\(\sqrt{225}\) = 15

HT

a, \(\sqrt{x^2-6x+9}=4\Leftrightarrow\sqrt{\left(x-3\right)^2}=4\Leftrightarrow\left|x-3\right|=4\)

TH1 : \(x-3=4\Leftrightarrow x=7\)

TH2 : \(x-3=-4\Leftrightarrow x=-1\)

b, \(\sqrt{x^2+2x+1}=3\Leftrightarrow\sqrt{\left(x+1\right)^2}=3\Leftrightarrow\left|x+1\right|=3\)

TH1 : \(x+1=3\Leftrightarrow x=2\)

TH2 : \(x+1=-3\Leftrightarrow x=-4\)

c, \(\sqrt{4x^2-4x+1}=5\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\Leftrightarrow\left|2x-1\right|=5\)

TH1 : \(2x-1=5\Leftrightarrow x=3\)

TH2 : \(2x-1=-5\Leftrightarrow x=-2\)

\(a.x=25\Rightarrow A=\frac{3\sqrt{25}-4}{\sqrt{25}-1}=\frac{15-4}{5-1}=\frac{11}{4}\)

\(b.B=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

\(c.B=\frac{1}{4}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\Leftrightarrow\sqrt{x}+1=4\left(\sqrt{x}-1\right)\Leftrightarrow\sqrt{x}=\frac{5}{3}\Leftrightarrow x=\frac{25}{9}\)

\(d.P=A.B=\frac{3\sqrt{x}-4}{\sqrt{x}+1}=3-\frac{7}{\sqrt{x}+1}< 3\)

ta có 

\(C=444..4000..0+888..8+1=4.10^n\left(1+10+..+10^{n-1}\right)+8.\left(1+10+..+10^{n-1}\right)+1\)

\(=4.10^n\frac{10^n-1}{9}+8\frac{10^n-1}{9}+1=\frac{4.10^{2n}+4.10^n+1}{9}=\left(\frac{2.10^n+1}{3}\right)^2\)

rõ ràng C là số tự nhiên nên \(\frac{2.10^n+1}{3}\) là số tự nhiên, vậy ta có đpcm

minh quang ơi bạn giải thích chi tiết ra đc không

mình nghĩ đề là ABC vuông tại A nhé, vì mình thử đủ mọi cách r ;))) 

Xét tam giác ABC vuông tại A, đường cao AH

cosB = \(\frac{AB}{BC}=\frac{3}{5}\Rightarrow AB=\frac{3}{5}.30=18\)cm 

cosB = \(\frac{BH}{AB}=\frac{3}{5}\Rightarrow BH=\frac{3}{5}.18=\frac{54}{5}\)cm 

Ta có : cosB = \(\frac{BH}{AB}=\frac{3}{5}\Rightarrow\frac{BH}{3}=\frac{AB}{5}\Rightarrow\frac{BH^2}{9}=\frac{AB^2}{25}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{AB^2}{25}=\frac{BH^2}{9}=\frac{AB^2-BH^2}{16}=\frac{AH^2}{16}\)

\(\Rightarrow\frac{AB^2}{25}=\frac{AH^2}{16}\Rightarrow AH^2=\frac{18^2.16}{25}=\frac{5184}{25}\Rightarrow AH=\frac{72}{5}\)cm 

ta có 

\(A=111..1000..0+222..2+3=10^{2007}\left(1+10+..+10^{2004}\right)+2.\left(1+10+..+10^{2006}\right)+3\)

\(=10^{2007}.\frac{10^{2005}-1}{9}+2.\frac{10^{2007}-1}{9}+3=\frac{10^{2.2006}-10.10^{2006}+25}{9}=\left(\frac{10^{2006}-5}{3}\right)^2\)

rõ ràng Alà số tự nhiên nên \(\left(\frac{10^{2006}-5}{3}\right)\) là số tự nhiên, vậy ta có đpcm

a, Thay x = 49 ta được : \(M=\frac{49+7}{7-5}=\frac{56}{2}=28\)

b, Với \(x\ge0;x\ne25\)

\(N=\frac{\sqrt{x}}{\sqrt{x}+5}+\frac{2\sqrt{x}-3}{\sqrt{x}-5}-\frac{2x-3\sqrt{x}-15}{x-25}\)

\(=\frac{x-5\sqrt{x}+\left(2\sqrt{x}-3\right)\left(\sqrt{x}+5\right)-2x+3\sqrt{x}+15}{x-25}\)

\(=\frac{-x-2\sqrt{x}+15+2x+7\sqrt{x}-15}{x-25}=\frac{x+5\sqrt{x}}{x-25}=\frac{\sqrt{x}}{\sqrt{x}-5}\)

c, Ta có : \(P=\frac{M}{N}\Rightarrow\frac{x+7}{\sqrt{x}-5}:\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{x+7}{\sqrt{x}}=\sqrt{x}+\frac{7}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{7}{\sqrt{x}}}=2\sqrt{7}\)

Dấu ''='' xảy ra khi \(\sqrt{x}=\frac{7}{\sqrt{x}}\Rightarrow x=7\)riêng ý c thì mình ko chắc :(( 

a.\(x=49\Rightarrow M=\frac{49+7}{\sqrt{49}-5}=\frac{56}{2}=28\)

\(N=\frac{\sqrt{x}\left(\sqrt{x}-5\right)+\left(2\sqrt{x}-3\right)\left(\sqrt{x}+5\right)-2x+3\sqrt{x}+15}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x-5\sqrt{x}+2x+7\sqrt{x}-15-2x+3\sqrt{x}+15}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{x+5\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}}{\sqrt{x}-5}\)

\(c.P=M:N=\frac{x+7}{\sqrt{x}-5}:\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{x+7}{\sqrt{x}}=\sqrt{x}+\frac{7}{\sqrt{x}}\ge2\sqrt{7}\)