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Bài 1.
\(B=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\div\frac{x}{x-\sqrt{x}}\)với \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
a) \(B=\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)
\(B=\left(\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)
\(B=\left(\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)
\(B=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\div\frac{x}{x-\sqrt{x}}\)
\(B=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x}\)
\(B=\frac{4\sqrt{x}\cdot\sqrt{x}}{\left(\sqrt{x}+1\right)x}=\frac{4x}{\left(\sqrt{x}+1\right)x}=\frac{4}{\sqrt{x}+1}\)
b) Để B > 1
=> \(\frac{4}{\sqrt{x}+1}>0\)( với \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\))
Vì 4 > 0
=> \(\sqrt{x}+1>0\)
<=> \(\sqrt{x}>-1\)( luôn luôn đúng \(\forall\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)) ( theo ĐKXĐ )
Vậy \(\forall\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)thì B > 1
Chưa chắc lắm ... Còn câu 2 thì tí nữa mình làm cho
Bài 2.
\(A=2\sqrt{5}-1\)
\(B=\frac{2}{x-1}\cdot\sqrt{\frac{x^2-2x+1}{4x^2}}\)( x > 0 )
a) \(B=\frac{2}{x-1}\cdot\frac{\sqrt{x^2-2x+1}}{\sqrt{4x^2}}\)
\(B=\frac{2}{x-1}\cdot\frac{\sqrt{\left(x-1\right)^2}}{\sqrt{\left(2x\right)^2}}\)
\(B=\frac{2}{x-1}\cdot\frac{\left|x-1\right|}{\left|2x\right|}\)
\(B=\frac{2}{x-1}\cdot\frac{x-1}{2x}=\frac{1}{x}\)( vì x > 0 )
b) Để A + B = 0
=> \(\left(2\sqrt{5}-1\right)+\frac{1}{x}=0\)( ĐKXĐ : \(x\ne0\))
<=> \(\frac{1}{x}=-\left(2\sqrt{5}-1\right)\)
<=> \(\frac{1}{x}=1-2\sqrt{5}\)
<=> \(x\times\left(1-2\sqrt{5}\right)=1\)
<=> \(x=\frac{1}{1-2\sqrt{5}}\)( tmđk )
Vậy \(x=\frac{1}{1-2\sqrt{5}}\)
mình thêm 1 vài bước nữa , thiếu rồi xin lỗi bạn nhé !
\(\frac{2\left(x+\sqrt{x}\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}=\frac{2\left[\sqrt{x}\left(\sqrt{x}+1\right)\right]^2}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}=\frac{2x.\left(\sqrt{x}+1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}\)
\(=\frac{2x}{x-1}\)(gọn rồi đấy)
không biết làm gì ngoài nhân chéo :((
\(\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\left(ĐKXĐ:x\ge0;x\ne1\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(x-1\right)-\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+1\right)}{\left(x+2\sqrt{x}+1\right)\left(x-1\right)}\left(x+\sqrt{x}\right)\)
\(=\frac{x\sqrt{x}-\sqrt{x}+2x-2-x\sqrt{x}-2x-\sqrt{x}+2x+4\sqrt{x}+2}{\left(x+2\sqrt{x}+1\right)\left(x-1\right)}.\left(x+\sqrt{x}\right)\)
\(=\frac{x\sqrt{x}-x\sqrt{x}-\sqrt{x}-\sqrt{x}+4\sqrt{x}+2x-2x+2x-2+2}{\left(x+2\sqrt{x}+1\right)\left(x-1\right)}.\left(x+\sqrt{x}\right)\)
\(=\frac{2\left(x+\sqrt{x}\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}\)
xong nhé :v bạn làm được tiếp thì làm
a) \(P=\dfrac{A}{B}=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-1}\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{x-1}{\sqrt{x}+1}=\dfrac{\left(x-1\right)^2}{\sqrt{x}\left(x-1\right)}=\dfrac{x-1}{\sqrt{x}}\)
b) \(P\sqrt{x}=m+\sqrt{x}\)
\(\Leftrightarrow\dfrac{x-1}{\sqrt{x}}.\sqrt{x}=m+\sqrt[]{x}\)
\(\Leftrightarrow x-1=m+\sqrt{x}\)
\(\Leftrightarrow m=x-\sqrt{x}-1\)
a) \(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}=\frac{1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}=\frac{2\sqrt{x}}{x-1}\)( x > 0 ; x ≠ 1 )
b) \(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)( x > 0 ; x ≠ 4 )
a) Với \(x>0\)và \(x\ne1\)ta có:
\(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}+1\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1+\sqrt{x}-1+x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Với \(x>0\)và \(x\ne4\)ta có:
\(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)-2\left(\sqrt{x}+2\right)+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)
\(B=\frac{x}{\sqrt{x}-1}-\frac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{x}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x-1}}\)\(=\frac{-\sqrt{x}-1}{\sqrt{x}-1}\)
để B<0 \(\Leftrightarrow\frac{-\sqrt{x}-1}{\sqrt{x}-1}< 0\)
có \(-\sqrt{x}-1< 0\Rightarrow\sqrt{x}-1>0\)
\(\Leftrightarrow x>1\)
kết hợp với điều kiện đề bài ta được x>1
Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc đề dễ hiểu hơn bạn nhé.
a) \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)
b) \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)
d) \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
Bài 3:
b: ĐKXĐ: \(\begin{cases}1-x^2\ge0\\ x+1\ge0\end{cases}\Rightarrow\begin{cases}x^2\le1\\ x\ge-1\end{cases}\Rightarrow\begin{cases}x=-1\\ x\ge1\end{cases}\)
\(\sqrt{1-x^2}+\sqrt{1+x}=0\)
=>\(\sqrt{1+x}\left(\sqrt{1-x}+1\right)=0\)
=>\(\sqrt{1+x}=0\)
=>x+1=0
=>x=-1(nhận)
c: Sửa đề: \(x+y+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)
ĐKXĐ: x>=2; y>=3; z>=5
\(x+y+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)
=>\(x+y+4-2\sqrt{x-2}-4\sqrt{y-3}-6\sqrt{z-5}=0\)
=>\(x-2-2\sqrt{x-2}+1+y-3-4\sqrt{y-3}+4+z-5-6\sqrt{z-5}+9=0\)
=>\(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
=>\(\begin{cases}x-2=1\\ y-3=4\\ z-5=9\end{cases}\Rightarrow\begin{cases}x=3\\ y=7\\ z=14\end{cases}\) (nhận)
d: \(x^2+2x-\sqrt{x^2+2x+1}-5=0\)
=>\(x^2+2x+1-\sqrt{x^2+2x+1}-6=0\)
=>\(\left(\left|x+1\right|\right)^2-\left|x+1\right|-6=0\)
=>(|x+1|-3)(|x+1|+2)=0
=>|x+1|-3=0
=>|x+1|=3
=>\(\left[\begin{array}{l}x+1=3\\ x+1=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-4\end{array}\right.\)
Bài 2:
a: DKXĐ: x>=0
\(\sqrt{x+4\sqrt{x}+4}=5x+2\)
=>\(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)
=>\(5x+2=\sqrt{x}+2\)
=>\(5x-\sqrt{x}=0\)
=>\(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)
=>\(\left[\begin{array}{l}\sqrt{x}=0\\ 5\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}=\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\frac{1}{25}\left(nhận\right)\end{array}\right.\)
b: ĐKXĐ: x∈R
\(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)
=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=4\)
=>|x+2|+|x-1|=4(1)
TH1: x<-2
=>x+2<0; x-1<0
(1) sẽ trở thành: -x-2+1-x=4
=>-2x-1=4
=>-2x=5
=>\(x=-\frac52\) (nhận)
TH2: -2<=x<1
=>x+2>=0; x-1<0
(1) sẽ trở thành: x+2+1-x=4
=>3=4(loại)
TH3: x>=1
=>x+2>0; x-1>=0
(1) sẽ trở thành: x+2+x-1=4
=>2x=3
=>x=3/2(nhận)
c: ĐKXĐ: x>=1
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\)
=>\(\left|\sqrt{x-1}-1\right|=2-\sqrt{x-1}-1=1-\sqrt{x-1}\)
=>\(\sqrt{x-1}-1\le0\)
=>\(\sqrt{x-1}\le1\)
=>x-1<=1
=>x<=2
=>1<=x<=2
\(B=\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}-1}\right)\cdot\dfrac{\sqrt{x}-1}{4\sqrt{x}+4}=\dfrac{1}{2}\)