Ôn tập phép nhân và phép chia đa thức - Hỏi đáp

Câu 1 :

Ta có : \(Q=x^2-2x+7=x^2-2x+1+6=\left(x-1\right)^2+6\)

Ta thấy : \(\left(x-1\right)^2\ge0\forall x\)

=> \(\left(x-1\right)^2+6\ge6\forall x\)

Vậy Min_Q = 6 <=> x = 1 .

Ta có : \(2x^2-3x+4=\left(x\sqrt{2}\right)^2-\frac{2.x.\sqrt{2}.3\sqrt{2}}{4}+\frac{9}{8}+\frac{23}{8}\)

\(=\left(x\sqrt{2}-\frac{3\sqrt{2}}{4}\right)^2+\frac{23}{8}\)

Ta thấy : \(\left(x\sqrt{2}-\frac{3\sqrt{2}}{4}\right)^2\ge0\forall x\)

=> \(\left(x\sqrt{2}-\frac{3\sqrt{2}}{4}\right)^2+\frac{23}{8}\ge\frac{23}{8}\forall x\)

Vậy Min_P = \(\frac{23}{8}\) <=> \(x=\frac{3}{4}\) .

Câu 2 :

Ta có : \(A=-4x^2+\frac{2.x.2.1}{2}-\frac{1}{4}+\frac{21}{4}=-\left(4x^2-2x+\frac{1}{4}\right)+\frac{21}{4}\)

\(=-\left(2x-\frac{1}{2}\right)^2+\frac{21}{4}\)

Ta thấy : \(\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)

=> \(-\left(2x-\frac{1}{2}\right)^2+\frac{21}{4}\le\frac{21}{4}\forall x\)

Vậy Max_A = 21/4 <=> x = 1/4 .

Câu 3 :

- Thay y = x + 5 vào biểu thức M ta được :

\(M=x\left(x+5\right)=x^2+5x=x^2+\frac{2.x.5}{2}+\frac{25}{4}-\frac{25}{4}\)

\(=\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\)

Ta thấy : \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\)

=> \(\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\ge-\frac{25}{4}\forall x\)

Vậy Min_M = -25/4 <=> x = -5/2 .

Ta có: B=1+2+3+...+100

=(1+100)+(2+99)+...+(50+51)

\(=101\cdot50\)

Ta có: \(A=1^3+2^3+3^3+...+100^3\)

\(=\left(1^3+100^3\right)+\left(2^3+99^3\right)+...+\left(50^3+51^3\right)\)

\(=\left(1+100\right)\cdot\left(1-100+100^2\right)+\left(2+99\right)\left(4-198+99^2\right)+...+\left(50+51\right)\left(2500+50\cdot51+51^2\right)\)

\(=101\cdot\left(1-100+100^2+4-198+99^2+...+50^2-50\cdot51+51^2\right)⋮101\)

Ta có: \(A=1^3+2^3+3^3+...+100^3\)

\(=\left(1^3+99^3\right)+\left(2^3+98^3\right)+...50^3+100^3\)

\(=\left(1+99\right)\left(1-99+99^2\right)+\left(2+98\right)\cdot\left(4-196+98^2\right)+...+50^3+50^3\cdot2^3⋮50\)

mà (50,101)=1

nên \(A⋮50\cdot101=B\)

hay \(A⋮B\)(đpcm)

Ta có : \(x^2-y=y^2-x\Rightarrow\left(x^2-y^2\right)+\left(x-y\right)=0\)

\(\Rightarrow\left(x-y\right)\left(x+y\right)+\left(x+y\right)=0\)

\(\Rightarrow\left(x-y\right)\left(x+y+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-y=0\\x+y+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x=-1-y\end{matrix}\right.\)

Xét x = y ta có :

\(A=2x^3+3x^2\left(2x^2\right)+6x^4\left(2x\right)=2x^3+6x^4+12x^5\)

Xét x = -1 - y ta có :

\(A=y^3+\left(-1-y\right)^3+3y\left(-1-y\right)\left[\left(-1-y\right)^2+y^2\right]+6\left(-1-y\right)^2y^2\left(-1-y-y\right)\)

Rút gọn ta được :

\(A=-6y^4-24y^3-42y^2-30y-7\)

\(a,\left(x+1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)\(\Leftrightarrow x^3+3x^2+3x+1+8-x^3+3x^2+6x-17=0\)\(\Leftrightarrow6x^2+9x-8=0\)

\(\Leftrightarrow x^2+\dfrac{3}{2}x-\dfrac{4}{3}=0\)

\(\Leftrightarrow\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{16}-\dfrac{4}{3}=0\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right)^2=\dfrac{91}{48}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\sqrt{\dfrac{91}{48}}\\x+\dfrac{3}{4}=-\sqrt{\dfrac{91}{48}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{91}{48}}-\dfrac{3}{4}\\x=-\sqrt{\dfrac{91}{48}}-\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{273}}{12}\\x=-\dfrac{9+\sqrt{273}}{12}\end{matrix}\right.\)

b, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow x^3+8-x^3+2x-15=0\)

\(\Leftrightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)

Câu d)

x(x + 5)(x - 5) - (x + 2)(x² - 2x + 4) = 3

<=> x³ + 5x² - 5x² - 25x - x³ + 2x² - 4x - 2x² + 4x - 8 = 3

<=> -25x - 8 = 3

<=> -25x = 3 + 8 = 11

=> x = -11/25

Chúc bạn học tốt!

Lời giải:

a) Có: \(A=x^2-4x-7=(x^2-4x+4)-11\)

\(=(x-2)^2-11\)

Vì \((x-2)^2\geq 0, \forall x\in\mathbb{R}\Rightarrow A=(x-2)^2-11\geq -11\)

Suy ra GTNN của A là $-11$ khi $x=2$

b) \(B=6x-x^2=9-(x^2-6x+9)\)

\(=9-(x-3)^2\)

Vì \((x-3)^2\geq 0, \forall x\Rightarrow B=9-(x-3)^2\leq 9-0=9\)

Vậy GTLN của B là $9$ khi $x=3$

Gọi a; a+1; a+2; a+3 là 4 số tự nhiên liên tiếp (\(a\in N\)*)

Ta có tích của 4 số tự nhiên cộng thêm 1 là: \(a\left(a+1\right)\left(a+2\right)\left(a+3\right)+1\)

\(=\left[a\left(a+3\right)\right]\left[\left(a+1\right)\left(a+2\right)\right]+1=\left(a^2+3a\right)\left(a^2+3a+2\right)+1\)

\(=\left(a^2+3a\right)^2+2\left(a^2+3a\right)+1=\left(a^2+3a+1\right)^2\)

=> đpcm

Câu 1:

\(A=x^2-3x+9\\ =x^2-3x+\dfrac{9}{4}+\dfrac{27}{4}\\ =\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{27}{4}\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\\ Do\text{ }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\ge0\forall x\\ \text{Dấu “=” xảy ra khi: }\\ \left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\text{ }A_{\left(Min\right)}=\dfrac{27}{4}\text{ }khi\text{ }x=\dfrac{3}{2}\)

\(B=9x^2-6x+2\\ =9x^2-6x+1+1\\ =\left(9x^2-6x+1\right)+1\\ =\left(3x-1\right)^2+1\\ Do\text{ }\left(3x-1\right)^2\ge0\forall x\\ \Rightarrow B=\left(3x-1\right)^2+1\ge1\forall x\\ \text{Dấu “=” xảy ra khi: }\\ \left(3x-1\right)^2=0\\ \Leftrightarrow3x-1=0\\ \Leftrightarrow3x=1\\ \Leftrightarrow x=\dfrac{1}{3}\\ Vậy\text{ }B_{\left(Min\right)}=1\text{ }khi\text{ }x=\dfrac{1}{3}\)

\(C=-x^2+2x+4\\ =-x^2+2x-1+5\\ =-\left(x^2-2x+1\right)+5\\ =-\left(x-1\right)^2+5\\ Do\text{ }\left(x-1\right)^2\ge0\forall x\\ \Rightarrow-\left(x-1\right)^2\le0\forall x\\ \Rightarrow C=-\left(x-1\right)^2+5\le5\forall x\\ \text{ Dấu “=” xảy ra khi: }\\ \left(x-1\right)^2=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\\ \text{Vậy }C_{\left(Max\right)}=5\text{ }khi\text{ }x=1\)

\(D=-x^2+4x\\ =-x^2+4x-4+4\\ =-\left(x^2-4x+4\right)+4\\ =-\left(x-2\right)^2+4\\ \\ Do\text{ }\left(x-2\right)^2\ge0\forall x\\ \Rightarrow-\left(x-2\right)^2\le0\forall x\\ \Rightarrow C=-\left(x-2\right)^2+4\le4\forall x\\ \text{ Dấu “=” xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }C_{\left(Max\right)}=4\text{ }khi\text{ }x=2\)

Câu 2:

\(\text{Ta có : }x+y=2\\ \Rightarrow\left(x+y\right)^2=2^2\\ \Rightarrow x^2+2xy+y^2=4\\ Thay\text{ }x^2+y^2=10\text{ }vào\\ \Rightarrow2xy+10=4\\ \Rightarrow2xy=-6\\ \Rightarrow xy=-3\\ \text{Ta lại có : }x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\\ Thay\text{ }x^2+y^2=10;x+y=2;xy=-3\text{ }ta\text{ }được:\\ x^3+y^3=2\cdot\left(10+3\right)=26\)

Vậy \(x^3+y^3=26\text{ }tại\text{ }x+y=2;x^2+y^2=10\)

Đề bài sai hoặc thiếu 1 điều kiện nào đó

Bạn xem lại

\(a^2+b^2+c^2\le abc\Rightarrow\frac{a^2+b^2+c^2}{abc}\le1\)

\(VT=\frac{a}{a^2+bc}+\frac{b}{b^2+ac}+\frac{c}{c^2+ab}\le\frac{a}{2\sqrt{a^2bc}}+\frac{b}{2\sqrt{b^2ac}}+\frac{c}{2\sqrt{c^2ab}}\)

\(VT\le\frac{1}{2}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}}\right)\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{c}\right)\)

\(VT\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{2}\left(\frac{ab+bc+ca}{abc}\right)\le\frac{1}{2}\left(\frac{a^2+b^2+c^2}{abc}\right)\le\frac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Ngaos lone à

đề đâu bn

\(\frac{x+1}{x+2}:\frac{x+2}{x+3}:\frac{x+3}{x+1}\)

\(=\frac{x+1}{x+2}\cdot\frac{x+3}{x+2}\cdot\frac{x+1}{x+3}=\frac{\left(x+1\right)^2}{\left(x+2\right)^2}\)

Làm

Ta có: f(x) chia cho x+1 dư 1 => f(-1)=4 (1) (Định lí Bơ-du)

Ta có : f(x)chia x²+1 dư 2x+3 => f(x)= (x²+1)g(x) + 2x+3 (2)

Khi chia f(x) cho đa thức (x+1)(x²+1) bậc 3 thì dư sẽ có dạng ax²+bx+c

=> f(x)= (x+1)(x²+1)k(x)+ax²+bx+c (4)

=> f(x)= (x+1)(x²+1)k(x) +a(x²+1)+bx+c-a

=>f(x) = (x²+1) [(x+1)(x²+1)k(x)+a] +bx+c-3 (3)

(2)(3)=> 2x+3= bx+c-a với mọi x

=> \(\left\{{}\begin{matrix}c-a=3\\b=2\end{matrix}\right.\)

(1)(4)=> a+c=6 mà c-a =3 \(\Rightarrow\left\{{}\begin{matrix}a=\frac{3}{2}\\c=\frac{9}{2}\end{matrix}\right.\)

Vậy đa thức dư là \(\frac{3}{2}x^2+2x+\frac{9}{2}\)