a) Tìm số nguyên âm x để đa thức:
f(x)= -x4+2x2-3x+5 chia hết cho g(x)= x-1
b) Tìm giá trị nhỏ nhất của biểu thức:
B= x2+5y2-2xy+4x-4y-2020
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Bài 3:
a) Ta có: \(A=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)
d) Ta có: \(D=x^2-2x+2\)
\(=x^2-2x+1+1\)
\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)
Bài 1:
a) Ta có: \(A=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
b) Ta có: \(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
b: \(x^2-x+1=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
c: \(A=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi x=3
d: \(B=-\left(x^2-4x+5\right)=-\left(x^2-4x+4+1\right)=-\left(x-2\right)^2-1\le-1\forall x\)
Dấu '=' xảy ra khi x=2
\(a,A=\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25+\left(y^2-2y+1\right)+2\\ A=\left(x-2y\right)^2+10\left(x-2y\right)+5+\left(y-1\right)^2+2\\ A=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=2y-5\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
\(b,\Leftrightarrow3x^3+10x^2-5+n=\left(3x+1\right)\cdot a\left(x\right)\)
Thay \(x=-\dfrac{1}{3}\Leftrightarrow3\left(-\dfrac{1}{27}\right)+10\cdot\dfrac{1}{9}-5+n=0\)
\(\Leftrightarrow-\dfrac{1}{9}+\dfrac{10}{9}-5+n=0\\ \Leftrightarrow-4+n=0\Leftrightarrow n=4\)
\(c,\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\\ \Leftrightarrow2n\left(n-2\right)+5\left(n-2\right)+3⋮n-2\\ \Leftrightarrow n-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow n\in\left\{-1;1;3;5\right\}\)
a) \(2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{4}\)
b) \(5x-x^2+4=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\)
\(ĐTXR\Leftrightarrow x=\dfrac{5}{2}\)
c) \(x^2+5y^2-2xy+4y+3=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)
\(ĐTXR\Leftrightarrow\)\(x=y=-\dfrac{1}{2}\)
b: ta có: \(-x^2+5x+4\)
\(=-\left(x^2-5x-4\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}\right)\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{41}{4}\le\dfrac{41}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)
b: \(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)
\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow n\in\left\{0;-1;1\right\}\)
Đáp án C
G T ⇔ x 2 + y − 3 x + y 2 − 4 y + 4 = 0 y 2 + x − 4 y + x 2 − 3 x + 4 = 0
có nghiệm ⇔ Δ x ≥ 0 Δ y ≥ 0 ⇔ 0 ≤ x ≤ 4 3 1 ≤ y ≤ 7 3
Và:
x y = 3 x + 4 y − x 2 − y 2 − 4 ⇒ P = 3 x 3 + 18 x 2 + 45 x − 8 ⏟ f x + − 3 y 3 + 3 y 2 + 8 y ⏟ g y
Xét hàm số f x = 3 x 3 + 18 x 2 + 45 x − 8 trên 0 ; 4 3 ⇒ max 0 ; 4 3 f x = f 4 3 = 820 9
Xét hàm số g x = − 3 y 3 + 3 y 2 + 8 y trên 1 ; 7 3 ⇒ max 1 ; 7 3 g x = f 4 3 = 80 9
Vật P ≤ max 0 ; 4 3 f x + max 1 ; 7 3 g x = 100
Dấu “=” xảy ra khi x = y = 4 3
b: Ta có: f(x):g(x)
\(=\dfrac{x^3-2x^2+3x+a}{x+1}\)
\(=\dfrac{x^3+x^2-3x^2-3x+6x+6+a-6}{x+1}\)
\(=x^2-3x+6+\dfrac{a-6}{x+1}\)
Để f(x):g(x) là phép chia hết thì a-6=0
hay a=6
a: Thay a=3 vào f(x), ta được:
\(f\left(x\right)=x^3-2x^2+3x+3\)
\(\dfrac{f\left(x\right)}{g\left(x\right)}=\dfrac{x^3-2x^2+3x+3}{x+1}\)
\(=\dfrac{x^3+x^2-3x^2-3x+6x+6-3}{x+1}\)
\(=x^2-3x+6-\dfrac{3}{x+1}\)
a)\(f\left(x\right)=2x^2-x-3+5=\left(x+1\right)\left(2x-3\right)+5\)
Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(x+1\right)\left(2x-3\right)+5⋮\left(x+1\right)\)
\(\Leftrightarrow5⋮\left(x+1\right)\)
mà \(x+1\in Z\Rightarrow x+1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)
\(\Leftrightarrow x\in\left\{-2;0;4;-6\right\}\)
Vậy...
b) \(f\left(x\right)=3x^2-4x+6=\left(3x^2-4x+1\right)+5=\left(3x-1\right)\left(x-1\right)+5\)
Để \(f\left(x\right)⋮g\left(x\right)\Leftrightarrow\left(3x-1\right)\left(x-1\right)+5⋮\left(3x-1\right)\)
\(\Leftrightarrow5⋮\left(3x-1\right)\) mà \(3x-1\in Z\Rightarrow3x-1\in U\left(5\right)=\left\{-1;1;5;-5\right\}\)
\(\Leftrightarrow x\in\left\{0;\dfrac{2}{3};2;-\dfrac{4}{3}\right\}\) mà x nguyên\(\Rightarrow x\in\left\{0;2\right\}\)
Vậy...
c)\(f\left(x\right)=\left(-2x^3-7x^2-5x+2\right)+3\)\(=\left(-2x^3-4x^2-3x^2-6x+x+2\right)+3\)\(=\left[-2x^2\left(x+2\right)-3x\left(x+2\right)+\left(x+2\right)\right]+3\)
\(=\left(x+2\right)\left(-2x^2-3x+1\right)+3\)
Làm tương tự như trên \(\Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Leftrightarrow x\in\left\{-5;-3;-1;1\right\}\)
Vậy...
d)\(f\left(x\right)=x^3-3x^2-4x+3=x\left(x^2-3x-4\right)+3=x\left(x+1\right)\left(x-4\right)+3\)
Làm tương tự như trên \(\Rightarrow x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow x\in\left\{-4;-2;0;2\right\}\)
Vậy...
\(a,f\left(x\right)⋮g\left(x\right)\\ \Leftrightarrow\dfrac{-x^4+2x^2-3x+5}{x-1}\in Z\\ \Leftrightarrow\dfrac{-x^4+x^3-x^3+x^2+x^2-x-2x+2+3}{x-1}\in Z\\ \Leftrightarrow\dfrac{-x^3\left(x-1\right)-x^2\left(x-1\right)+x\left(x-1\right)-2\left(x-1\right)+3}{x-1}\in Z\\ \Leftrightarrow-x^3-x^2+x-2+\dfrac{3}{x-1}\in Z\\ \Leftrightarrow3⋮x-1\\ \Leftrightarrow x-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\\ Mà.x< 0\\ \Leftrightarrow x=-2\\ b,B=\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4+4y^2-2024\\ B=\left(x-y\right)^2+4\left(x-y\right)+4+4y^2-2024\\ B=\left(x-y-2\right)^2+4y^2-2024\ge-2024\\ B_{min}=-2024\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)