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Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
Đặt biểu thức trên bằng A. ĐK: \(x\ge0;x\ne1\)
\(A=\frac{1+\sqrt{x}+1-\sqrt{x}}{1-x}:\frac{1+\sqrt{x}-1+\sqrt{x}}{1-x}+\frac{1}{1-\sqrt{x}}\)
\(A=\frac{1}{1-x}.\frac{1-x}{2\sqrt{x}}+\frac{1}{1-\sqrt{x}}\)
\(A=\frac{1}{2\sqrt{x}}+\frac{1}{1-\sqrt{x}}\)
\(A=\frac{1-\sqrt{x}+2\sqrt{x}}{2\sqrt{x}\left(1-\sqrt{x}\right)}=\frac{1+\sqrt{x}}{2\sqrt{x}-2x}\)
\(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}\right)^3-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)+\left(x+1\right)}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
Để xem nào ...
Ta có HĐT : \(\hept{\begin{cases}a\sqrt{a}+b\sqrt{b}=\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\\a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\end{cases}\left(a,b\ge0\right)}\)
\(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
ĐKXĐ : x > 0 ; x khác 1
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
Bài 1:
a, (Xin được sửa đề bài) \(C=\sqrt{x-2-2\sqrt{x-3}}-\sqrt{x+1-4\sqrt{x-3}}\)
\(=\sqrt{x-3-2\sqrt{x-3}+1}-\sqrt{x-3-4\sqrt{x-3}+4}\)
\(=\sqrt{\left(\sqrt{x-3}-1\right)^2}-\sqrt{\left(\sqrt{x-3}-2\right)^2}\)
\(=\sqrt{x-3}-1-\sqrt{x-3}+2=1\)
b, \(D=\sqrt{m^2}-\sqrt{m^2-10m+25}\)
\(=m-\sqrt{\left(m-5\right)^2}\)
\(=m-m+5=5\)
Bài 2:
a, \(VT=\sqrt{x+2\sqrt{x-2}-1}.\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)\)
\(=\sqrt{x-2+2\sqrt{x-2}+1}.\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{x-2}+1\right)^2}.\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)\)
\(=\left(\sqrt{x-2}-1\right)\left(\sqrt{x-2}+1\right):\left(\sqrt{x}-\sqrt{3}\right)\)
\(=\left(x-3\right):\left(\sqrt{x}-\sqrt{3}\right)\)
\(=\sqrt{x}+\sqrt{3}=VP\)
b, \(VT=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a+1-2\sqrt{a}}\)
\(=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\left(\frac{\sqrt{a}-1+\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)^2}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)^2}:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}=VP\)
bài nay rút gọn biểu thức à
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