^3-3y^4\)
\(\Rightarrow K=\left(5x^2-7x^2y^3+3y^4\right)-\left(3x^2-7x^2y^3-3y^4\right)\)
\(\Rightarrow K=5x^2-7x^2y^3+3y^4-3x^2+7x^2y^3+3y^4\)
\(\Rightarrow K=2x^2+6y^4\)
________________
\(3x^2-8x+5-K=-2K+4x-6+x^2\)
\(\Rightarrow-K+2K=\left(4x-6+x^2\right)-\left(3x^2-8x+5\right)\)
\(\Rightarrow K=4x-6+x^2-3x^2+8x-5\)
\(\Rightarrow K=-2x^2+12x-11\)
\(\Leftrightarrow5x^2-3x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
a) \(x^2-x+x=4\)
\(x^2=4\)
\(x=\pm2\)
b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)
\(\left(x-5\right)\left(3x-2\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c) Ta có: \(a+b+c=5-3-2=0\)
\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)
d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :
\(t^2-11t+18=0\)
\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)
\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)
\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)
\(A=\left(4x^2-4xy+y^2\right)+\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{21}{4}\\ A=\left(2x-y\right)^2+\left(x+\dfrac{3}{2}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\\ A_{min}=-\dfrac{21}{4}\Leftrightarrow\left\{{}\begin{matrix}2x=y\\x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-3\end{matrix}\right.\)
\(B=\left[\left(x-1\right)\left(x+2\right)\right]\left[x\left(x+1\right)\right]=\left(x^2+x-2\right)\left(x^2+x\right)\\ B=\left(x^2+x\right)^2-2\left(x^2+x\right)\\ B=\left(x^2+x\right)^2-2\left(x^2+x\right)+1-1=\left(x^2+x-1\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow x^2+x-1=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}=0\\ \Leftrightarrow\left(x+\dfrac{1-\sqrt{5}}{2}\right)\left(x+\dfrac{1+\sqrt{5}}{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)
a:Ta có: \(x\left(x-1\right)+x=4\)
\(\Leftrightarrow x^2-x+x=4\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
b: Ta có: \(3x\left(x-5\right)-2x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c: Ta có: \(5x^2-3x-2=0\)
\(\Leftrightarrow5x^2-5x+2x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d: Ta có: \(x^4-11x^2+18=0\)
\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
a) x(x-1)+x=4
⇔x2=4⇔\(x=\pm2\)
b)3x(x-5)-2x+10=0
⇔3x(x-5)-2(x-5)=0
⇔(x-5)(3x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
c)5x2-3x-2=0
⇔ 5x(x-1)+2(x-1)=0
⇔ (x-1)(5x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d)x4-11x2+18=0
⇔ x2(x2-2)-9(x2-2)=0
⇔ (x2-2)(x2-9)=0
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)
a) Ta có: B(x)-M(x)=A(x)
nên M(x)=B(x)-A(x)
\(=x^4-2x^3+5x^2+x+10-x^4-2x^3+5x^2+3x+6\)
\(=-4x^3+10x^2+4x+16\)
5x2 = 2 + 3x
<=> 5x2 - 3x - 2 = 0
<=> 5x2 - 5x + 2x - 2 = 0
<=> 5x( x - 1 ) + 2( x - 1 ) = 0
<=> ( x - 1 )( 5x + 2 ) = 0
<=> x - 1 = 0 hoặc 5x + 2 = 0
<=> x = 1 hoặc x = -2/5
\(5x^2=2+3x\Leftrightarrow5x^2-3x-2=0\Leftrightarrow\left(5x+2\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{5}\\x=1\end{cases}}\)