P = \(\frac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}\) + \(\frac{\sqrt{2x}-2}{x-2}\)
tính P
Cho:
\(P=\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
a) Rút gọn P
b) Tính P khi \(x=\frac{1}{2}\left(3+2\sqrt{2}\right)\)
Mình rút gọn như thế này đúng không nhỉ?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
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 \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
Cho:
\(A=\left(\frac{x\sqrt{x}+x-2}{x-1}-\frac{\sqrt{x}+2}{x+3\sqrt{x}+2}\right).\left(\frac{\sqrt{x}-1}{2x+\sqrt{x}-2}\right)\)
a) Rút gọn A
b) Tính A khi \(\frac{x}{4}=\sqrt{\frac{1009+\sqrt{2017}}{2}}-\sqrt{\frac{1009-\sqrt{2017}}{2}}\)
\(\frac{2x-1}{MTC}+\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{MTC}-\frac{\left(\sqrt{2x}+\sqrt{x}\left(\sqrt{2x}+1\right)\right)}{MTC}\)
=\(\frac{2x-1+x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-2x-\sqrt{2x}-x\sqrt{2}-\sqrt{x}}{MTC}\)
=\(\frac{-2\sqrt{x}-2}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x+1}\right)}\)
P=\(\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}-\sqrt{x}}{\sqrt{2x}-1}\right)\)
=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)}{\left(\sqrt{2x}+1\right)\left(\sqrt{2x}-1\right)}+\frac{\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}+1\right)}{MTC}-\frac{2x-1}{MTC}\)
=\(\frac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}-2x+1}{MTC}\)
=\(\frac{2x\sqrt{2}+2\sqrt{2x}}{MTC}\)
Cho \(P=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right).\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a) Rút gọn P
b) Tính giá trị của P với \(x=3-2\sqrt{2}\)
ĐKXĐ : \(x\ge1;x\ne2;x\ne3\)
a) P = \(\left[\frac{\sqrt{x}+\sqrt{x-1}}{1}-\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-3}\right].\frac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)( trục căn thức ở mẫu thứ nhất )
\(P=\left(\sqrt{x}-\sqrt{2}\right).\frac{\left(\sqrt{x}-\sqrt{2}\right)}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)
b) \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}-1\)
\(P=\frac{\sqrt{2}-\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1\)
giải pt
a) \(3\sqrt{x}+\frac{3}{2\sqrt{x}}=2x+\frac{1}{2x}-7\)
b) \(5\sqrt{x}+\frac{5}{2\sqrt{x}}=2x+\frac{1}{2x}+4\)
c) \(\sqrt{2x^2+8x+5}+\sqrt{2x^2-4x+5}=6\sqrt{x}\)
d) \(x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\)
e) \(x^2+2x\sqrt{x-\frac{1}{x}}=3x+1\)
f) \(x^2-6x+x\sqrt{\frac{x^2-6}{x}}-6=0\)
g) \(\frac{3x^2}{3+\sqrt{x}}+6+2\sqrt{x}=5x\)
h) \(\frac{x^2}{4-3\sqrt{x}}+8=3\left(x+2\sqrt{x}\right)\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
e/ ĐKXĐ: ...
\(\Leftrightarrow x^2-1+2x\sqrt{\frac{x^2-1}{x}}=3x\)
Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:
\(\frac{x^2-1}{x}+2\sqrt{\frac{x^2-1}{x}}=3\)
Đặt \(\sqrt{\frac{x^2-1}{x}}=a\ge0\)
\(a^2+2a=3\Leftrightarrow a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\frac{x^2-1}{x}}=1\Leftrightarrow x^2-1=x\Leftrightarrow x^2-x-1=0\)
f/ ĐKXĐ: ...
\(\Leftrightarrow x^2-6+x\sqrt{\frac{x^2-6}{x}}-6x=0\)
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{x^2-6}{x}+\sqrt{\frac{x^2-6}{x}}-6=0\)
Đặt \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)
\(a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\frac{x^2-6}{x}}=2\Leftrightarrow x^2-4x-6=0\)
A = \(\frac{x-4\sqrt{x}+2}{\sqrt{x}-2}\) (\(x\ge0;x\ne4\))
B = \(\frac{x\sqrt{x}-1}{x-1}\) (\(x\ge0;x\ne1\))
C = \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\) ( \(x>0;x\ne1\))
D = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) (\(x\ge2\))
E = \(\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2}-2x}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2}-2x}\)
giải phương trình
a) \(\left(x+\frac{5-x}{\sqrt{x}+1}\right)^2+\frac{16\sqrt{x}\left(5-x\right)}{\sqrt{x}+1}-16\)\(=0\)
b) \(\sqrt{2x-\frac{3}{x}}+\sqrt{\frac{6}{x}-2x}=1+\frac{3}{2x}\)
c) \(\sqrt{2x+1}+\frac{2x-1}{x+3}-\left(2x-1\right)\sqrt{x^2+4}-\sqrt{2}=0\)
d) \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)
P=\(\left(\frac{2\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\right).\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
rút gọn P
tính P khi x =\(8+2\sqrt{10}\)
Giả sứ nếu x là số lớn nhất trong 3 chữ số. Ta sẽ lấy x, y để so sánh tạm nhé...
Từ đề bài ta có;
\(\sqrt{x+2011}+\sqrt{y+2012}+\sqrt{z+2013}=\sqrt{z+2011}+\sqrt{x+2012}+\)
\(\sqrt{y+2013}\)
\(\Leftrightarrow\sqrt{x+2012}-\sqrt{x+2011}+\sqrt{y+2013}-\sqrt{y+2012}=\sqrt{z+2012}-\)
\(\sqrt{z+2011}+\sqrt{z+2013}-\sqrt{z-2012}\)
\(\Leftrightarrow\frac{1}{\sqrt{x+2012}+\sqrt{x+2011}}+\frac{1}{\sqrt{x+2013}+\sqrt{x+2012}}=\)
\(\frac{1}{\sqrt{z+2012}+\sqrt{z+2011}}+\frac{1}{\sqrt{z+2013}+\sqrt{z+2012}}\)
Ta lại có
\(\frac{1}{\sqrt{x+2012}+\sqrt{x+2011}}\ge\frac{1}{\sqrt{z+2012}+\sqrt{z+2011}}\)
\(\frac{1}{\sqrt{y+2013}+\sqrt{y+2012}}\ge\frac{1}{\sqrt{z+2013}+\sqrt{z+2012}}\)
P/s; Mình ko chắc đâu nhé