tính \(\sqrt{18-\sqrt{128}}\)
tìm đk B=\(\dfrac{2x+3}{x^2+5x-6}\)
giải pt
\(\sqrt{5x+2}=\sqrt{3-x}\)
\(\sqrt{4x-1}=\sqrt{2x-7}\)
\(\sqrt{3+x}+\sqrt{6-x}-\sqrt{\left(3+x\right)\left(6-x\right)}=3\)
CM
\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{2}\)
xét biểu thức
P=\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\)
a. rút gọn P
b. CMR nếu 0<x<1 thì P>0
Câu đầu tiên: \(\sqrt{18-\sqrt{128}}=\sqrt{16-2\sqrt[]{16}\sqrt{2}+2}=\sqrt{\left(\sqrt{16}-\sqrt{2}\right)^2}=\sqrt{16}-\sqrt{2}=4-\sqrt{2}\)
CM\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=2\)
Biến đổi vế trái ta có:
\(VT^2=\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(\sqrt{4-\sqrt{7}}\right)}+4-\sqrt{7}=8-2\sqrt{16-7}=8-2\sqrt{9}=8-2.3=2\Rightarrow VT=\sqrt{2}\)
Câu 4:
a: \(P=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b: Để P>0 thì \(\sqrt{x}\left(\sqrt{x}-1\right)< 0\)
=>căn x-1<0
=>0<x<1
