\(2x\left(x-5-x\right)=10\\ 2x.\left(-5\right)=10\\ 2x=-2\\ x=\dfrac{-2}{2}=-1\)

Bài 5:

a: 3x+2(5-x)=0

=>3x+10-2x=0

=>x+10=0

=>x=-10

b: 2x(x+3)+2(x+3)=0

=>(x+3)(2x+2)=0

=>(x+1)(x+3)=0

=>\(\left[\begin{array}{l}x+1=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ x=-3\end{array}\right.\)

Bài 6:

\(A=2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)

\(=-6x^5+4x^4+2x^3-2x^2+2x^3-6x^2+2x\)

\(=-6x^5+4x^4+4x^3-8x^2+2x\)

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

bạn ơi có câu c không bạn

TA có: \(2x+7\vdots2x^2+7\)

=>\(\left(2x+7\right)\left(2x-7\right)\vdots2x^2+7\)

=>\(4x^2-49\vdots2x^2+7\)

=>\(4x^2+14-63\vdots2x^2+7\)

=>\(-63\vdots2x^2+7\)

=>\(2x^2+7\in\left\lbrace7;9;21;63\right\rbrace\)

=>\(2x^2\in\left\lbrace0;2;14;56\right\rbrace\)

=>\(x^2\in\left\lbrace0;1;7;28\right\rbrace\)

mà x nguyên

nên x∈{0;1;-1}

Ta có số x² ≥ x với mọi x nguyên

=> 2x² + 7 ≥ 2x + 7 (1)

Mà 2x + 7 ≥ 2x² + 7 do 2x + 7 ⋮ 2x² + 7 (2)

Từ (1) và (2) suy ra 2x + 7 = 2x² + 7

=> 2x = 2x²

=> x = x²

=> x ϵ { 1;0 }

a) cho A(x) = 0

\(=>2x^2-4x=0\)

\(x\left(2-4x\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\4x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)

b)\(B\left(y\right)=4y-8\)

cho B(y) = 0

\(4y-8=0\Rightarrow4y=8\Rightarrow y=2\)

c)\(C\left(t\right)=3t^2-6\)

cho C(t) = 0

\(=>3t^2-6=0=>3t^2=6=>t^2=2\left[{}\begin{matrix}t=\sqrt{2}\\t=-\sqrt{2}\end{matrix}\right.\)

d)\(M\left(x\right)=2x^2+1\)

cho M(x) = 0

\(2x^2+1=0\Rightarrow2x^2=-1\Rightarrow x^2=-\dfrac{1}{2}\left(vl\right)\)

vậy M(x) vô nghiệm

e) cho N(x) = 0

\(2x^2-8=0\)

\(2\left(x^2-4\right)=0\)

\(2\left(x^2+2x-2x-4\right)=0\)

\(2\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)