\(1,ĐK:x\ge2\\ PT\Leftrightarrow\sqrt{3x-6}+x-2-\left(\sqrt{2x-3}-1\right)=0\\ \Leftrightarrow\dfrac{3\left(x-2\right)}{\sqrt{3x-6}}+\left(x-2\right)-\dfrac{2\left(x-2\right)}{\sqrt{2x-3}+1}=0\\ \Leftrightarrow\left(x-2\right)\left(\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1=0\left(1\right)\end{matrix}\right.\)

Với \(x>2\Leftrightarrow-\dfrac{2}{\sqrt{2x-3}+1}>-\dfrac{2}{1+1}=-1\left(3x-6\ne0\right)\)

\(\Leftrightarrow\left(1\right)>0-1+1=0\left(vn\right)\)

Vậy \(x=2\)

\(2,ĐK:x\ge-1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow a^2+b^2=x^2+2\)

\(PT\Leftrightarrow2a^2+2b^2-5ab=0\\ \Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\)

Với \(a=2b\Leftrightarrow x+1=4x^2-4x+4\left(vn\right)\)

Với \(b=2a\Leftrightarrow4x+4=x^2-x+1\Leftrightarrow x^2-5x-3=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\left(tm\right)\\x=\dfrac{5-\sqrt{37}}{2}\left(tm\right)\end{matrix}\right.\)

Vậy ...

Thấy gì đâu??

._.?

Ai giúp mình zới, khó quá huhuhuhuhuh

Bài 22:

a: \(\frac{x}{x-1}-\frac{2}{x+1}-\frac{2}{x^2-1}\)

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

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

\(=\frac{x^2-3x}{x^2-1}\)

b: \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{3x-6}{4-9x^2}\)

\(=\frac{1}{3x-2}-\frac{4}{3x+2}+\frac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{3x+2-4\left(3x-2\right)+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\frac{6x-4-12x+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\frac{-2\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=\frac{-2}{3x+2}\)

c: \(\frac{x^2}{x^2-x}-\frac{x^2}{x+1}-\frac{2x}{x^2-1}\)

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

\(=\frac{x\left(x+1\right)-x^2\left(x-1\right)-2x}{\left(x-1\right)\left(x+1\right)}=\frac{x^2+x-x^3+x^2-2x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{-x^3+2x^2-x}{\left(x-1\right)\left(x+1\right)}=\frac{-x\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

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

d: \(\frac{4x^2-3x+5}{x^3-1}-\frac{1-2x}{x^2+x+1}-\frac{6}{x-1}\)

\(=\frac{4x^2-3x+5+\left(2x-1\right)\left(x-1\right)-6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{4x^2-3x+5+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12x}{x^3-1}\)

e: \(\frac{5}{2x^2+6x}-\frac{4-3x^2}{x^2-9}-3\)

\(=\frac{5}{2x\left(x+3\right)}+\frac{3x^2-4}{\left(x-3\right)\left(x+3\right)}-3\)

\(=\frac{5\left(x-3\right)+2x\left(3x^2-4\right)-3\cdot2x\left(x-3\right)\left(x+3\right)}{2x\left(x-3\right)\left(x+3\right)}\)

\(=\frac{5x-15+6x^3-8x-6x\left(x^2-9\right)}{2x\left(x-3\right)\left(x+3\right)}=\frac{6x^3-3x-15-6x^3+54x}{2x\left(x-3\right)\left(x+3\right)}\)

\(=\frac{51x-15}{2x\left(x-3\right)\left(x+3\right)}\)

f: \(\frac{5}{x+1}-\frac{10}{x-x^2-1}-\frac{15}{x^3+1}\)

\(=\frac{5}{x+1}+\frac{10}{x^2-x+1}-\frac{15}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{5\left(x^2-x+1\right)+10\left(x+1\right)-15}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{5x}{x^2-x+1}\)

Bài 24:

\(\frac{1}{x}-\frac{1}{x+1}\)

\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

a: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}\)

\(=\frac{1}{x}-\frac{1}{x+4}=\frac{x+4-x}{x\left(x+4\right)}=\frac{4}{x\left(x+4\right)}\)

b: \(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x+5}\)

\(=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)

\(=\frac{1}{x}\)

a, Vì ABCD là hbh nên AB//CD

Do đó \(\widehat{A}+\widehat{D}=180^0\Rightarrow3\widehat{D}=180^0\Rightarrow\widehat{D}=60^0\Rightarrow\widehat{A}=120^0\)

Mà ABCD là hbh nên \(\left\{{}\begin{matrix}\widehat{A}=\widehat{C}=120^0\\\widehat{D}=\widehat{B}=60^0\end{matrix}\right.\)

b, Vì CE=CB nên tam giác CEB cân tại C

Do đó \(\widehat{B}=\widehat{CEB}\)

\(\Rightarrow\widehat{D}=\widehat{CEB}\left(1\right)\)

Mà ABCD là hbh nên AB//CD hay AE//CD

Do đó AECD là hình thang

Kết hợp (1) ta được AECD là hthang cân

\(\Delta ABC\) đều cạnh là mấy a ? 

Đề không cho ạ

Đổi 100ml = 0,1 lít

Ta có: \(n_{HCl}=2.0,1=0,2\left(mol\right)\)

a. PTHH: \(AgNO_3+HCl--->AgCl\downarrow+HNO_3\)

Theo PT: \(n_{AgNO_3}=n_{HCl}=0,2\left(mol\right)\)

Đổi 200ml = 0,2 lít

=> \(C_{M_{AgNO_3}}=\dfrac{0,2}{0,2}=1M\)

b. Ta có: \(m_{dd_{HNO_3}}=0,1\left(lít\right)\)

Theo PT: \(n_{HNO_3}=n_{HCl}=0,2\left(mol\right)\)

=> \(C_{M_{HNO_3}}=\dfrac{0,2}{0,1}=2M\)

Cảm ơn ạ