vì 10.10=100; 49.1.2=98

100-98=2( thừa ra 2 ô vuông)

nên có thể làm được

\(\frac{1}{2}.x^2+x=0\)

\(\Leftrightarrow x.\left(\frac{1}{2}x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

Ta xét :

\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)

\(=\left(1+1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=2-\frac{1}{99}-\frac{1}{100}\)

Mà \(2-\frac{1}{99}-\frac{1}{100}< 2\)

\(\RightarrowĐPCM\)

 1.2−12! +2.3−13! +3.4−14! +....+99.100−1100=2 suy ra 1.2−12! +2.3−13! +3.4−14! +....+99.100−1100<2

a, Có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\)

Có: \(\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\Leftrightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

b, Co: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\Rightarrow\frac{ab}{cd}\)

Lại có:\(\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\left(1\right)\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\left(2\right)\)

Tu (1)&(2),có: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

đề bài sai bn oi

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