\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$

$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

a: P(x)=3x^4+6x^2-5x-2

Q(x)=-2x^6+2x^4+4x^2-5x-4

b: H(x)=P(x)-Q(x)

=3x^4+6x^2-5x-2+2x^6-2x^4-4x^2+5x+4

=2x^6+x^4+2x^2+2

c: H(x)=x^2(2x^4+x^2+2)+2>=2>0 với mọi x

=>H(x) ko có nghiệm

a) \(A=\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}\) có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}-5\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}\ne5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)

Khi \(x=16\Rightarrow A=\dfrac{\sqrt[]{16}+2}{\sqrt[]{16}-5}=\dfrac{4+2}{4-5}=-6\)

b) \(B=\dfrac{3}{\sqrt[]{x}+5}+\dfrac{20-2\sqrt[]{x}}{x-25}\)

B có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x-25\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)

\(\Leftrightarrow B=\dfrac{3\left(\sqrt[]{x}-5\right)+20-2\sqrt[]{x}}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)

\(\Leftrightarrow B=\dfrac{3\sqrt[]{x}-15+20-2\sqrt[]{x}}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)

\(\Leftrightarrow B=\dfrac{\sqrt[]{x}+5}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)

\(\Leftrightarrow B=\dfrac{1}{\sqrt[]{x}-5}\left(dpcm\right)\)

c) \(A=\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}\in Z\left(x\in Z\right)\)

\(\Leftrightarrow\sqrt[]{x}+2⋮\sqrt[]{x}-5\)

\(\Leftrightarrow\sqrt[]{x}+2-\left(\sqrt[]{x}-5\right)⋮\sqrt[]{x}-5\)

\(\Leftrightarrow\sqrt[]{x}+2-\sqrt[]{x}+5⋮\sqrt[]{x}-5\)

\(\Leftrightarrow7⋮\sqrt[]{x}-5\)

\(\Leftrightarrow\sqrt[]{x}-5\in U\left(7\right)=\left\{-1;1;-7;7\right\}\)

\(\Leftrightarrow x\in\left\{16;36;144\right\}\)

d) \(A>B\left(2\sqrt[]{x}+5\right)\)

\(\Leftrightarrow\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}>\dfrac{1}{\sqrt[]{x}-5}\left(2\sqrt[]{x}+5\right)\)

\(\Leftrightarrow\sqrt[]{x}+2>2\sqrt[]{x}+5\)

\(\Leftrightarrow\sqrt[]{x}< -3\)

mà \(\sqrt[]{x}\ge0\)

\(\Leftrightarrow x\in\varnothing\)

\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

a. PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PTHH: \(n_{Fe}=n_{H_2}=1,5\left(mol\right)\)

\(\Rightarrow m_{Fe}=56\cdot1,5=84\left(g\right)\)

b. Đổi: \(500ml=0,5l\)

\(CM_{H_2SO_4}=\dfrac{1,5}{0,5}=3M\)

c. \(2H_2+O_2\rightarrow2H_2O\)

Theo PTHH: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot1,5=0,75\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,75\cdot22,4=16,8\left(l\right)\)

c, \(2H_2+O_2 \rightarrow2H_2O\)

\(n_{H_2}=\dfrac{33,6}{22,4}=1,5(mol) \Rightarrow n_{O_2}=0,75(mol)\)

\(V_{O_2}=22,4.0,75=16,8(l)\)

a) Fe+H2SO4---->FeSO4+H2(1)

nH2=33,6:22,4=1,5 mol

Theo PT(1):nFe=nH2=1,5 mol

=>mFe=56.1,5=84g

b)TheoPT(1):nH2SO4=nH2=1,5 mol

=>CM(H2SO4)=1,5:0,5=3 M

c)2H2+O2---->2H2O(2)

TheoPT(2):nO2=1/2nH2=1/2.1,5=0,75mol

=>VO2=22,4.0,75=16,8l

giúp mình câu c câu d với

08:43 :vvvv

Vì \(\widehat{MIA}=90^0\left(\text{góc nt chắn nửa đường tròn}\right)\) nên \(MI\perp IA\)

Xét \(\Delta MBP\) có \(\left\{{}\begin{matrix}PK\perp MB\left(PK\perp MN\right)\\MI\perp PB\left(MI\perp IA\right)\\\left\{H\right\}=PK\cap MI\end{matrix}\right.\) nên H là trực tâm 

Do đó \(HB\perp PM\)

Mà \(AM\perp PM\Rightarrow HB\text{//}AM\)

Vì \(HB\text{//}OA\Rightarrow\dfrac{PB}{PA}=\dfrac{HB}{OA}\)

Ta có \(\sin MPB=\sin MPA=\dfrac{MA}{PA}=\dfrac{2OA}{PA}\)

\(\Rightarrow\dfrac{1}{2}BP\cdot\sin MPB=\dfrac{PB\cdot\dfrac{2OA}{PA}}{2}=\dfrac{PB\cdot2OA}{2PA}=\dfrac{PB}{PA}\cdot OA=\dfrac{HB}{OA}\cdot OA=HB\left(đpcm\right)\)

\(a,A=\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}\\ A=\dfrac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x}{x+3}\\ b,\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x-2=1\\2-x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\\ \Leftrightarrow A=\dfrac{-3\cdot1}{1+3}=\dfrac{-3}{4}\\ c,A=\dfrac{-3\left(x+3\right)+9}{x+3}=-3+\dfrac{9}{x+3}\in Z\\ \Leftrightarrow x+3\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\\ \Leftrightarrow x\in\left\{-12;-6;-4;-2;0;6\right\}\left(tm\right)\)

12.

a)

\(P=\dfrac{\sqrt{x}-1+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\left(x>0;x\ne1\right)\\ P=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)