\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

    \(=\left[\dfrac{x+2}{\sqrt{x^3}-1}+\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x^3}-1}-\dfrac{x+\sqrt{x}+1}{\sqrt{x^3}-1}\right]:\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

    \(=\left(\dfrac{x+2+x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}-1}{\sqrt{x^3}-1}\right):\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

    \(=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)

    \(=1\)

\(P=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right).\dfrac{1}{2a\sqrt{a}}\)

    \(=\left[\dfrac{\left(\sqrt{a}+1\right)^2}{a-1}-\dfrac{\left(\sqrt{a}-1\right)^2}{a-1}+\dfrac{4\sqrt{a}\left(a-1\right)}{a-1}\right].\dfrac{1}{2a\sqrt{a}}\)

\(=\left(\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}\right).\dfrac{1}{2a\sqrt{a}}\)

\(=\dfrac{4a\sqrt{a}}{a-1}.\dfrac{1}{2a\sqrt{a}}\)

\(=\dfrac{2}{a-1}\)

bạn chịu khó đánh máy ra đc không? Chứ khó nhìn đề lắm

Ta có: \(\dfrac{AB}{AC}=\dfrac{2}{3}\). Gọi \(AB=2x\left(cm\right),AC=3x\left(cm\right)\)

Áp dụng định lý Pytago trong tam giác vuông ABC:

\(BC^2=AB^2+AC^2=4x^2+9x^2=13x^2\)

\(\Rightarrow BC=\sqrt{13}x\)

Xét tam giác ABC vuông tại A có đường cao AH:

\(AH.BC=AB.AC\)(hệ thức lượng trong tam giác vuông)

\(\Rightarrow6\sqrt{13}x=6x^2\)

\(\Rightarrow x^2-\sqrt{13}x=0\)

Vì x > 0

\(\Rightarrow x=\sqrt{13}\left(cm\right)\)

\(\Rightarrow\left\{{}\begin{matrix}AB=2x=2\sqrt{13}\left(cm\right)\\AC=3x=3\sqrt{13}\left(cm\right)\\BC=\sqrt{13}x=13\left(cm\right)\end{matrix}\right.\)

Ta có: \(\dfrac{AB}{AC}=\dfrac{2}{3}\)

nên \(\dfrac{HB}{HC}=\dfrac{4}{9}\)

\(\Leftrightarrow HB=\dfrac{4}{9}HC\)

Ta có: \(AH^2=HB\cdot HC\)

\(\Leftrightarrow HC^2\cdot\dfrac{4}{9}=36\)

\(\Leftrightarrow HC^2=16\)

\(\Leftrightarrow HC=4\left(cm\right)\)

\(\Leftrightarrow HB=9\left(cm\right)\)

Ta có: BH+HC=BC

nên BC=4+9=13(cm)

Áp dụng hệ thức lượng trong tam giác vuông vào ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC, ta được:

\(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=2\sqrt{13}\left(cm\right)\\AC=3\sqrt{13}\left(cm\right)\end{matrix}\right.\)

\(a,\Delta'=\left[-\left(m+1\right)\right]^2-\left(m^2-4m+3\right)\\ =m^2+2m+1-m^2+4m-3\\ =6m-2\)

Để pt vô nghiệm thì \(6m-2< 0\Leftrightarrow m< \dfrac{1}{3}\)

Để pt có nghiệm kép thì \(6m-2=0\Leftrightarrow m=\dfrac{1}{3}\)

Để pt có 2 nghiệm phân biệt thì \(6m-2>0\Leftrightarrow m>\dfrac{1}{3}\)

\(b,\Delta=\left(m-3\right)^2-4.\left(-3m\right)\\ =m^2-6m+9+12m\\ =m^2+6m+9\\ =\left(m+3\right)^2\ge0\)

Suy ra pt luôn không vô nghiệm với mọi m

PT có nghiệm kép khi \(\left(m+3\right)^2=0\Leftrightarrow m=-3\)

PT có 2 nghiệm phân biệt khi \(\left(m+3\right)^2>0\Leftrightarrow m\ne-3\)

85: =3-căn 2-căn 2+1=4-2căn 2

86: =căn 8+căn 5-căn 8+căn 5=2căn 5

93: =(căn 3+căn 5)(căn 5-căn 2)

=căn 15-căn 6+5-căn 10

94: =(căn 7-căn 3)(căn 7+căn 3)=7-3=4

\(53,\sqrt{\left(a-2b\right)^2}\left(a\le2b\right)\)

\(=\left|a-2b\right|=-a+2b\)

\(54,\sqrt{4x^2-4xy+y^2}\left(2x\ge y\right)\)

\(=\sqrt{\left(2x-y\right)^2}=\left|2x-y\right|=2x-y\)

\(55,\sqrt{\left(2x-1\right)^2}\left(x\ge\dfrac{1}{2}\right)\)

\(=\left|2x-1\right|=2x-1\)

\(56,\sqrt{\left(3a-2\right)^2}\left(3a\le2\right)\)

\(=\left|3a-2\right|=-3a+2\)

\(57,\sqrt{\left(6-9x\right)^2}\left(3x\ge2\right)\)

\(=\left|6-9x\right|=-6+9x\)

\(58,\sqrt{25a^2-10a+1}\left(5a\le1\right)\)

\(=\sqrt{\left(5a-1\right)^2}=\left|5a-1\right|=-5a+1\)

\(59,\sqrt{m^2+4mn+4n^2}\left(m\ge-2n\right)\)

\(=\sqrt{\left(m+2n\right)^2}=\left|m+2n\right|=m+2n\)

\(60,\sqrt{9x^2-24xy+16y^2}\left(3x\le4y\right)\)

\(=\sqrt{\left(3x-4y\right)^2}=\left|3x-4y\right|=-3x+4y\)

Bài 3:

53. \(\sqrt{\left(a-2b\right)^2}=\left|a-2b\right|=2b-a\)

54. \(\sqrt{4x^2-4xy+y^2}=\sqrt{\left(2x-y\right)^2}=\left|2x-y\right|=2x-y\)

55. \(\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)

56. \(\sqrt{\left(3a-2\right)^2}=\left|3a-2\right|=2-3a\)

57. \(\sqrt{\left(6-9x\right)^2}=\left|6-9x\right|=6-9x\)

58. \(\sqrt{25a^2-10a+1}=\sqrt{\left(5a-1\right)^2}=\left|5a-1\right|=1-5a\)

59. \(\sqrt{m^2+4mn+4n^2}=\sqrt{\left(m+2n\right)^2}=\left|m+2n\right|=m+2n\)

60. \(\sqrt{9x^2-24xy+16y^2}=\sqrt{\left(3x-4y\right)^2}=\left|3x-4y\right|=4y-3x\)