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a: Ta có: \(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\)

\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{x-1+3x}{\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{4x-1}{\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{\left(2x-2\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)

3 tháng 10 2021

Câu a đã làm: F=(2√x/2√x-1     -    1/√x) ( √x+1/√x-1    +       3x/x-2√x+1) với x >0, x khác 1, x khác 1/4 a) rút gọn F - Hoc24

\(b,F=2\Leftrightarrow\dfrac{\left(2\sqrt{x}+1\right)\left(2x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}=2\\ \Leftrightarrow2\sqrt{x}\left(x-2\sqrt{x}+1\right)=2x\sqrt{x}-4x+2\sqrt{x}+2x-2\sqrt{x}+1\\ \Leftrightarrow2x\sqrt{x}-4x+2\sqrt{x}=2x\sqrt{x}-2x+1\\ \Leftrightarrow2x-2\sqrt{x}+1=0\\ \Leftrightarrow2\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{2}=0\\ \Leftrightarrow2\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{2}=0\\ \Leftrightarrow x\in\varnothing\)

 

2 tháng 10 2021

\(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\left(x>0;x\ne1;x\ne\dfrac{1}{4}\right)\\ F=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{x-1+3x}{\left(\sqrt{x}-1\right)^2}\\ F=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\\ F=\dfrac{\left(2\sqrt{x}+1\right)\left(2x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)

a: Ta có: \(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\)

\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{4x-1}{\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{\left(2x-2\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)

11 tháng 8 2021

d, \(\frac{3x}{x+2}=\frac{3\left(x+2\right)-6}{x+2}=3-\frac{6}{x+2}\)

\(\Rightarrow x+2\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

x + 21-12-23-36-6
x-1-30-41-54-4

e, \(C=\frac{A}{B}>0\Rightarrow\frac{3x}{x+2}.\frac{x+2}{x^2+2}=\frac{3x}{x^2+2}>0\)

\(\Rightarrow3x>0\Rightarrow x>0\)vì \(x^2+2>0\)

Kết hợp với đk vậy \(x>0;x\ne\pm2\)

11 tháng 8 2021

f, vừa hỏi thầy, nên quay lại làm nốt :> 

f, Để \(\left|C\right|>C\Rightarrow C< 0\)vì \(\left|C\right|\ge0\)

\(\Rightarrow C=\frac{3x}{x^2+2}< 0\Rightarrow3x< 0\Leftrightarrow x< 0\)

21 tháng 10 2023

a: \(f\left(x\right)=\sqrt{x^2-6x+9}=\sqrt{\left(x-3\right)^2}=\left|x-3\right|\)

\(f\left(-1\right)=\left|-1-3\right|=4\)

\(f\left(5\right)=\left|5-3\right|=\left|2\right|=2\)

b: f(x)=10

=>\(\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=-7\end{matrix}\right.\)

c: \(A=\dfrac{f\left(x\right)}{x^2-9}=\dfrac{\left|x-3\right|}{\left(x-3\right)\left(x+3\right)}\)

TH1: x<3 và x<>-3

=>\(A=\dfrac{-\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-1}{x+3}\)

TH2: x>3

\(A=\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x+3}\)

2 tháng 9 2017

a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)

b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)

=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)

=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)

15 tháng 10 2023

 

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;\dfrac{25}{9};\dfrac{9}{4}\right\}\end{matrix}\right.\)

a: \(C=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}-\dfrac{5}{2\sqrt{x}-3}\right):\left(3-\dfrac{2}{\sqrt{x}-1}\right)\)

\(=\dfrac{2\sqrt{x}-5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}:\dfrac{3\sqrt{x}-3-2}{\sqrt{x}-1}\)

\(=\dfrac{2\sqrt{x}-5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-1}{3\sqrt{x}-5}\)

\(=-\dfrac{1}{2\sqrt{x}-3}\)

b: \(x=\dfrac{2}{2-\sqrt{3}}=2\left(2+\sqrt{3}\right)=4+2\sqrt{3}\)

Khi \(x=4+2\sqrt{3}\) thì \(C=-\dfrac{1}{2\left(\sqrt{3}+1\right)-3}=\dfrac{-1}{2\sqrt{3}-1}=\dfrac{-2\sqrt{3}-1}{11}\)

c: C=-1

=>\(2\sqrt{x}-3=1\)

=>\(\sqrt{x}=2\)

=>x=4(nhận)

d: C>0

=>\(2\sqrt{x}-3< 0\)

=>\(\sqrt{x}< \dfrac{3}{2}\)

=>\(0< =x< \dfrac{9}{4}\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< \dfrac{9}{4}\\x< >1\end{matrix}\right.\)

 

a: Ta có: \(A=\left(\dfrac{2}{x-\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x-4}{x\sqrt{x}+\sqrt{x}-2x}\)

\(=\dfrac{2-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{x-4}\)

\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

\(=\dfrac{-\sqrt{x}+1}{\sqrt{x}+2}\)