\(A=\frac{x}{x-1}-\frac{2x-\sqrt{x}}{x-\sqrt{x}}\) (x>0; x≠1)
a)rút gọn
b) tính A với \(x=3+2\sqrt{2}\)
giúp mk vs ah
Những câu hỏi liên quan
giải phương trìnha) left(x+frac{5-x}{sqrt{x}+1}right)^2+frac{16sqrt{x}left(5-xright)}{sqrt{x}+1}-160b) 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-1right)sqrt{x^2+4}-sqrt{2}0d) sqrt{x+2+3sqrt{2x-5}}+sqrt{x-2-sqrt{2x-5}}2sqrt{2}
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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) \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)
A=\(\left(\frac{1}{1-\sqrt{X}}-\frac{1}{\sqrt{X}}\right):\left(\frac{2X+\sqrt{X}-1}{1-X}+\frac{2X\sqrt{X}+X-\sqrt{X}}{1+X\sqrt{X}}\right)\) với X >0 , X khác \(\frac{1}{4}\), X khác 1
Bleft(frac{asqrt{a}+1}{sqrt{a}+1}right):left(a-1right)+frac{2a+sqrt{a}+1}{sqrt{a}+1}-frac{sqrt{a}}{a-1}vớia1 Cleft(frac{X-1}{sqrt{X}-1}+frac{sqrt{X^3}-1}{1-X}right)-left(frac{left(X-1right)^2+sqrt{X}}{sqrt{X}+1}right)vớiX0,Xne1Dfrac{x^2-sqrt{x}}{x+sqrt{x}+1}-frac{2x+sqrt{x}}{sqrt{x}}+frac{2left(x-1right)}{sqrt{x}-1}vớix0,xne1
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\(B=\left(\frac{a\sqrt{a}+1}{\sqrt{a}+1}\right):\left(a-1\right)+\frac{2a+\sqrt{a}+1}{\sqrt{a}+1}-\frac{\sqrt{a}}{a-1}vớia>1\)
\(C=\left(\frac{X-1}{\sqrt{X}-1}+\frac{\sqrt{X^3}-1}{1-X}\right)-\left(\frac{\left(X-1\right)^2+\sqrt{X}}{\sqrt{X}+1}\right)vớiX>0,X\ne1\)
\(D=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}vớix>0,x\ne1\)
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
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Mấy cái này chỉ có nhân lên rồi rút gọn thôi ah. Nên mình cho bạn đáp án để kiểm tra lại thôi ah
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24 tháng 3 2020 lúc 8:55
Giải phương trình:
a) \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}=2\)
b) \(2\sqrt[3]{\frac{2x-3}{1-x}}+\sqrt[3]{\frac{1-x}{2x-3}}=3\)
c) \(x+\frac{1}{x}+4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+6=0\)
a) \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}=2\)
Ta có: \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}\ge2\sqrt{\sqrt{\frac{2x-1}{x+1}}\cdot\sqrt{\frac{x+1}{2x-1}}}=2\) (BĐT Cô-si)
Mà \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}=2\) (theo đề bài)
Suy ra dấu bằng phải xảy ra \(\Rightarrow\sqrt{\frac{2x-1}{x+1}}=\sqrt{\frac{x+1}{2x-1}}\) \(\Leftrightarrow\frac{2x-1}{x+1}=\frac{x+1}{2x-1}\) \(\Leftrightarrow\left(2x-1\right)^2=\left(x+1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+1\\2x-1=-x-1\end{matrix}\right.\Leftrightarrow\) \(x=2\) (tmđkxđ) hoặc \(x=0\) (không tmđkxđ)
Vậy \(S=\left\{2\right\}\).
Bạn đừng quên tự tìm ĐKXĐ cho câu a nhé bạn.
c) \(x+\frac{1}{x}+4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+6=0\) ĐKXĐ: \(x>0\)
Vì \(x>0\Rightarrow x+\frac{1}{x}+4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+6>0\)
Vậy \(S=\varnothing\).
A frac{x-4sqrt{x}+2}{sqrt{x}-2} (xge0;xne4)
B frac{xsqrt{x}-1}{x-1} (xge0;xne1)
C frac{xsqrt{x}-1}{x-sqrt{x}}-frac{xsqrt{x}+1}{x+sqrt{x}}+frac{x+1}{sqrt{x}} ( x0;xne1)
D sqrt{x+2sqrt{2x-4}}+sqrt{x-2sqrt{2x-4}} (xge2)
E frac{x+sqrt{x^2-2x}}{x-sqrt{x^2}-2x}-frac{x-sqrt{x^2-2x}}{x+sqrt{x^2}-2x}
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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 = \(\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2}-2x}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2}-2x}\)
\(N=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}+\frac{x+\sqrt{x}}{x-1}\right)\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
với x\(\ge\)0 ;x\(\ne1\)
A frac{x-4sqrt{x}+2}{sqrt{x}-2} left(xge0;xne4right)
B frac{xsqrt{x}-1}{x-1} left(xge0;xne1right)
C frac{xsqrt{x}-1}{x-sqrt{x}}-frac{xsqrt{x}+1}{x+sqrt{x}}+frac{x+1}{sqrt{x}} left(x0;xne1right)
D sqrt{x+2sqrt{2x-4}}+sqrt{x-2sqrt{2x-4}} left(xge2right)
E frac{x+sqrt{x^2}-2x}{x-sqrt{x^2-2x}}-frac{x-sqrt{x^2-2x}}{x+sqrt{x^2-2x}}
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A= \(\frac{x-4\sqrt{x}+2}{\sqrt{x}-2}\) \(\left(x\ge0;x\ne4\right)\)
B= \(\frac{x\sqrt{x}-1}{x-1}\) \(\left(x\ge0;x\ne1\right)\)
C= \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\) \(\left(x>0;x\ne1\right)\)
D= \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) \(\left(x\ge2\right)\)
E= \(\frac{x+\sqrt{x^2}-2x}{x-\sqrt{x^2-2x}}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2-2x}}\)
B=\(\frac{x\sqrt{x}-1}{x-1}\)(x>0,x≠1)
=\(\frac{\sqrt{x^3}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
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26 tháng 10 2019 lúc 22:59
giải pt
a) 3sqrt{x}+frac{3}{2sqrt{x}}2x+frac{1}{2x}-7
b) 5sqrt{x}+frac{5}{2sqrt{x}}2x+frac{1}{2x}+4
c) sqrt{2x^2+8x+5}+sqrt{2x^2-4x+5}6sqrt{x}
d) x+1+sqrt{x^2-4x+1}3sqrt{x}
e) x^2+2xsqrt{x-frac{1}{x}}3x+1
f) x^2-6x+xsqrt{frac{x^2-6}{x}}-60
g) frac{3x^2}{3+sqrt{x}}+6+2sqrt{x}5x
h) frac{x^2}{4-3sqrt{x}}+83left(x+2sqrt{x}right)
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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 \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)
\(a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\frac{x^2-6}{x}}=2\Leftrightarrow x^2-4x-6=0\)
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Chứng minh các biểu thức sau không phụ thuộc vào biến:
a) A frac{1}{x}.left(frac{sqrt{x+1}+sqrt{x-1}}{sqrt{x+1}-sqrt{x-1}}+frac{sqrt{x+1}-sqrt{x-1}}{sqrt{x+1}+sqrt{x-1}}right) với x1
b) B frac{2x}{x+3sqrt{x}+2}+frac{5sqrt{x}+1}{x+4sqrt{x}+3}+frac{sqrt{x}+10}{x+5sqrt{x}+6} với x 0
c) C frac{sqrt{a^3}+a}{a^2+sqrt{a^5}}.left(frac{b^2}{a-sqrt{a^2-b^2}}+frac{b^2}{a+sqrt{a^2-b^2}}right) với a0 và |a| |b|
d) D frac{a+bsqrt{a}}{b-a}.sqrt{frac{ab+a^2-2sqrt{a^3b}}{b^2+2bsqrt{a}+a}}:frac{a}{sqrt{a}+s...
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Chứng minh các biểu thức sau không phụ thuộc vào biến:
a) A = \(\frac{1}{x}.\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}+\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\right)\) với x>1
b) B = \(\frac{2x}{x+3\sqrt{x}+2}+\frac{5\sqrt{x}+1}{x+4\sqrt{x}+3}+\frac{\sqrt{x}+10}{x+5\sqrt{x}+6}\) với x>= 0
c) C = \(\frac{\sqrt{a^3}+a}{a^2+\sqrt{a^5}}.\left(\frac{b^2}{a-\sqrt{a^2-b^2}}+\frac{b^2}{a+\sqrt{a^2-b^2}}\right)\) với a>0 và |a| > |b|
d) D = \(\frac{a+b\sqrt{a}}{b-a}.\sqrt{\frac{ab+a^2-2\sqrt{a^3b}}{b^2+2b\sqrt{a}+a}}:\frac{a}{\sqrt{a}+\sqrt{b}}\) với b>a>0
Tính: a) A= \(\frac{1+2a}{1+\sqrt{1+2a}}+\frac{1-2a}{1-\sqrt{1-2a}}\)với a= \(\frac{\sqrt{3}}{4}\)
b) B= \(\frac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}-\frac{\sqrt{x}-4}{\sqrt{x}+1}+\frac{\sqrt{x}+8}{4-\sqrt{x}}\)với 0 < x khác 16)