\(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)=> \(a+b+c=\frac{ab+bc+ac}{abc}=ab+bc+ac\)

Ta có \(\left(a-1\right)\left(b-1\right)\left(c-1\right)=\left(abc-1\right)+a+b+c-ab-bc-ac=0\)

=> có ít nhất 1 trong 3 số a,b,c bằng 1

Vậy có ít nhất 1 trong 3 số a,b,c bằng 1

Sử dụng hệ thức lượng trong tam giác vuông:

\(AB^2=BH.BC\Rightarrow BC=\frac{AB^2}{BH}=\frac{6^2}{3}=12\)

=> \(HC=BC-BH=12-3=9\)

=> \(AH^2=BH.CH=3.9=27\Rightarrow AH=3\sqrt{3}\)

Áp dụng định lí pi-ta-go

\(AC^2=BC^2-AB^2=12^2-6^2=108\)

=> \(AC=6\sqrt{3}\)

\(A,ĐKXĐ:x;y\ge0\)

\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)

\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)

\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)

\(ĐKXĐ:x;y\ge0\)

\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)

\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)

\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)

Xét  ▲ ABH vuông tại H :

ADĐL pi- ta - go ta có:

AB² = AH^{2 +} BH²

=> AH² = AB² - BH²

     AH² = 6² - 3²

     AH² = 27

      AH = \(\sqrt{27}\)

AC , HC bn tự tính nốt nhé....

a, \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)

\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\frac{\sqrt{x-1}}{\sqrt{64}}=-17\)

\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+\frac{24\sqrt{x-1}}{8}=-17\)

\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Rightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)

\(\Rightarrow\sqrt{x-1}.-1=-17\)

\(\Rightarrow\sqrt{x-1}=17\)

\(\Rightarrow x-1=289\)

\(\Rightarrow x=290\)

b, \(3x-7\sqrt{x}+4=0\)

\(\Rightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)

\(\Rightarrow3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=0\)

\(\Rightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}-1=0\\3\sqrt{x}-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}\sqrt{x}=1\\3\sqrt{x}=4\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{16}{9}\end{cases}}}\)

c, \(-5x+7\sqrt{x}+12=0\)

\(\Rightarrow-5x-5\sqrt{x}+12\sqrt{x}+12=0\)

\(\Rightarrow-5\sqrt{x}\left(\sqrt{x}+1\right)+12\left(x+1\right)=0\)

\(\Rightarrow\left(\sqrt{x}+1\right)\left(-5\sqrt{x}+12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}+1=0\\-5\sqrt{x}+12=0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}=-1VN\\-5\sqrt{x}=-12\end{cases}}\Rightarrow\orbr{\begin{cases}\\\sqrt{x}=\frac{12}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}\\x=\frac{144}{25}\end{cases}}}\)

1) ĐK: \(x-1\ge0\Leftrightarrow x\ge1\)

pt \(\Leftrightarrow\frac{1}{2}\sqrt{x-1}-\frac{3}{2}.3\sqrt{x-1}+\frac{24}{8}\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=17^2=289\Leftrightarrow x=290\left(tm\right)\)

b) \(3x-7\sqrt{x}+4=0\)

ĐK: \(x\ge0\)

Đặt \(\sqrt{x}=t\left(t\ge0\right)\Leftrightarrow t^2=x\)

Ta có phương trình ẩn t: 

\(3t^2-7t+4=0\)( giải đen ta)

\(\Leftrightarrow\orbr{\begin{cases}t=1\\t=\frac{4}{3}\end{cases}}\)

Với t=1 ta có: \(\sqrt{x}=1\Leftrightarrow x=1\) (tm)

Với t=4/3 ta có: \(\sqrt{x}=\frac{4}{3}\Leftrightarrow x=\frac{16}{9}\) (tm)

Câu c em làm tương tự  câu b nhé!

a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)

b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)

a) Ta có:

5√15+12√20+√5515+1220+5

=√52.15+√(12)2.20+√5=√25.15+√14.20+√5=√255+√204+√5=√5+√5+√5=(1+1+1)√5=3√5=52.15+(12)2.20+5=25.15+14.20+5=255+204+5=5+5+5=(1+1+1)5=35

b)  Ta có: 

√12+√4,5+√12,512+4,5+12,5

=√12+√92+√252=√12+√9.12+√25.12=√12+√32.12+√52.12=√12+3√12+5√12=(1+3+5).√12=9√12=91√2=9.√22=9√22=12+92+252=12+9.12+25.12=12+32.12+52.12=12+312+512=(1+3+5).12=912=912=9.22=922

c) Ta có:

√20−√45+3√18+√72=√4.5−√9.5+3√9.2+√36.2=√22.5−√32.5+3√32.2+√62.2=2√5−3√5+3.3√2+6√2=2√5−3√5+9√2+6√2=(2√5−3√5)+(9√2+6√2)=(2−3)√5+(9+6)√2=−√5+15√2=15√2−√520−45+318+72=4.5−9.5+39.2+36.2=22.5−32.5+332.2+62.2=25−35+3.32+62=25−35+92+62=(25−35)+(92+62)=(2−3)5+(9+6)2=−5+152=152−5

d) Ta có:

0,1√200+2√0,08+0,4.√50=0,1√100.2+2√0,04.2+0,4√25.2=0,1√102.2+2√0,22.2+0,4√52.2=0,1.10√2+2.0,2√2+0,4.5√2=1√2+0,4√2+2√2=(1+0,4+2)√2=3,4√2

Bạn giải bài đâu vậy? Kiếm điểm hỏi đáp hở, Boy anime?

\(x^2+5x+1=\left(x+5\right)\sqrt{x^2+1}\)

<=> \(\left(x+5\right)\left(x-\sqrt{x^2+1}\right)=-1\)

Nhân liên hợp ta có

\(x+5=x+\sqrt{x^2+1}\)

=> \(x^2+1=25\)

=> \(x=\pm2\sqrt{6}\)

Vậy \(x=\pm2\sqrt{6}\)

nhân liên hợp là j ạ