d: Xét ΔABC có

BK,CH là đường cao

BK cắt CH tại I

=>I là trực tâm

=>AI vuông góc BC

mà HF vuông góc BC

nên AI//HF
e: Xét ΔABC cân tại A có góc BAC=60 độ

nên ΔABC đều

Xét ΔABC đều có I là trực tâm

nên I là tâm đường tròn ngoại tiếp ΔABC

=>IA=IB=IC

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

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) \(A=\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}\) 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 \(\sqrt[]{x}\ge0\)

\(\Leftrightarrow x\in\varnothing\)

a) \(N\left(x\right)=9-x^3+4x^3-7x+3x^2+x^2\)

\(N\left(x\right)=-\left(x^3-4x^3\right)+\left(3x^2+x^2\right)-7x+9\)

\(N\left(x\right)=3x^3+4x^2-7x+9\)

\(M\left(x\right)=4+6x^2+3x+5x^3-2x^3-2x^2\)

\(M\left(x\right)=\left(5x^3-2x^3\right)+\left(6x^2-2x^2\right)+3x+4\)

\(M\left(x\right)=3x^3+4x^2+3x+4\)

b) \(P\left(x\right)=N\left(x\right)-M\left(x\right)\)

\(P\left(x\right)=\left(3x^3+4x^2-7x+9\right)-\left(3x^2+4x^2+3x+4\right)\)

\(P\left(x\right)=3x^3+4x^2-7x+9-3x^3-4x^2-3x-4\)

\(P\left(x\right)=-10x+5\)

\(Q\left(x\right)=N\left(x\right)+M\left(x\right)\)

\(Q\left(x\right)=\left(3x^3+4x^2-7x+9\right)+\left(3x^3+4x^2+3x+4\right)\)

\(Q\left(x\right)=3x^3+4x^2-7x+9+3x^3+4x^2+3x+4\)

\(Q\left(x\right)=6x^3+8x^2-4x+13\)

c) Nghiệm của đa thức \(P\left(x\right)\)

\(P\left(x\right)=-10x+5=0\)

\(\Rightarrow-10x=-5\)

\(\Rightarrow10x=5\)

\(\Rightarrow x=\dfrac{5}{10}=\dfrac{1}{2}\)

Nghiệm của đa thức \(Q\left(x\right)\)

Vì: \(Q\left(x\right)=6x^3+8x^2+4x+13\ge0\)

\(\Rightarrow Q\left(x\right)\ge0\)

Vậy đa thức vô nghiệm

d) \(Q\left(x\right)\left(1-2x\right)\)

\(=\left(6x^3+8x^2-4x+13\right)\left(1-2x\right)\)

\(=6x^3+8x^2-4x+13-12x^4-16x^3+8x^2-26x\)

\(=-12x^4-10x^3+16x^2-30x+13\)

d. \(\dfrac{\pi}{2}< a;b< \pi\Rightarrow sina>0;sinb>0\)

\(sina=\sqrt{1-cos^2a}=\dfrac{4}{5}\Rightarrow tana=\dfrac{sina}{cosa}=-\dfrac{4}{3}\)

\(sinb=\sqrt{1-cos^2b}=\dfrac{5}{13}\Rightarrow tanb=-\dfrac{5}{12}\)

Vậy:

\(sin\left(a-b\right)=sina.cosb-cosa.sinb=\dfrac{4}{5}.\left(-\dfrac{12}{13}\right)-\left(-\dfrac{3}{5}\right)\left(\dfrac{5}{13}\right)=...\)

\(cos\left(a-b\right)=cosa.cosb-sina.sinb=...\) (bạn tự thay số bấm máy)

\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana.tanb}=...\)

\(cot\left(a+b\right)=\dfrac{1}{tan\left(a+b\right)}=\dfrac{1-tana.tanb}{tana+tanb}=...\)

e.

\(0< y< \dfrac{\pi}{2}\Rightarrow cosy>0\Rightarrow cosy=\sqrt{1-sin^2y}=\dfrac{4}{5}\)

\(\Rightarrow tany=\dfrac{siny}{cosy}=\dfrac{3}{4}\)

Vậy: \(tan\left(x+y\right)=\dfrac{tanx+tany}{1-tanx.tany}=...\)

\(cot\left(x-y\right)=\dfrac{1}{tan\left(x-y\right)}=\dfrac{1+tanx.tany}{tanx-tany}=...\)

no

c) 108(12 + 13 ) + 25 . 92

= 2700 + 2300

= 5000

d) 2.169.12 - 3.68.8 - 24 

= ( 2 .12 ) . 169 - (3.8) . 68 - 24

= 24 . 169 - 24 . 68 - 24

= 24(169 - 68 ) - 24

= 2424 - 24 = 2400

e) 2.56.24 - 3.36.16 + 4.12.95 + 6.3.8.5

= ( 2 . 24 ) . 56 - ( 3.16 ) . 36 + (4.12) .95 + (6.8) . 3 . 5

= 48 . 56 - 48 . 36 + 48 . 95 + 48 . 15

= 48(56 - 36 + 95 + 15 )

= 6240

c)   108.12+25.92+13.108

=(108+12)+(25.13)+(108+92)

=120+325+200

=645

\(c,\Rightarrow\left|x-\dfrac{1}{9}\right|=-\dfrac{4}{5}\\ \Rightarrow x\in\varnothing\left(\left|x-\dfrac{1}{9}\right|\ge0>-\dfrac{4}{5}\right)\\ d,\Rightarrow\left\{{}\begin{matrix}3x-2=0\\4y-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{7}{4}\end{matrix}\right.\\ e,\Rightarrow\left\{{}\begin{matrix}2x+1=0\\x-y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=y=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow x=y=-\dfrac{1}{2}\)