`4)x^2-5x+6`

`=x^2-2x-3x+6`

`=x(x-2)-3(x-2)=(x-2)(x-3)`

`5)x^2+7x+10`

`=x^2+5x+2x+10`

`=x(x+5)+2(x+5)=(x+5)(x+2)`

`6)x+7\sqrt{x}+10`    `ĐK: x >= 0`

`=(\sqrt{x})^2+5\sqrt{x}+2\sqrt{x}+10`

`=\sqrt{x}(\sqrt{x}+5)+2(\sqrt{x}+5)=(\sqrt{x}+5)(\sqrt{x}+2)`

`7)3x^4+7x^2+4`

`=3x^4+3x^2+4x^2+4`

`=3x^2(x^2+1)+4(x^2+1)=(x^2+1)(3x^2+4)`

`8)x^2-x-2`

`=x^2-2x+x-2`

`=x(x-2)+(x-2)=(x-2)(x+1)`

`9)x^6-x^3-2`

`=x^6+x^3-2x^3-2`

`=x^3(x^3+1)-2(x^3+1)`

`=(x^3+1)(x^3-2)`.

Phân tích đa thức thành nhân tử:

Đặt \(\hept{\begin{cases}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{cases}}\)

\(\Rightarrow a^2+4b^2=5ab\)

\(\Leftrightarrow\left(b-a\right)\left(4b-a\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\a=4b\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\sqrt[3]{65+x}=\sqrt[3]{65-x}\\\sqrt[3]{65+x}=4\sqrt[3]{65-x}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}65+x=65-x\\65+x=4\left(65-x\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=39\end{cases}}\)

a) \(\dfrac{x+43}{57}+\dfrac{x+46}{54}=\dfrac{x+49}{51}+\dfrac{x+52}{48}\)

\(\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}\right)\)

\(\dfrac{x+43+57}{57}+\dfrac{x+46+54}{54}-\dfrac{x+49+51}{51}-\dfrac{x+52+48}{48}=0\)

\(\dfrac{x+100}{57}+\dfrac{x+100}{54}-\dfrac{x+100}{51}-\dfrac{x+100}{48}=0\)

\(\left(x+100\right)\left(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\right)=0\)

Mà \(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\ne0\)

Nên: \(x+100=0\)

\(x=-100\)

1) \(x^2+x-6=x\left(x-2\right)+3\left(x-2\right)=\left(x+3\right)\left(x-2\right)\)

2) \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)

3) \(x^2+2x-48=\left(x-6\right)\left(x+8\right)\)

4) \(x^2-2x-48=\left(x-8\right)\left(x+6\right)\)

5) \(x^2+x-42=\left(x-6\right)\left(x+7\right)\)

6) \(x^2-x-42=\left(x-7\right)\left(x+6\right).\)

bài này tìm x hay tìm cực trị vậy

số đó là

177

ai k mình

mình j lại cho

số đó là

177

ai k mình

mình j lại cho

Câu 20: \(x^2-2x+1<9\)

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

=>(x-4)(x+2)<0

=>-2<x<4

=>Chọn A

Câu 19: (x-4)(9-x)>=0

=>(x-4)(x-9)<=0

=>4<=x<=9

=>Chọn C

Câu 17: \(x^2-4x+4\le0\)

=>\(\left(x-2\right)^2\le0\)

mà \(\left(x-2\right)^2\ge0\forall x\)

nên x-2=0

=>x=2

Câu 15: \(\left(2x+1\right)\left(x^2-4\right)>0\)

TH1: \(\begin{cases}2x+1>0\\ x^2-4>0\end{cases}\)

=>x>-1/2 và x^2>4

=>(x>-1/2) và (x>2 hoặc x<-2)

=>x>2

TH2: \(\begin{cases}2x+1<0\\ x^2-4<0\end{cases}\)

=>\(\begin{cases}x<-\frac12\\ x^2<4\end{cases}\Rightarrow\begin{cases}x<-\frac12\\ -2

Câu 14: (x-1)(x+2)>0

TH1: \(\begin{cases}x-1>0\\ x+2>0\end{cases}\)

=>\(\begin{cases}x>1\\ x>-2\end{cases}\Rightarrow x>1\)

TH2: \(\begin{cases}x-1<0\\ x+2<0\end{cases}\Rightarrow\begin{cases}x<1\\ x<-2\end{cases}\)

=>x<-2

Câu 12: (x+2)(x-5)<0

TH1: \(\begin{cases}x+2>0\\ x-5<0\end{cases}=>\begin{cases}x>-2\\ x<5\end{cases}=>-2

TH2: \(\begin{cases}x+2<0\\ x-5>0\end{cases}\Rightarrow\begin{cases}x<-2\\ x>5\end{cases}\)

=>x∈∅

Câu 10: \(-2x^2+11x-15>0\)

=>\(2x^2-11x+15<0\)

=>\(2x^2-6x-5x+15<0\)

=>(x-3)(2x-5)<0

=>5/2<x<3

mà x nguyên

nên x∈∅

Câu 8: (x-2)(x-5)>0

TH1: \(\begin{cases}x-2>0\\ x-5>0\end{cases}\Rightarrow\begin{cases}x>2\\ x>5\end{cases}\Rightarrow x>5\)

TH2: \(\begin{cases}x-2<0\\ x-5<0\end{cases}\Rightarrow\begin{cases}x<2\\ x<5\end{cases}\)

=>x<2

Câu 7: (x+5)(7-2x)>0

=>(x+5)(2x-7)<0

=>-5<x<7/2

mà x nguyên

nên x∈{-4;-3;-2;-1;0;1;2;3}

=>Có 8 giá trị x nguyên thỏa mãn

Câu 6: \(\left(2x-3\right)\left(x^2+1\right)\le0\)

mà \(x^2+1\ge1>0\forall x\)

nên 2x-3<=0

=>2x<=3

=>x<=3/2

ĐKXĐ : \(\forall x\)

Ta có : \(\dfrac{x^2}{x^2+2x+2}+\dfrac{x^2}{x^2-2x+2}-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)

\(\Leftrightarrow\dfrac{x^2\left(x^2-2x+2\right)+x^2\left(x^2+2x+2\right)-4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)

\(\Leftrightarrow\dfrac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)

\(\Leftrightarrow\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)

\(\Leftrightarrow65\left(2x^4+20\right)=322\left(x^4+4\right)\)

\(\Leftrightarrow130x^4+1300=322x^4+1288\)

\(\Leftrightarrow192x^4-12=0\)

\(\Leftrightarrow x^4=\dfrac{12}{192}\)

\(\Leftrightarrow x^4=\dfrac{1}{16}\)

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

Vậy ...

\(\dfrac{x^2}{x^2+2x+2}+\dfrac{x^2}{x^2-2x+2}-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)

\(\Leftrightarrow x^2\left(\dfrac{1}{x^2+2+2x}+\dfrac{1}{x^2+2-2x}\right)-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)

\(\Leftrightarrow x^2\left(\dfrac{2x^2+4}{x^4+4}\right)-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)

\(\Leftrightarrow\dfrac{2x^4+4x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)

\(\Leftrightarrow\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)

\(\Leftrightarrow65x^4+650=161x^4+644\)

\(\Leftrightarrow96x^4=6\)

\(\Leftrightarrow x^4=\dfrac{1}{16}\)

\(\Rightarrow x=\pm\dfrac{1}{2}\)