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1.
a, ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
b,
\(M=(\dfrac{\sqrt{x}}{\sqrt{x}-2}\times\dfrac{\sqrt{x}}{\sqrt{x}+2})\times\dfrac{x-4}{\sqrt{4x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\times\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2+\sqrt{x}-2\right)}{x-4}\times\dfrac{x-4}{2\sqrt{x}}\)
\(=(\sqrt{x}\times2\sqrt{x})\times\dfrac{1}{2\sqrt{x}}\)
\(=\sqrt{x}\)
c,
\(M>3\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
Bài 2:
a: \(A=\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}:\dfrac{3+\sqrt{1-a^2}}{\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}=\sqrt{1-a}\)
b: Để A=căn A thì A=1 hoặc A=0
=>A=1
=>1-a=1
=>a=0
c: Thay \(a=\dfrac{\sqrt{3}}{2+\sqrt{3}}=\sqrt{3}\left(2-\sqrt{3}\right)=2\sqrt{3}-3\) vào A, ta được:
\(A=\sqrt{1-2\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)
điều kiện xác định : \(x\ge0;x\ne1\)
a) ta có : \(A=\left(\dfrac{1}{1-\sqrt{x}}+\dfrac{1}{1+\sqrt{x}}\right):\left(\dfrac{1}{1-\sqrt{x}}-\dfrac{1}{1+\sqrt{x}}\right)+\dfrac{1}{1-\sqrt{x}}\)
\(\Leftrightarrow A=\left(\dfrac{2}{1-x}\right):\left(\dfrac{2\sqrt{x}}{1-x}\right)+\dfrac{1}{1-\sqrt{x}}\)
\(\Leftrightarrow A=\left(\dfrac{2}{1-x}\right)\left(\dfrac{1-x}{2\sqrt{x}}\right)+\dfrac{1}{1-\sqrt{x}}=\dfrac{1}{\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}\)ta có : \(x=7+4\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
\(\Rightarrow A=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{1-2-\sqrt{3}}=\dfrac{5-3\sqrt{3}}{2}\)
b) áp dụng cauchuy-schwarz dạng engel ta có :
\(A=\dfrac{1}{\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}\ge4\)
dấu "=" xảy ra khi : \(\sqrt{x}=1-\sqrt{x}\Leftrightarrow2\sqrt{x}=1\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)
vậy ....................................................................................................................
a/ \(N=\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}\right).\left(-1\right)\)
\(N=-\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}\right)\)
\(N=-\left(\dfrac{1}{x-2}-\dfrac{2x}{2^2-x^2}\right)\)
\(N=-\left(\dfrac{1}{x-2}-\dfrac{2x}{\left(2-x\right)\left(2+x\right)}\right)\)
\(N=-\left(-\dfrac{1}{2+x}-\dfrac{2x}{\left(2-x\right)\left(2+x\right)}\right)\)
\(N=-\left(\dfrac{-\left(2-x\right)-2x}{\left(2-x\right)\left(2+x\right)^2}\right)\)
\(N=-\left(\dfrac{-2+x-2x}{\left(2-x\right)\left(2+x\right)^2}\right)\)
\(N=-\left(\dfrac{-2-x}{\left(2-x\right)\left(2+x\right)^2}\right)\)
\(N=-\left(\dfrac{-\left(2+x\right)}{\left(2-x\right)\left(2+x\right)^2}\right)\)
\(N=-\left(-\dfrac{1}{\left(2-x\right)\left(2+x\right)}\right)\)
\(N=\dfrac{1}{4-x^2}\)
b/ ĐK của N: \(4-x^2\ne0\Leftrightarrow x\ne\pm2\)
\(2x^2+x=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Tới đây bạn thay vào tính.
c/ Thay N = 1/2 vào:
\(\Rightarrow\dfrac{1}{2}=\dfrac{1}{4-x^2}\)
\(\Leftrightarrow4-x^2=2\)
\(\Leftrightarrow x^2=2\)
\(\Rightarrow x=\pm\sqrt{2}\left(loai\right)\)
a/ ĐKXĐ: \(x\ne\pm2\)
N= \(\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}+\dfrac{1}{2+x}\right).\left(\dfrac{2}{x}-1\right)\)
= \(\left(\dfrac{-1}{2-x}-\dfrac{2x}{\left(2-x\right)\left(2+x\right)}+\dfrac{1}{2+x}\right).\dfrac{2-x}{x}\)
= \(\dfrac{-\left(2+x\right)-2x+2-x}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x}\)
= \(\dfrac{-4x\left(2-x\right)}{x\left(2-x\right)\left(2+x\right)}\)
= \(\dfrac{-4}{2+x}\) (1)
b/ Ta có:
\(2x^2+x=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{2}\end{matrix}\right.\) (t/m đk)
Thay x=0 vào (1) ta được:
\(N=\dfrac{-4}{2+0}=\dfrac{-4}{2}=-2\)
Vậy khi x=0 thì N=-2
Thay \(x=\dfrac{-1}{2}\) vào (1) ta được:
\(N=\dfrac{-4}{2-\dfrac{1}{2}}=\dfrac{-4}{\dfrac{3}{2}}=\dfrac{-8}{3}\)
Vậy khi \(x=\dfrac{-1}{2}\) thì \(N=\dfrac{-8}{3}\)
c/ Với \(x\ne\pm2\)
Để \(N=\dfrac{1}{2}\) \(\Leftrightarrow\dfrac{-4}{2+x}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{-8}{2\left(2+x\right)}=\dfrac{2+x}{2\left(2+x\right)}\)
\(\Leftrightarrow-8-2-x=0\)
\(\Leftrightarrow x=-10\) (t/m đk)
Vậy để \(N=\dfrac{1}{2}\) thì x=-10
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
a) \(A=\dfrac{4}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+2}+\dfrac{4\sqrt{x}+6}{4-x}\)
\(A=\dfrac{4}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+2}-\dfrac{4\sqrt{x}+6}{x-4}\)
\(A=\dfrac{4}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+2}-\dfrac{4\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\dfrac{4\left(\sqrt{x}+2\right)+2\left(\sqrt{x}-2\right)-\left(4\sqrt{x}+6\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\dfrac{4\sqrt{x}+8+2\sqrt{x}-4-4\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\dfrac{2\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\left(\dfrac{-\left(2+\sqrt{x}\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4x+2\sqrt{x}-4}{\sqrt{x}^2-2^2}\right):\left(\dfrac{2}{2-\sqrt{x}}-\dfrac{\sqrt{x}+3}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(P=\left(\dfrac{-\left(2+\sqrt{x}\right)^2+\sqrt{x}\left(\sqrt{x}-2\right)-4x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(P=\left(\dfrac{-4-4\sqrt{x}-x+x-2\sqrt{x}-4x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\right)\)
\(P=\dfrac{-4x\left(\sqrt{x}\left(2-\sqrt{x}\right)\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{-4x\left(-\sqrt{x}\left(\sqrt{x}-2\right)\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{\sqrt{16x^3}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
Có j bạn xem lại coi có sai xót chỗ nào ko nhé, mk ko chắc là đúng 100% đâu.
cảm ơn c nhiều
a: \(P=1:\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{3x}{2\left(x-4\right)}+\dfrac{2}{2\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{1}{4-2\sqrt{x}}\)
\(=1:\left(\dfrac{2\left(\sqrt{x}-2\right)-3x+2\sqrt{x}+4}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{1}{2\left(2-\sqrt{x}\right)}\)
\(=1:\dfrac{2\sqrt{x}-4-3x+2\sqrt{x}+4}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{2\left(2-\sqrt{x}\right)}\)
\(=\dfrac{2\left(x-4\right)}{-3x+4\sqrt{x}}\cdot\dfrac{1}{2\left(2-\sqrt{x}\right)}\)
\(=\dfrac{\sqrt{x}+2}{3x-4\sqrt{x}}\)
b: Để P=20 thì \(\sqrt{x}+2=60x-80\sqrt{x}\)
\(\Leftrightarrow60x-81\sqrt{x}-2=0\)
Đặt \(\sqrt{x}=a\)
Pt sẽ là \(60a^2-81a-2=0\)
\(\text{Δ}=\left(-81\right)^2-4\cdot60\cdot\left(-2\right)=7041>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}a_1=\dfrac{81-\sqrt{7041}}{120}\left(loại\right)\\a_2=\dfrac{81+\sqrt{7041}}{120}\left(nhận\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\left(\dfrac{81+\sqrt{7041}}{120}\right)^2\)
`a)`\(M=A.B\)
\(M=\left(\dfrac{x+2}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{x+2}\right):\left(\dfrac{x^2-3x}{2x^2-x^3}\right)\)
\(M=\left(\dfrac{-\left(x+2\right)^2-4x^2-\left(2-x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x-3}{2x-x^2}\right)\)
\(M=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}.\dfrac{2x-x^2}{x-3}\)
\(M=\dfrac{-4x^2-8x}{\left(x-2\right)\left(x+2\right)}.\dfrac{2x-x^2}{x-3}\)
\(M=\dfrac{-4x\left(2x-x^2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(M=\dfrac{4x^2\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(M=\dfrac{4x^2}{x-3}\)
`b)`\(\left|x-7\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=4\\x-7=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)
`@` Với `x=11`\(\Rightarrow M=\dfrac{121}{2}\)
`@` Với `x=3` `=>` `M` không xác định
`c)`\(A>0\)
\(\Leftrightarrow\dfrac{4x^2}{x-3}>0\)
`<=>x-3>0`
`<=>x>3`
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