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1/ a/ Ta có:
\(P\left(2\right)=m.2^2+\left(2m+1\right).2-10=16\)
\(\Leftrightarrow m-3=0\)
\(\Leftrightarrow m=3\)
b/ Theo câu a thì
\(P\left(x\right)=3x^2+7x-10=0\)
\(\Leftrightarrow\left(3x^2-3x\right)+\left(10x-10\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{10}{3}\end{cases}}\)
2/ Tương tự a phân tích nhân tử hộ thôi nha
a/ \(1-5x=0\)
b/ \(x^2\left(x+2\right)=0\)
c/ \(\left(x-1\right)\left(2x-3\right)=0\)
d/ \(\left(x-2\right)^2+4x^{2018}\ge0\) vì dấu = không xảy ra nên đa thức vô nghiệm
\(\left(2x-1\right)^{2018}=\left(2x-1\right)^{2016}\)
\(\Rightarrow\left(2x-1\right)^{2018}-\left(2x-1\right)^{2016}=0\)
\(\Rightarrow\left(2x-1\right)^{2016}\left[\left(2x-1\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-1\right)^{2016}=0\\\left(2x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(2x-1\right)^{2016}=0\\\left(2x-1\right)^2=1\end{cases}}\)
TH 1 : \(\left(2x-1\right)^{2016}=0\Rightarrow2x-1=0\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)
TH 2 : \(\left(2x-1\right)^2=1\Rightarrow\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}2x=2\\2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{2};1;0\right\}\)
_Chúc bạn học tốt_
( 2x - 1 )2018 = ( 2x - 1 )2016
( 2x - 1 )2 = 0
( 2x )2 - 1 = 0
4x2 = 1
x2 = 1 / 4 \(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
1) Tìm số nguyên x, biết :
a) 3x = 94/ 273
3x = 1/3
3x = 3-1
=> x = -1
b) 3x = 98 / 273 . 812
3x = 37.38
3x = 315
=> x = 15
c) 2x - 3 / 410 = 83
2x - 3 = 83.410
2x - 3 = 226
=> x - 3 = 26
=> x = 29
d) 22x - 3 / 410 = 83 . 165
22x - 3 / 410 = 269
22x - 3 = 269 . 410
22x - 3 = 289
=> 2x - 3 = 89
2x = 91
x = 91/2
e) 35 / 3x = 310
3x = 35 : 310
3x = 3-5
=> x = -5
a)\(A=x^2-1\)
\(Nx:\)\(x^2\ge0\)
\(\Rightarrow A_{Min}=0-1=-1\Leftrightarrow x=0\)
b) \(B=x^2-2x+3\)
\(=x\left(x-2\right)+3\)
\(Nx:x\left(x-2\right)\ge0\)
\(\Rightarrow B_{Min}=3\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow x=0\)
c) \(C=\left|2x+1\right|-5\)
\(Nx:\left|2x+1\right|\ge0\Rightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)
\(\Rightarrow C_{Min}=-5\Leftrightarrow x=\frac{-1}{2}\)
d) \(D=3x^2+6x-7\)
\(=3\left(x^2+2x\right)-7\)
\(Nx:Min_{x^2+2x}=-1\Leftrightarrow x=-1\)
\(D_{Min}=-8\Leftrightarrow x=-1\)
B(x)=5x2+x-5
=>2B(x)=2(5x2+x-5)
=>2B(x)=10x2+2x-10
+)Ta có : C(x)-2B(x)=A(x)
=>C(x)=A(x)+2B(x)
A(x)+2B(x)=(3x3+3x2+2x-1)+(10x2+2x-10)
A(x)+2B(x)=3x3+3x2+2x-1+10x2+2x-10
A(x)+2B(x)=3x3+(3x2+10x2)+(2x+2x)+(-1-10)
A(x)+2B(x)=3x3+13x2+4x-11
=> C(x)=3x3+13x2+4x-11
\(A\left(x\right)=3x^3+3x^2+2x-1\)
\(B\left(x\right)=5x^2+x-5\)
Ta có : \(C\left(x\right)-2B\left(x\right)=A\left(x\right)\)
\(\Leftrightarrow C\left(x\right)-10x^2+2x-10=3x^3+3x^2+2x-1\)
\(\Leftrightarrow C\left(x\right)=-10x^2+2x-10-3x^3-3x^2-2x+1=0\)
\(\Leftrightarrow C\left(x\right)=-13x^2-9-3x^3=0\)
Vậy \(C\left(x\right)=-13x^2-9-3x^3\)
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