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

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

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

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

\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

=> A = \(\frac{\sqrt{x}}{x+\sqrt{x}+1}:\frac{\sqrt{x}+3}{x+\sqrt{x}+1}=\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}+3}\)

5050 nhé bạn ơi

Kết quả bằng 5050

a,\(\Delta\)ABC vuông tại A ta có:

            AC²=HC.BC             (ĐL1)

\(\Rightarrow\)HC= AC²/BC=44²/55=35,2 (cm)

\(\Delta\)ABC vuông tại A , đường cao AH ta có:

       AH²=BH.HC

\(\Rightarrow\)AH=\(\sqrt{\left(BC-HC\right).HC}\)= \(\sqrt{\left(55-35,2\right).35,2}\)=26,4(cm)

b, sinCAH=\(\frac{35,2}{44}\)\(\approx\)53^o

chào bạn

hello hello

1 + 2 = 3 = - 143 + 146 

@Cỏ

#Forever

\(\hept{\begin{cases}x\ne0\\x\ne\frac{1\pm\sqrt{3}}{2}\end{cases}}\)

\(\overline{abcd}\left(a\ne0\right)\)

\(\overline{abcd}+a+b+c+d=2023\)

\(\left(=\right)a.1000+b.100+c.10+d+a+b+c+d=2023\)

\(\left(=\right)1001a+101b+11c+2d=2023\)

Có \(2023=2.1001+0.101+1.11+2.5\)

\(\Rightarrow a=2;b=0;c=1;d=5\)

\(\Rightarrow abcd=2015\)

Đó là 2015 b nhé !!
#Học tốt

\(P=\sqrt{3-x}+\sqrt{4-x}\)(ĐK: \(x\le3\))

\(\le\sqrt{3-3}+\sqrt{4-3}=1\)

Dấu \(=\)khi \(x=3\).