ĐKXĐ: \(x\ge0;x\ne9\)

Ta có: \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}=-3\Rightarrow2\sqrt{x}-1=-3\sqrt{x}+9\)

\(\Leftrightarrow5\sqrt{x}-10=0\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\) (thoả mãn ĐKXĐ)

Với \(x \ge 0,x \ne 1\) có:

\(A=\dfrac{x+3\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-1)}-\dfrac{x+\sqrt{x}}{x-1}\)

\(A=\dfrac{(x+3\sqrt{x}+2)(\sqrt{x}+1)-(x+\sqrt{x})(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)(\sqrt{x}+1)}\)

\(A=\dfrac{x\sqrt{x}+x+3x+3\sqrt{x}+2\sqrt{x}+2-x\sqrt{x}-2x-x-2\sqrt{x}}{(\sqrt{x}+2)(\sqrt{x}-1)(\sqrt{x}+1)}\)

\(A=\dfrac{x+3\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-1)(\sqrt{x}+1)}\)

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

\(A=\dfrac{1}{\sqrt{x}+1}\)

Đoạn cuối là HC nha mn giúp mik vs

Với \(x \ge 0,x \ne 9\) có:

\(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}-21}{9-x}+\dfrac{1}{\sqrt{x}+3}\)

\(B=\dfrac{\sqrt{x}(\sqrt{x}+3)-\sqrt{x}+21+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}\)

\(B=\dfrac{x+3\sqrt{x}-\sqrt{x}+21+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}\)

\(B=\dfrac{x+3\sqrt{x}+18}{x-9}\)

Viết latex để mình biết rõ đề nhé bạn

b) Áp dụng hệ thức lượng vào tam giác vuông ABC là: 

\(AH^2=HB.HC=9.12=108\Rightarrow AH=\sqrt{108}=6\sqrt{3}\)

\(AB^2=BH.BC=BH.\left(BH+HC\right)=9.\left(9+12\right)=189\Rightarrow AB=\sqrt{189}=3\sqrt{21}\)

\(AC^2=HC.BC=12\left(12+9\right)=252\Rightarrow AC=\sqrt{252}=6\sqrt{7}\)

a)

Áp dụng hệ thức lượng trong tam giác ABC vuông tại A, ta có : 

$AH^2 = BH.HC = 9.12 = 108 \Rightarrow AH = $ \(6\sqrt{3}\) (cm)

Áp dụng định lí Pitago, ta có : 

$AB^2 = AH^2 + HB^2 = 189 \Rightarrow AB = $ \(3\sqrt{21}\) (cm)

$AC^2 = AH^2 + HC^2 = 252 \Rightarrow AC = $ \(6\sqrt{7}\) (cm)

b)

Trong tứ giác ADHE có góc DAE = góc AEH = góc ADH = 90^o

Do đó ADHE là hình chữ nhật 

Suy ra : 

DE = AH = \(6\sqrt{3}\) cm

Ta có : \(\dfrac{1}{HE^2}=\dfrac{1}{AH^2}+\dfrac{1}{HC^2}=\dfrac{1}{\left(6\sqrt{3}\right)^2}+\dfrac{1}{12^2}\Rightarrow HE=\dfrac{12\sqrt{21}}{7}\)cm

Ta có : \(\dfrac{1}{DH^2}=\dfrac{1}{AH^2}+\dfrac{1}{BH^2}=\dfrac{1}{\left(6\sqrt{3}\right)^2}+\dfrac{1}{9^2}\Rightarrow DH=\dfrac{18\sqrt{7}}{7}\)cm

\(S_{ADHE}=HE.DH=\dfrac{12\sqrt{21}}{7}.\dfrac{18\sqrt{7}}{7}=53,44\left(cm^2\right)\)

\(2\left(3n-1\right)-2\left(n+1\right)>4\)

\(\Leftrightarrow6n-2-2n-2>4\)

\(\Leftrightarrow4n>8\Leftrightarrow n>2\)

\(5\left(n+2\right)-2\left(2n-1\right)\le15\)

\(\Leftrightarrow5n+10-4n+2\le15\)

\(\Leftrightarrow n\le3\)

n = 3 thoả mãn 2 BPT

ĐKXĐ: \(\forall x\in R\)

Ta có: \(2x^2-4x+11+\sqrt{3x^4-6x^2+28}=-3x^2+6x+5\)

\(\Leftrightarrow\sqrt{3x^4-6x^2+28}=-3x^2+6x+5-2x^2+4x-11\)

\(\Leftrightarrow\sqrt{3x^4-6x^2+28}=-5x^2+10x-6\)

\(\Leftrightarrow3x^4-6x^2+28=\left(-5x^2+10x-6\right)^2\)

\(\Leftrightarrow3x^4-6x^2+28=25x^4-100x^3+160x^2-120x+36\)

\(\Leftrightarrow22x^4-100x^3+166x^2-120x+8=0\) (Vô nghiệm)

Ta có: 2x^2-4x+11+\sqrt{3x^4-6x^2+28}=-3x^2+6x+52x2−4x+11+3x4−6x2+28​=−3x2+6x+5

\Leftrightarrow\sqrt{3x^4-6x^2+28}=-3x^2+6x+5-2x^2+4x-11⇔3x4−6x2+28​=−3x2+6x+5−2x2+4x−11

\Leftrightarrow\sqrt{3x^4-6x^2+28}=-5x^2+10x-6⇔3x4−6x2+28​=−5x2+10x−6

\Leftrightarrow3x^4-6x^2+28=\left(-5x^2+10x-6\right)^2⇔3x4−6x2+28=(−5x2+10x−6)2

\Leftrightarrow3x^4-6x^2+28=25x^4-100x^3+160x^2-120x+36⇔3x4−6x2+28=25x4−100x3+160x2−120x+36

\Leftrightarrow22x^4-100x^3+166x^2-120x+8=0⇔22x4−100x3+166x2−120x+8=0 (Vô nghiệm)