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1) \(\left(x-3\right)\left(x-5\right)+2\)
\(=x^2-8x+15+2\)
\(=\left(x^2-8x+16\right)+1\)
\(=\left(x-4\right)^2+1\)
Vì \(\left(x-4\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x-4\right)^2+1\ge1>0;\forall x\)
Vậy....
2) tương tự
\(1.\left(x-3\right)\left(x-5\right)+2\)
\(=x^2-8x+15+2\)
\(=x^2-2.4x+16+1\)
\(=\left(x-4\right)^2+1\)
Do \(\left(x-4\right)^2\ge0\)nên \(\left(x-4\right)^2+1\ge1\)
hay \(\left(x-3\right)\left(x-5\right)+2>0\)

c) Ta có: \(P=2x+\dfrac{1}{x+1}\)
\(\Leftrightarrow\dfrac{-x}{x+1}=2x+\dfrac{1}{x+1}\)
\(\Leftrightarrow\dfrac{-x}{x+1}=\dfrac{2x\left(x+1\right)+1}{x+1}\)
Suy ra: \(2x^2+2x+1=-x\)
\(\Leftrightarrow2x^2+3x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: Để \(P=2x+\dfrac{1}{x+1}\) thì \(x=-\dfrac{1}{2}\)
`a)F=((x+1)/(1-x)-(1-x)/(x+1)-(4x^2)/(x^2-1)):(4x^2-4)/(x^2-2x+1)`
`đk:x ne +-1`
`F=((-(x+1)^2+(x-1)^2-4x^2)/(x^2-1)):(4(x-1)(x+1))/(x-1)^2`
`=(-x^2-2x-1+x^2-2x+1-4x^2)/(x^2-1):(4(x+1))/(x-1)`
`=(-4x^2-4x)/((x-1)(x+1)).(x-1)/(4(x+1))`
`=(-4(x-1))/((x-1)(x+1)).(x-1)/(4(x+1))`
`=-4/(x+1).(x-1)/(4(x+1)`
`=(1-x)/(x+1)^2`
`F<-1`
`<=>(1-x-(x+1)^2)/(x+1)^2<0`
Vì `(x+1)^2>0`
`=>1-x-(x+1)^2<0`
`<=>(x+1)^2+x-1>0`
`<=>x^2+2x+1+x-1>0`
`<=>x^2+3x>0`
`<=>x(x+3)>0`
`<=>` $\left[ \begin{array}{l}x>0\\x<-3\end{array} \right.$

Gọi số ly trà sữa là x
=>Số ly trà đào là 210-x
Theo đề, ta có: 27000x=2*18000(210-x)
=>27000x-36000(210-x)=0
=>27000x-7560000+36000x=0
=>x=120
=>Số ly trà đào là 90 ly


Bài 2:
a: \(\left(-\frac13x^2y\right)\cdot2xy^3=\left(-\frac13\cdot2\right)\cdot x^2\cdot x\cdot y\cdot y^3=-\frac23x^3y^4\)
b: \(\left(-\frac34x^2y\right)\cdot\left(-xy\right)^3=\left(-\frac34\right)\cdot\left(-1\right)\cdot x^2\cdot x^3\cdot y\cdot y^3=\frac34x^5y^4\)
c: \(\frac35\cdot x^2y^5\cdot x^3y^2\cdot\frac{-2}{3}=\left(\frac35\cdot\frac{-2}{3}\right)\cdot x^2\cdot x^3\cdot y^5\cdot y^2=-\frac25x^5y^7\)
d: \(\left(\frac34x^2y^3\right)\cdot\left(2\frac25x^4\right)=\frac34x^2y^3\cdot\frac{12}{5}x^4=\frac34\cdot\frac{12}{5}\cdot x^2\cdot x^4\cdot y^3=\frac95x^6y^3\)
e: \(\left(\frac{12}{15}x^4y^5\right)\cdot\left(\frac59x^2y\right)=\frac45\cdot\frac59\cdot x^4\cdot x^2\cdot y^5\cdot y=\frac49x^6y^6\)
f: \(\left(-\frac17x^2y\right)\left(-\frac{14}{5}x^4y^5\right)=\frac17\cdot\frac{14}{5}\cdot x^2\cdot x^4\cdot y\cdot y^5=\frac25x^6y^6\)
Bài 1: Các đơn thức là \(x^2y;-13;\left(-2\right)^3xy^7\)

\(a,9x^2-1=0\)
\(\left(3x\right)^1-1=0\)
\(\left(3x-1\right)\cdot\left(3x+1\right)=0\)
\(\hept{\begin{cases}3x-1=\\3x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}}\)
\(b,x\cdot\left(x+5\right)-x-5=0\)
\(x\cdot\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\cdot\left(x-1\right)=0\)
\(\hept{\begin{cases}x+5=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-5\\x=1\end{cases}}}\)

\(A=3\left(x-3\right)\left(x+7\right)+\left(x+4\right)^2+48\)
\(A=3\left(x^2-4x-21\right)+\left(x^2+8x+16\right)+48\)
\(A=\left(3x^2+x^2\right)-\left(12x-8x\right)-\left(21-16-48\right)\)
\(A=4x^2-4x+43\)
\(A=\left(4x^2-4x+1\right)+42\)
\(A=\left(2x+1\right)^2+42\)
Thay \(x=\frac{1}{2}\) vao A ta duoc:
\(A=\left(2\cdot\frac{1}{2}+1\right)^2+42=46\)
\(A=3\left(x-3\right)\left(x-7\right)+\left(x+4\right)^2+48\)
\(=3x^2-13x+63+x^2+8x+16+48\)
\(=4x^2-5x+127\)
\(4\cdot0,25-5\cdot0,5+127=1-1+127=127\)

\(A=\left(x+5\right)\left(2x-3\right)-2x\left(x+3\right)-\left(x-15\right)\)
\(=\left(2x^2-3x+10x-15\right)-2x^2-6x-x+15\)
\(=2x^2-3x+10x-15-2x^2-6x-x+15\)
\(=0\)
- Xét △OBC có: \(BC\)//\(AD\) (gt).
=>\(\dfrac{OD}{OC}=\dfrac{OA}{OB}\) (định lí Ta-let).
=>\(OD=\dfrac{OA}{OB}.OC=\dfrac{2,5}{2}.3=3,75\) (cm).
tớ cảm ơn ạ