((1)/(\sqrt(x)-1)+(x-\sqrt(x)+1)/(x+\sqrt(x)-2)):((\sqrt(x)+1)/(\sqrt(x)+2)-(x-\sqrt(x)-4)/(x+\sqrt(x)-2))
a,Rg
b,tìm min
1/ Tính: \(\sqrt[3]{54}-\sqrt[3]{16}\)
2/ so sánh các cặp số sau
a) \(3\sqrt{2}\) và \(2\sqrt{3}\)
b) 4.\(\sqrt[3]{5}\) và 5.\(\sqrt[3]{4}\)
3/ cho biểu thức A= \(_{\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)}\)\(\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
a) tìm điều kiện x để A có nghĩa
b) Rút gọn A
Rút gọn:
\(C=\left(\dfrac{1}{x+1}-\dfrac{x+3\sqrt{x}-4}{\left(x^2-1\right)\left(\sqrt{x}+4\right)}\right):\dfrac{\sqrt{x}+1}{x^2\sqrt{x}+x^2-\sqrt{x}-1}\)
\(\sqrt{x-1+2\sqrt{x-2}}-\sqrt{x-1-2\sqrt{x-2}=}1\)( ĐK : \(x\ge2\))
\(\Leftrightarrow\sqrt{x-2+2\sqrt{x-2}+1}+\sqrt{x-2-2\sqrt{x-2}+1}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}+1\right)^2}-\sqrt{\left(\sqrt{x-2}-1\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-2}+1\right|-\left|\sqrt{x-2}-1\right|=1\)
\(\Leftrightarrow\sqrt{x-2}=\left|\sqrt{x-2}-1\right|\)
\(\Leftrightarrow x-2=x-2-2\sqrt{x-2}+1\)
\(\Leftrightarrow\sqrt{x-2}=\frac{1}{2}\)
\(\Leftrightarrow x-2=\frac{1}{4}\)\(\Leftrightarrow x=\frac{9}{4}\)(TĐK)
Rút gọn:
\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{x-\sqrt{x}+1}{x+\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{x-\sqrt{x}-4}{x+\sqrt{x}-2}\right)\)
Giải phương trình: \(\frac{\left(x^6+3x^4\sqrt{x^2-x+1}\right)\left(3+x-x^2\right)}{4\left(2+\sqrt{x^2-x+1}\right)\left(x^2-x+1\right)}=\sqrt{x^2-x+1}\left(2-\sqrt{x^2-x+1}\right)\)
Giải phương trình: \(\frac{\left(x^6+3x^4\sqrt{x^2-x+1}\right)\left(3+x-x^2\right)}{4\left(2+\sqrt{x^2-x+1}\right)\left(x^2-x+1\right)}=\sqrt{x^2-x+1}\left(2-\sqrt{x^2-x+1}\right)\)
Giải phương trình
1) \(x^2+\sqrt{x+5}=5\)
2)\(\sqrt{x}+\sqrt{1-x}+2\sqrt{x\left(1-x\right)}-2\sqrt[4]{x\left(1-x\right)}=1\)
3) \(2x^2+2x+1=\sqrt{4x+1}\)
4)\(13\sqrt{x-1}+9\sqrt{x+1}=16x\)
5)\(x^2+\sqrt{x+1}=1\)
mọi người zải nhanh zùm e cảm ơn
\(\sqrt{\left(x-2\right)\left(4-x\right)}+\sqrt{x-2}+\sqrt{4-x}=\frac{x-1}{2}+\sqrt{x+1}\)
Rút gọn:
\(C=\dfrac{\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+2+4\sqrt{x-2}}}{\sqrt{\dfrac{4}{x^2}-\dfrac{4}{x}+1}}\)