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\(\text{f(1)=}2.1^2+1=3\)
\(\text{f(-1)=}2.\left(-1\right)^2+1=3\)
\(\text{f(2)=}2.2^2+1=9\)
\(\text{f(0)=}2.0^2+1=1\)
\(\text{f(-3)=}=2.\left(-3\right)^2+1=19\)
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Câu 3:
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x+y}{3+2}=\dfrac{90}{5}=18\)
Do đó: x=54; y=36
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a: Xét ΔBID và ΔBIC có
BI chung
\(\widehat{BID}=\widehat{BIC}\)
BD=BC
Do đó: ΔBID=ΔBIC
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\(=\dfrac{3\left(\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{\dfrac{1}{100}\left(\dfrac{100}{1\cdot99}+\dfrac{100}{3\cdot97}+...+\dfrac{100}{99\cdot1}\right)}\)
\(=\dfrac{3\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{\dfrac{1}{100}\cdot\left(\dfrac{1}{1}+\dfrac{1}{99}+\dfrac{1}{3}+\dfrac{1}{97}+...+\dfrac{1}{99}+\dfrac{1}{1}\right)}\)
\(=\dfrac{3\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}{\dfrac{1}{50}\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)}=3:\dfrac{1}{50}=150\)
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d) \(\frac{x+2}{7}+\frac{x+3}{6}=\frac{x+4}{5}+\frac{x+5}{4}\)
\(\Leftrightarrow\frac{x+2}{7}+1+\frac{x+3}{6}+1=\frac{x+4}{5}+1+\frac{x+5}{4}+1\)
\(\Leftrightarrow\frac{x+9}{7}+\frac{x+9}{6}=\frac{x+9}{5}+\frac{x+9}{4}\)
\(\Leftrightarrow\left(x+9\right)\left(\frac{1}{7}+\frac{1}{6}-\frac{1}{5}-\frac{1}{4}\right)=0\)
\(\Leftrightarrow x+9=0\)
\(\Leftrightarrow x=-9\)
e) Tương tự d).
f) \(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\Leftrightarrow\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
\(\Leftrightarrow416-x=0\)
\(\Leftrightarrow x=416\).