sách nâng cao và phát triển toán 8 tập 2 bài 326

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

=>\(\left(x+\sqrt{x-1}\right)\left(4x-4\sqrt{x-1}\right)=0\)

từ đay tự giải nốt

\(\frac{2x}{5}+\frac{3-2x}{3}\ge\frac{3x+2}{2}\)

\(\Rightarrow\frac{12x}{30}+\frac{10\left(3-2x\right)}{30}-\frac{15\left(3x+2\right)}{30}\ge0\)

\(\Rightarrow12x+30-20x-45x-30\ge0\)

\(\Rightarrow-53x\ge0\)\(\Leftrightarrow x\le0\)\(\left(1\right)\)

\(\frac{x}{2}+\frac{3-2x}{5}\ge\frac{3x-5}{6}\)

\(\Rightarrow\frac{15x}{30}+\frac{6\left(3-2x\right)}{30}-\frac{5\left(3x-5\right)}{30}\ge0\)

\(\Rightarrow15x+18-12x-15x+25\ge0\)

\(\Rightarrow-12x\ge-43\)\(\Rightarrow12x\le43\Leftrightarrow x\le\frac{43}{12}\)\(\left(2\right)\)

Từ ( 1 ) và ( 2 ) ta có tập nghiệm chung của cả hai phương trình là \(x\le0\)

a, \(x^3+2\sqrt{2}x^2+2x=0\)

\(x\left(x^2+2\sqrt{2}x+2\right)+0\)

\(x\left(x+\sqrt{2}\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+\sqrt{2}=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-\sqrt{2}\end{cases}}\)

Vậy x = 0 ; x = \(-\sqrt{2}\)

b,vì  \(n^2+n+1\)là số chính phương nên đặt \(n^2+n+1=a^2\)với \(a\in N\)

\(n^2+n+1=a^2\)

\(\Leftrightarrow4n^2+4n+4=4a^2\)

\(\Leftrightarrow4n^2+4n+1+3=4a^2\)

\(\Leftrightarrow\left(2n+1\right)^2+3=4a^2\)

\(\Leftrightarrow4a^2-\left(2n+1\right)^2=3\)

\(\Leftrightarrow\left(2a-2n-1\right)\left(2a+2n+1\right)=3\)

Ta thấy \(\hept{\begin{cases}2a-2n-1=1\\2a+2n+1=3\end{cases}}\) Vì \(\left(2a+2n+1>2a-2n-1>0\right)\)

\(\Leftrightarrow\hept{\begin{cases}2\left(a-n\right)=2\\2\left(a+n\right)=2\end{cases}\Leftrightarrow}\hept{\begin{cases}a-n=1\\a+n=1\end{cases}}\)

\(a-n=1\Rightarrow a=1+n\)

\(\Rightarrow1+n+n=1\)

\(\Leftrightarrow2n=1-1\)

\(\Leftrightarrow2n=0\)

\(\Leftrightarrow n=0\)

\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)

_Minh ngụy_

\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )

\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)

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

\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )

\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)

_Minh ngụy_

\(\frac{4}{x+2}+\frac{-3}{x-2}+\frac{12}{x^2-4}.\)

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

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

\(=\frac{x-4}{x^2-4}\)

\(\frac{4}{x+2}+\frac{\left(-2\right)}{x-2}+\frac{12}{x^2-4}\)

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

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

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

\(=\frac{1}{x+2}\)

VÌ hình  thang cân 

=> AC= BD 

Kẻ đường cao BK của hình thang ta co

HK=AB= 14cm

=> KD=CH=(24-14):2=5 cm

Tam giác ACH vuông tại H có 

\(AC^2=CH^2+AH^2\) ( định lý Py- ta -go )

\(AC^2=5^2+12^2\)

AC=13cm

Chu vi hình thang là AB+BD+AC+DC =14+24+13+13=64cm

Diện tích hình thang là 

S=\(\frac{\left(14+24\right)\times12}{2}=228cm^2\)

=180/1111

2/11 - 2/101 

= 180/1111

Học tốt

Bài 2 mình ghi sau nha đây mới đúng nè.                               B = \(\frac{x+1}{x^2-1}-\frac{x^2+2}{x^3-1}-\frac{x-1}{x^2+x+1}\)

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

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

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

 \(=3+\frac{1}{x+2}+\frac{1}{x-1}\)  

b) Vì A>3\(\Rightarrow3+\frac{1}{x+2}+\frac{1}{x-1}>3\Rightarrow\frac{1}{x+2}+\frac{1}{x-1}>0\)  

\(\Rightarrow\frac{1}{x+2}>0\Rightarrow x+2>0\Rightarrow x>-2\) 

và \(\frac{1}{x-1}>0\Rightarrow x-1>0\Rightarrow x>1\) 

\(\Rightarrow x>1\) 

Vậy để B>3 thì x>1