a) \(x^2+x=0\)

\(\Rightarrow x.\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

b) \(x^2-5x=0\)

\(\Rightarrow x.\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

c) \(x.\left(x-2\right)+x-2=0\)

\(\Rightarrow x.\left(x-2\right)+\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right).\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

d) \(x.\left(x-3\right)-x+3=0\)

\(\Rightarrow x.\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-1\right).\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

e) \(x^2-4=0\)

\(\Rightarrow x^2=4\)

\(\Rightarrow x=\pm2\)

\(A=\left(2x-1\right)^2-5x\left(x-1\right)+2\left(x+1\right)\left(x-2\right)\)

\(=4x^2-4x+1-5x^2+5x+2\left(x^2-x-2\right)\)

\(=x^2-x-3\)

24. 

Ta có: \(3k^2+3k+1=k^3+3k^2+3k+1-k^3=\left(k+1\right)^3-k^3\)

Do đó \(a_k=\frac{\left(k+1\right)^3-k^3}{\left(k^2+k\right)^3}=\frac{\left(k+1\right)^3-k^3}{k^3.\left(k+1\right)^3}=\frac{1}{k^3}-\frac{1}{\left(k+1\right)^3}\)

Áp dụng ta được: 

\(P=a_1+a_2+...+a_9\)

\(=\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{2^3}-\frac{1}{3^3}+...+\frac{1}{9^3}-\frac{1}{10^3}\)

\(=1-\frac{1}{10^3}=\frac{999}{1000}\)

23. Ta có: 

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

\(=\frac{1.1.3.3.5.5.....\left(2n+1\right)\left(2n+1\right)}{\left(1.3\right).\left(3.5\right).\left(5.7\right).....\left[\left(2n+1\right)\left(2n+3\right)\right]}\)

\(=\frac{\left[1.3.5.....\left(2n+1\right)\right].\left[1.3.5.....\left(2n+1\right)\right]}{\left[1.3.5.....\left(2n+1\right)\right].\left[3.5.7.....\left(2n+3\right)\right]}\)

\(=\frac{1}{2n+3}\)

a,Do AE////DC nên FEFDFEFD=AFFCAFFC(1)

Do AD////CG nên FDFGFDFG=AFFGAFFG(2)

Từ (1) và 2 ) =>

FDFGFDFG=FEFDFEFD

=>FD²=EF .FG

b,Ta có E là trung điểm của AB nên AE=EB=18
Ta có: AE^2+AD^2=DE^2
nên 1872=DE^2 nên DE=30
lại có t/g AED=T/G BEG(G.C.G) nên EG=DE nên DG=2DE=60

Trả lời:

a, \(5\left(2x-1\right)-4\left(8-3x\right)=-5\)

\(\Leftrightarrow10x-5-32+12x=-5\)

\(\Leftrightarrow22x-37=-5\)

\(\Leftrightarrow22x=32\)

\(\Leftrightarrow x=\frac{16}{11}\)

Vậy x = 16/11 là nghiệm của pt.

b, \(\left(x+1\right)\left(x+2\right)-\left(x-3\right)\left(x+4\right)=6\)

\(\Leftrightarrow x^2+3x+2-\left(x^2+x-12\right)=0\)

\(\Leftrightarrow x^2+3x+2-x^2-x+12=0\)

\(\Leftrightarrow2x+14=0\)

\(\Leftrightarrow2x=-14\)

\(\Leftrightarrow x=-7\)

Vậy x = - 7 là nghiệm của pt.

b) \(\left(5x-2\right)\left(x+1\right)-10x^2+4x=0\)

\(\Leftrightarrow\left(5x-2\right)\left(x+1\right)-2x\left(5x-2\right)=0\)

\(\Leftrightarrow\left(5x-2\right)\left(1-x\right)=0\)

\(\Leftrightarrow x=\frac{2}{5},x=1\)