Bài 5: ĐKXĐ: x>=9/5

\(K=\sqrt{5x+6\cdot\sqrt{5x-9}}+\sqrt{5x-6\cdot\sqrt{5x-9}}\)

\(=\sqrt{5x-9+2\cdot\sqrt{5x-9}\cdot3+9}+\sqrt{5x-9-2\cdot\sqrt{5x-9}\cdot3+9}\)

\(=\sqrt{\left(\sqrt{5x-9}+3\right)^2}+\sqrt{\left(\sqrt{5x-9}-3\right)^2}=\left|\sqrt{5x-9}+3\right|+\left|\sqrt{5x-9}-3\right|\)

=>\(K\ge\left|\sqrt{5x-9}+3-\sqrt{5x-9}+3\right|=6\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{5x-9}+3\right)\left(\sqrt{5x-9}-3\right)\le0\)

=>\(\sqrt{5x-9}-3\le0\)

=>\(\sqrt{5x-9}\le3\)

=>5x-9<=9

=>5x<=18

=>x<=3,6

=>1,8<=x<=3,6

Bài 4:

1:

BC=BH+HC

=>HC=8-2=6(cm)

Xét ΔABC vuông tại A có AH là đường cao

nên \(BA^2=BH\cdot BC=2\cdot8=16=4^2\)

=>BA=4(cm)

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(AC^2=8^2-4^2=64-16=48\)

=>\(AC=\sqrt{48}=4\sqrt3\) (cm)

2: Xét ΔABK vuông tại A có AD là đường cao

nên \(BD\cdot BK=BA^2\)

=>\(BD\cdot BK=BH\cdot BC\)

Bài 3:

1: Thay x=1,44 vào A, ta được:

\(A=\frac{1,44+7}{\sqrt{1,44}}=\frac{8.44}{1.2}=\frac{844}{120}=\frac{211}{30}\)

2: \(B=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-1}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\)

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

\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\sqrt{x}}{\sqrt{x}-3}\)